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\(a.\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\\\Leftrightarrow 9x^2+12x+4-9x^2+12x-4=5x+38\\ \Leftrightarrow24x-5x=38\\ \Leftrightarrow19x=38\\\Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
\(b.3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\\\Leftrightarrow 3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\\ \Leftrightarrow3x^2-3x^2-12x+9x-3x=-12+9-9\\ \Leftrightarrow-6x=-12\\\Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
\(c.\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x-2\right)\\ \Leftrightarrow x^3-3x^2+3x-1-x\left(x^2+2x+1\right)=10x-5x^2-11x+22\\ \Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x+22\\\Leftrightarrow x^3-x^3-3x^2-2x^2+5x^2+3x-x-10x+11x=1+22\\ \Leftrightarrow3x=23\\\Leftrightarrow x=\frac{23}{3}\)
Vậy nghiệm của phương trình trên là \(\frac{23}{3}\)
\(d.\left(x+3\right)^2-\left(x-3\right)^2=6x+18\\ \Leftrightarrow x^2+6x+9-x^2+6x-9=6x+18\\ \Leftrightarrow12x-6x=18\\ \Leftrightarrow6x=18\\ \Leftrightarrow x=3\)
Vậy nghiệm của phương trình trên là \(3\)
\(e.\left(x+1\right)\left(x^2-x+1\right)-2x=x\left(x-1\right)\left(x+1\right)\\\Leftrightarrow x^3+1-2x=x\left(x^2-1\right)\\\Leftrightarrow x^3+1-2x=x^3-x\\ \Leftrightarrow x^3-x^3-2x+x=-1\\ \Leftrightarrow-x=-1\\ \Leftrightarrow x=1\)
Vậy nghiệm của phương trình trên là \(1\)
\(f.\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\\\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\\ \Leftrightarrow x^3-x^3-6x^2+9x^2-3x^2+12x-3x=8+1+1\\ \Leftrightarrow9x=10\\ \Leftrightarrow x=\frac{10}{9}\)
Vậy nghiệm của phương trình trên là \(\frac{10}{9}\)
1: \(A=\left(-x+5\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)\)
\(=-x^2+2x+5x-10+x^2-49=7x-59\)
\(B=\left(3x+1\right)^2-\left(3x-2\right)\left(3x+2\right)\)
\(=9x^2+6x+1-9x^2+4=6x+5\)
=>7x-59=6x+5
=>x=64
2: \(A=\left(5x-1\right)\left(x+1\right)-2\left(x-3\right)^2\)
\(=5x^2+5x-x-1-2x^2+12x-9\)
\(=3x^2+16x-10\)
\(B=\left(x+2\right)\left(3x-1\right)-\left(x+4\right)^2+x^2-x\)
\(=3x^2-x+6x-2-x^2-8x-16+x^2-x\)
\(=3x^2-4x-18\)
=>16x-10=-4x-18
=>20x=-8
hay x=-2/5
ĐKXĐ : \(\hept{\begin{cases}x-2\ne0\\3-4x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne\frac{3}{4}\end{cases}}}\)
\(\frac{5}{x-2}+\frac{6}{3-4x}=0\)
\(\frac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\frac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)
\(15-20x+6x-12=0\)
\(3-14x=0\Leftrightarrow14x=3\Leftrightarrow x=\frac{3}{14}\)theo ĐKXĐ : x thỏa mãn
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
a)Ta có:
\(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\\ =x^2-4x+4-x^2+4x-3\\ =1\)
Vậy biểu thức \(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\)không phụ thuộc vào biến
b) Ta có:
\(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\\ =x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\\ =-8\)
Vậy.....
c) Ta có:
\(\left(x-3\right)\left(x+3\right)\left(x^2+9\right)-\left(x^2-2\right)\left(x^2+2\right)\\ =\left(x^2-9\right)\left(x^2+9\right)-x^4+4\\ =x^4-81-x^4+4=-77\)
Vậy....
d) Ta có: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2\\ =\left(3x+1-3x+5\right)^2\\ =6^2=36\)
Vậy....
d, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+10=0\) (Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\) ≠ 0)
\(\Leftrightarrow x=-10\)
Vậy x = -10 là nghiệm của phương trình.
