Question

Answer 1

dài vl

Answer 2

dòn thài lài

Answer 3

a.

\(x^4+x^2+6x-8=0\)

\(\Leftrightarrow\left(x^4-x^3+4x^2\right)+\left(x^3-x^2+4x\right)-\left(2x^2-2x+8\right)=0\)

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

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

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

\(\Leftrightarrow\left(x^2-x+4\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x^2-x+4\right)\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x-1=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Answer 4

b.

\(x^4-3x^3+2x^2-9x+9=0\)

Nhận thấy \(x=0\) ko phải nghiệm

Với \(x\ne0\) chia 2 vế cho \(x^2\) ta được:

\(x^2-3x+2-\dfrac{9}{x}+\dfrac{9}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\dfrac{9}{x^2}+6\right)-3\left(x+\dfrac{3}{x}\right)-4=0\)

\(\Leftrightarrow\left(x+\dfrac{3}{x}\right)^2-3\left(x+\dfrac{3}{x}\right)-4=0\)

Đặt \(x+\dfrac{3}{x}=t\)

\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{x}=-1\\x+\dfrac{3}{x}=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+3=-x\\x^2+3=4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+3=0\left(vn\right)\\x^2-4x+3=0\end{matrix}\right.\)

\(\Rightarrow x=\left\{1;3\right\}\)

 

Answer 5

c.

\(\left(x^2-2x\right)^2-6\left(x^2-2x\right)+9=0\)

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

\(\Leftrightarrow x^2-2x-3=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

d.

Đặt \(4x-5=t\) ta được:

\(t^2+7t-8=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-8\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x-5=1\\4x-5=-8\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

Answer 6

e.

\(\Leftrightarrow\left(x^2+6x+9\right)\left(x^2+6x+1\right)=9\)

\(\Leftrightarrow\left(x^2+6x+1+8\right)\left(x^2+6x+1\right)-9=0\)

Đặt \(x^2+6x+1=t\)

\(\Rightarrow\left(t+8\right).t-9=0\)

\(\Leftrightarrow t^2+8t-9=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+6x+1=1\\x^2+6x+1=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\\left(x+3\right)^2+1=0\left(vn\right)\end{matrix}\right.\)

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

Answer 7

f.

\(\Leftrightarrow2\left(8x^2-x\right)\left(8x^2-x+2\right)-126=0\)

Đặt \(8x^2-x=t\)

\(\Rightarrow2t\left(t+2\right)-126=0\)

\(\Leftrightarrow2t^2+4t-126=0\)

\(\Rightarrow\left[{}\begin{matrix}t=7\\t=-9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}8x^2-x=7\\8x^2-x=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}8x^2-x-7=0\\8x^2-x+9=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{8}\end{matrix}\right.\)