Câu hỏi của trần thị anh thư - Toán lớp 8 - Học toán với OnlineMath

   x4-30x2+31x-30=0
<=>x4+x-30x2+30x-30=0
<=>x(x3+1)-30(x2-x+1)=0
<=>x(x+1)(x2-x+1)-30(x2-x+1)=0
<=>(x2-x+1)(x2+x-30)=0
<=>(x2-x+1)(x2-5x+6x-30)=0
<=>(x2-x+1)[x(x-5)+6(x-5)]=0
<=>(x2-x+1)(x-5)(x+6)=0
Vì x2-x+1=x2-2x.1/2+1/4+3/4=(x-1/2)2+3/4>0 với mọi x
Do đó: <=>x-5 =0    <=> x=5
                x+6=0           x=-6
Vậy phương trình có tập nghiệm là S={5;-6}

P/S: kham khảo

a, Đặt \(x^2-5x=a\)

\(\Rightarrow\)\(a^2+10a+24=0\)

\(\Rightarrow a^2+4a+6a+24=0\)

\(\Rightarrow\left(a+4\right)\left(a+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a+4=0\\a+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2-5x+4=0\left(1\right)\\x^2-5x+6=0\left(2\right)\end{cases}}}\)

Giải pt (1) ta có : \(x^2-5x+4=0\)

\(\Rightarrow x^2-4x-x+4=0\)

\(\Rightarrow\left(x-4\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)

Giải pt (2) ta có : \(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

Vậy \(S=\left\{1;2;3;4\right\}\)

\(x^4-30x^2+31x-30=0\)

\(\Rightarrow x^4-30x^2+x+30x-30=0\)

\(\Rightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)

\(\Rightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)\)

\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)\)

\(\Rightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)

Mà \(x^2-x+1>0\)với \(\forall\)\(x\)

\(\Rightarrow x^2+x-30=0\)

\(\Rightarrow x^2-5x+6x-30=0\)

\(\Rightarrow x\left(x-5\right)+6\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)

Vậy \(S=\left\{5;-6\right\}\)

Hel me

= x^4+x^2+1-31x^2+31x-31

= (x^2+x+1)(x^2-x+1)-31(x^2-x+1)

= (x^2-x+1)(x^2+x+1-31)

= (x^2-x-1)(x^2+x-30)

 =  (x^2-x+1)(x^2+6x-5x-30)

=    (x^2-x+1)(x-5)(x+6)

vũ mạnh phi sai ở dấu = thứ  ấy là cộng 1 chớ ko phải trừ 1

\(\frac{4x^4-20x^3+13x^2+30x+9}{\left(4x^2-1\right)^2}\)

\(=\frac{4x^3\left(x-3\right)-8x^2\left(x-3\right)-11x\left(x-3\right)-3\left(x-3\right)}{\left(4x^2-1\right)^2}\)

\(=\frac{\left(x-3\right)\left(4x^3-8x^2-11x-3\right)}{\left(4x^2-1\right)^2}\)

\(=\frac{\left(x-3\right)\left[4x^2\left(x-3\right)+4x\left(x-3\right)+\left(x-3\right)\right]}{\left[\left(2x-1\right)\left(2x+1\right)\right]^2}\)

\(=\frac{\left(x-3\right)^2\left(4x^2+4x+1\right)}{\left(2x-1\right)^2\left(2x+1\right)^2}=\frac{\left(x-3\right)^2\left(2x+1\right)^2}{\left(2x-1\right)^2\left(2x+1\right)^2}=\frac{\left(x-3\right)^2}{\left(2x-1\right)^2}\)

\(\frac{-10x^2+30x+30}{x-4}=\frac{-10\left(x-4\right)\left(x+1\right)-10}{x-4}=-10\left(x+1\right)-\frac{10}{x-4}\inℤ\)

mà \(x\inℤ\)nên \(x-4\inƯ\left(10\right)=\left\{-10,-5,-2,-1,1,2,5,10\right\}\Leftrightarrow x\in\left\{-6,-1,2,3,5,6,9,14\right\}\).

Thx nha

\(x^4-30x^2+31x-30=0\)

\(\Leftrightarrow x^4-5x^3+5x^3-25x^2-5x^2+25x+6x-30=0\)

\(\Leftrightarrow x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)

\(\Leftrightarrow\left(x^3+5x^2-5x+6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x^3+6x^2-x^2-6x+x+6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]\left(x-5\right)=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\left(x+6\right)\left(x-5\right)=0\)

Vì \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=5\end{matrix}\right.\)

Vậy x = -6 hoặc x = 5

\(x^4-30x^2+31x-30=0\)

\(\Leftrightarrow x^4-30x^2+30x-30+x=0\)

\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)

\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+1=0\\x^2+x-30=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vl\right)\\\left(x-5\right)\left(x+6\right)=0\end{matrix}\right.\)

=> x = 5 hoặc x = -6.

p/s: ***** = vô lý :V

\(ĐKXĐ:x\ne49;x\ne50\)

Đặt \(x-49=u;x-50=v\)

Phương trình trở thành \(\frac{50}{u}+\frac{49}{v}=\frac{u}{50}+\frac{v}{49}\)

\(\Rightarrow\frac{50v+49u}{uv}=\frac{49u+50v}{2450}\)

\(\Rightarrow\orbr{\begin{cases}50v+49u=0\\uv=2450\end{cases}}\)

+) \(50v+49u=0\)

\(\Rightarrow50v=-49u\)

\(\Rightarrow\frac{v}{-49}=\frac{u}{50}=\frac{\left(x-50\right)-\left(x-49\right)}{-49-50}\)

\(=\frac{-1}{-99}=\frac{1}{99}\)

\(\Rightarrow\hept{\begin{cases}v=\frac{-49}{99}\\u=\frac{50}{99}\end{cases}}\Rightarrow x=\frac{4901}{99}\)(tm)

+) \(uv=2450\)

hay \(\left(x-49\right)\left(x-50\right)=2450\)

\(\Leftrightarrow x^2-99x+2450=2450\)

\(\Leftrightarrow x^2-99x=0\Leftrightarrow x\left(x-99\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=99\end{cases}}\left(tm\right)\)

Vậy phương trình có 3 nghiệm \(S=\left\{0;\frac{4901}{99};99\right\}\)

ok cảm ơn bn

MÌNH KHÔNG BIẾT ^_^

\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)

Đặt \(a=x^2+x\)

\(\Leftrightarrow a^2+4a=12\)

\(\Leftrightarrow a^2+4a-12=0\)

\(\Leftrightarrow a^2+6a-2a-12=0\)

\(\Leftrightarrow a\left(a+6\right)-2\left(a+6\right)=0\)

\(\Leftrightarrow\left(a+6\right)\left(a-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-6\\a=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+x=-6\\x^2+x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{23}{4}=0\\x^2+2x-x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\\\left(x+2\right)\left(x-1\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

Vậy....