Giải các phương trình
\(a,3x-2=2x-3\)
\(\Leftrightarrow3x-2x=-3+2\)
\(\Leftrightarrow x=-1\)
Vậy pt có tập nghiệm S = { - 1 }
\(b,2x+3=5x+9\)
\(\Leftrightarrow2x-5x=9-3\)
\(\Leftrightarrow-3x=6\)
\(\Leftrightarrow x=-2\)
Vậy pt có tập nghiệm S = { - 2 }
\(c,11x+42-2x=100-9x-22\)
\(\Leftrightarrow11x-2x+9x=100-22-42\)
\(\Leftrightarrow18x=36\)
\(\Leftrightarrow x=2\)
Vậy pt có tập nghiệm S = { - 2 }
\(d,2x-\left(3-5x\right)=4\left(x+3\right)\)
\(\Leftrightarrow2x-3+5x=4x+12\)
\(\Leftrightarrow2x+5x-4x=12+3\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
Vậy pt có tập nghiệm S = { - 5 }
\(e,\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2}{6}+\dfrac{2x.6}{6}\)
\(\Leftrightarrow9x+6-3x-1=10+12x\)
\(\Leftrightarrow9x-3x-12x=10-6+1\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy pt có tập nghiệm S = { - \(\dfrac{5}{6}\) }
f,\(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30x}{30}+\dfrac{4.30}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow6x-30x-10x+15x=30-24-120\)
\(\Leftrightarrow-19x=-114\)
\(\Leftrightarrow x=6\)
Vậy pt có tập nghiệm S = { - 6 }
\(g,\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { \(1;-\dfrac{1}{2}\) }
\(h,\left(x+\dfrac{2}{3}\right)\left(x-\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { \(-\dfrac{2}{3};\dfrac{1}{2}\) }
\(i,\left(3x-1\right)\left(2x-3\right)\left(2x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(2x-3\right)^2\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\2x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { \(\dfrac{1}{3};\dfrac{3}{2};-5\) }
\(k,3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3x-15=2x^2-10x\)
\(\Leftrightarrow-2x^2+3x+10x=15\)
\(\Leftrightarrow-2x^2+13x-15=0\)
\(\Leftrightarrow-2x^2+10x+3x-15=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { \(5;\dfrac{3}{2}\) }
\(m,\left|x-2\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { -1; 5 }
\(n,\left|x+1\right|=\left|2x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2x+3\\x+1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { \(-2;-\dfrac{4}{3}\) }
\(j,\dfrac{7x-3}{x-1}=\dfrac{2}{3}\) ĐKXĐ : x≠ 1
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow x=\dfrac{7}{19}\) ( t/m )
Vậy pt có tập nghiệm S = { \(\dfrac{7}{19}\) }
đ, ĐKXĐ : x ≠ - 1
\(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=1+x\)
\(\Leftrightarrow12-28x=1+x\)
\(\Leftrightarrow-29x=-11\)
\(\Leftrightarrow x=\dfrac{11}{29}\) ( t/m)
Vậy pt có tập nghiệm S = { \(\dfrac{11}{29}\) }
\(y,\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\) ĐKXĐ : \(\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{\left(x+5\right)^2-\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\dfrac{20}{\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow20x=20\)
\(\Leftrightarrow x=1\) ( t/m )
Vậy pt có tập nghiệm S = { 1 }
\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{x}{x^2-1}\) ĐKXĐ : \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x+1+2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow3x-1=x\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)( t/m)
Vậy pt có tập nghiệm S = { \(\dfrac{1}{2}\) }
1. \(\frac{x + 3}{x^{2} - 1} - \frac{1}{x^{2} - x}\)
