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a,\(Đkxđ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(=\frac{x+2}{x-2}+\frac{2}{x^2-2x}+\frac{1}{x}\)
\(=\frac{x+2}{x-2}+\frac{2}{x\left(x-2\right)}+\frac{1}{x}\)
\(=\frac{x\left(x+2\right)+2+x-2}{x\left(x-2\right)}\)
\(=\frac{x^2+2x+2+x-2}{x\left(x-2\right)}\)
\(=\frac{x^2+3x}{x\left(x-2\right)}\)\(=\frac{x\left(x+3\right)}{x\left(x-2\right)}=\frac{x+3}{x-2}\)
\(\Rightarrow\frac{x+3}{x-2}=0\Leftrightarrow x+3=0\Rightarrow x=-3\left(tm\right)\)
b, \(\left(x-3\right)\left(2x+1\right)=\left(x-3\right)\left(x-1\right)\)
\(\Rightarrow\left(x-3\right)\left(2x+1\right)-\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\left(x-3\right)\left(2x+1-x+1\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
bai 1
1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0
<=>(2x)^2-5^2=0
<=>(2x+5)*(2x-5)=0
<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự
1.\(\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{x^2-x-6}\)
\(\Leftrightarrow\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)
\(ĐK:x\ne3;-2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)+x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+2\right)+x\left(x-3\right)=x^2+6\)
\(\Leftrightarrow x^2+4x+4+x^2-3x-x^2-6=0\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)
Vậy \(S=\left\{1\right\}\)
b.\(\left(x+1\right)^2+\left|x-1\right|=x^2+4\)
\(\Leftrightarrow\) \(\left(x+1\right)^2+x-1=x^2+4\) hoặc \(\left(x+1\right)^2+1-x=x^2+4\)
Xét \(\left(x+1\right)^2+x-1=x^2+4\)
\(\Leftrightarrow x^2+2x+1+x-1-x^2-4=0\)
\(\Leftrightarrow3x-4=0\)
\(\Leftrightarrow x=\dfrac{4}{3}\)
Xét \(\left(x+1\right)^2+1-x=x^2+4\)
\(\Leftrightarrow x^2+2x+1+1-x-x^2-4=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{\dfrac{4}{3};2\right\}\)
2.\(1-\dfrac{x-1}{3}< \dfrac{x+3}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6-2\left(x-1\right)}{6}< \dfrac{2\left(x+3\right)-3\left(x-2\right)}{6}\)
\(\Leftrightarrow6-2\left(x-1\right)< 2\left(x+3\right)-3\left(x-2\right)\)
\(\Leftrightarrow6-2x+2< 2x+6-3x+6\)
\(\Leftrightarrow-x< 4\)
\(\Leftrightarrow x>4\)
Vậy \(S=\left\{x|x>4\right\}\)
0 4
a) \(\frac{x^2+2}{5}\ge0\)
\(\Rightarrow x^2+2\ge0\)( đúng với mọi x )
Vậy \(S=\left\{ℝ\right\}\)
b) \(\frac{x+2}{x-3}< 0\)( ĐKXĐ : \(x\ne3\))
Xét hai trường hợp :
1. \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\)( loại )
2. \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3\)
Vậy nghiệm của bất phương trình là -2 < x < 3
c) \(\frac{x-1}{x-3}>1\)( ĐKXĐ : \(x\ne3\))
\(\Leftrightarrow\frac{x-3+2}{x-3}>1\)
\(\Leftrightarrow1+\frac{2}{x-3}>1\)
\(\Leftrightarrow\frac{2}{x-3}>0\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\)
Vậy nghiệm của bất phương trình là x > 3
Nhờ bạn khác vẽ trục số nhé vì mình mới lên lớp 8
\(\left|x+2\right|-\left|x-1\right|< \dfrac{x-3}{2}\)
Lập bảng xét dấu:
x x+2 x-1 -2 1 0 0 _ + + _ _ +
+) Với \(x< -2\Leftrightarrow-\left(x+2\right)-\left(1-x\right)< \dfrac{x-3}{2}\)
\(\Leftrightarrow-x-2-1+x< \dfrac{x-3}{2}\\ \Leftrightarrow-3< \dfrac{x-3}{2}\\ \Leftrightarrow\dfrac{x-3}{2}>-3\\ \Leftrightarrow x-3>-6\\ \Leftrightarrow x>-3\\ \Leftrightarrow-3< x< -2\)
+) Với \(-2\le x< 1\Leftrightarrow\left(x+2\right)-\left(1-x\right)< \dfrac{x-3}{2}\)
\(\Leftrightarrow x+2-1+x< \dfrac{x-3}{2}\\ \Leftrightarrow2x+1< \dfrac{x-3}{2}\\ \Leftrightarrow4x+2< x-3\\ \Leftrightarrow4x-x< -3-2\\ \Leftrightarrow3x< -5\\ \Leftrightarrow x< -\dfrac{5}{3}\\ \Leftrightarrow-2\le x< -\dfrac{5}{3}\)
+) Với \(x\ge1\Leftrightarrow x+2-x+1< \dfrac{x-3}{2}\)
\(\Leftrightarrow3< \dfrac{x-3}{2}\\ \Leftrightarrow\dfrac{x-3}{2}>3\\ \Leftrightarrow x-3>6\\ \Leftrightarrow x>9\left(T/m\right)\\ \)
Vậy bất phương trình có tập nghiệm \(S=\left\{x|-3< x< -\dfrac{5}{3};x>9\right\}\)