\(\frac{x + 3}{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)} - \frac{1}{x \left(\right. x - 1 \left.\right)}\) \(= \frac{x \left(\right. x + 3 \left.\right) - \left(\right. x + 1 \left.\right)}{x \left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}\) \(= \frac{x^{2} + 3 x - x - 1}{x \left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}\) \(= \boxed{\frac{x^{2} + 2 x - 1}{x \left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}}\)
2. \(\frac{x}{x - 1} - \frac{1}{x^{2} - x}\)
\(= \frac{x}{x - 1} - \frac{1}{x \left(\right. x - 1 \left.\right)}\) \(= \frac{x^{2} - 1}{x \left(\right. x - 1 \left.\right)}\) \(= \boxed{\frac{x^{2} - 1}{x \left(\right. x - 1 \left.\right)}}\)
3. \(\frac{x + 1}{x - 5} + \frac{x - 11}{x - 5}\)
\(= \frac{x + 1 + x - 11}{x - 5}\) \(= \frac{2 x - 10}{x - 5}\) \(= \boxed{2}\)
4. \(\frac{2}{x + 3} - \frac{2}{3 - x} - \frac{3 x}{x^{2} - 9}\)
\(= \frac{2}{x + 3} + \frac{2}{x - 3} - \frac{3 x}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)}\) \(= \frac{2 \left(\right. x - 3 \left.\right) + 2 \left(\right. x + 3 \left.\right) - 3 x}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)}\) \(= \frac{x}{x^{2} - 9}\) \(= \boxed{\frac{x}{x^{2} - 9}}\)
5.
\(\frac{2 x^{2} + 1}{x^{3} + 1} - \frac{x - 1}{x^{2} - x + 1} - \frac{1}{x + 1}\) \(= \frac{2 x^{2} + 1}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)} - \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)} - \frac{x^{2} - x + 1}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)}\) \(= \frac{x \left(\right. x^{2} - x + 1 \left.\right)}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)}\) \(= \boxed{\frac{x}{x^{2} - x + 1}}\)
6.
\(\frac{5 - 3 x}{x + 1} - \frac{x - 1}{x^{2} - x + 1} - \frac{1}{x + 1}\) \(= \frac{5 - 3 x - 1}{x + 1} - \frac{x - 1}{x^{2} - x + 1}\) \(= \boxed{\frac{4 - 3 x}{x + 1} - \frac{x - 1}{x^{2} - x + 1}}\)
7. \(\frac{3}{x + 1} - \frac{2 + 3 x}{x^{3} + 1}\)
\(= \frac{3 \left(\right. x^{2} - x + 1 \left.\right) - \left(\right. 2 + 3 x \left.\right)}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)}\) \(= \frac{3 x^{2} - 6 x + 1}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)}\) \(= \boxed{\frac{3 x^{2} - 6 x + 1}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)}}\)
1. \(\frac{x + 3}{x^{2} - 1} - \frac{1}{x^{2} - x}\)
\(\frac{x + 3}{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)} - \frac{1}{x \left(\right. x - 1 \left.\right)}\) \(= \frac{x \left(\right. x + 3 \left.\right) - \left(\right. x + 1 \left.\right)}{x \left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}\) \(= \frac{x^{2} + 3 x - x - 1}{x \left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}\) \(= \boxed{\frac{x^{2} + 2 x - 1}{x \left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}}\)
2. \(\frac{x}{x - 1} - \frac{1}{x^{2} - x}\)
\(= \frac{x}{x - 1} - \frac{1}{x \left(\right. x - 1 \left.\right)}\) \(= \frac{x^{2} - 1}{x \left(\right. x - 1 \left.\right)}\) \(= \boxed{\frac{x^{2} - 1}{x \left(\right. x - 1 \left.\right)}}\)
3. \(\frac{x + 1}{x - 5} + \frac{x - 11}{x - 5}\)
\(= \frac{x + 1 + x - 11}{x - 5}\) \(= \frac{2 x - 10}{x - 5}\) \(= \boxed{2}\)
4. \(\frac{2}{x + 3} - \frac{2}{3 - x} - \frac{3 x}{x^{2} - 9}\)
\(= \frac{2}{x + 3} + \frac{2}{x - 3} - \frac{3 x}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)}\) \(= \frac{2 \left(\right. x - 3 \left.\right) + 2 \left(\right. x + 3 \left.\right) - 3 x}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)}\) \(= \frac{x}{x^{2} - 9}\) \(= \boxed{\frac{x}{x^{2} - 9}}\)
5.
\(\frac{2 x^{2} + 1}{x^{3} + 1} - \frac{x - 1}{x^{2} - x + 1} - \frac{1}{x + 1}\) \(= \frac{2 x^{2} + 1}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)} - \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)} - \frac{x^{2} - x + 1}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)}\) \(= \frac{x \left(\right. x^{2} - x + 1 \left.\right)}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)}\) \(= \boxed{\frac{x}{x^{2} - x + 1}}\)
6.
\(\frac{5 - 3 x}{x + 1} - \frac{x - 1}{x^{2} - x + 1} - \frac{1}{x + 1}\) \(= \frac{5 - 3 x - 1}{x + 1} - \frac{x - 1}{x^{2} - x + 1}\) \(= \boxed{\frac{4 - 3 x}{x + 1} - \frac{x - 1}{x^{2} - x + 1}}\)
7. \(\frac{3}{x + 1} - \frac{2 + 3 x}{x^{3} + 1}\)
\(= \frac{3 \left(\right. x^{2} - x + 1 \left.\right) - \left(\right. 2 + 3 x \left.\right)}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)}\) \(= \frac{3 x^{2} - 6 x + 1}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)}\) \(= \boxed{\frac{3 x^{2} - 6 x + 1}{\left(\right. x + 1 \left.\right) \left(\right. x^{2} - x + 1 \left.\right)}}\)
1: Ta có: \(\frac{x+3}{x^2-1}-\frac{1}{x^2-x}\)
\(=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x-1\right)}\)
\(=\frac{x\left(x+3\right)-x-1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x-1}{x\left(x+1\right)\left(x-1\right)}\)
2: Ta có: \(\frac{x}{x-1}-\frac{1}{x^2-x}\)
\(=\frac{x}{x-1}-\frac{1}{x\left(x-1\right)}\)
\(=\frac{x^2-1}{x\cdot\left(x-1\right)}=\frac{\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)}=\frac{x+1}{x}\)
3: TA có: \(\frac{x+1}{x-5}+\frac{x-11}{x-5}\)
\(=\frac{x+1+x-11}{x-5}\)
\(=\frac{2x-10}{x-5}=\frac{2\left(x-5\right)}{x-5}=2\)
4: Ta có: \(\frac{2}{x+3}-\frac{2}{3-x}-\frac{3x}{x^2-9}\)
\(=\frac{2}{x+3}+\frac{2}{x-3}-\frac{3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2\left(x-3\right)+2\left(x+3\right)-3x}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{2x-6+2x+6-3x}{\left(x+3\right)\left(x-3\right)}=\frac{x}{\left(x+3\right)\left(x-3\right)}\)
5: Ta có: \(\frac{2x^2+1}{x^3+1}-\frac{x-1}{x^2-x+1}-\frac{1}{x+1}\)
\(=\frac{2x^2+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{x-1}{x^2-x+1}-\frac{1}{x+1}\)
\(=\frac{2x^2+1-\left(x-1\right)\left(x+1\right)-x^2+x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^2+x-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x+1}{x^3-1}\)
6: Ta có: \(\frac{5-3x}{x+1}-\frac{x-1}{x^2-x+1}-\frac{1}{x+1}\)
\(=\frac{4-3x}{x+1}-\frac{x-1}{x^2-x+1}\)
\(=\frac{\left(-3x+4\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x+1\right)}{\left.\left(x+1\right)\right.\cdot\left(x^2-x+1\right)}\)
\(=\frac{-3x^3+3x^2-3x+4x^2-4x+4-x^2+1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{-3x^3+6x^2-7x+5}{\left(x+1\right)\left(x^2-x+1\right)}\)
7: \(\frac{3}{x+1}-\frac{2+3x}{x^3+1}\)
\(=\frac{3}{x+1}-\frac{3x+2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{3\left(x^2-x+1\right)-3x-2}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{3x^2-6x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)