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

\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)

\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+2y^2-4y+3=0\\2x^2+2x^2y^2-4y=0\left(1\right)\end{matrix}\right.\Rightarrow}x^3+2y^2-4y-2x^2-2x^2y^2+4y=0\Rightarrow x^3+1-2x^2y^2+2y^2-2x^2+2=0\Rightarrow\left(x+1\right)\left(x^2-x+1\right)-2y^2\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(x^2-x+1-2xy^2+2y^2-2x+2\right)=0\Rightarrow x=-1\)Thay x=-1 vào (1) ta được y²-2y+1=0⇒ (y-1)²=0⇒y-1=0⇒y=1

Do đó Q=x²+y²=(-1)²+1²=2

Ta có: x=2 

nên x-1=1

Ta có: \(B=\left(x+1\right)\left(x^7-x^6+x^5-x^4+x^3-x^2+x-1\right)\)

\(=\left(x+1\right)\left[x^6\left(x-1\right)+x^4\left(x-1\right)+x^2\left(x-1\right)+\left(x-1\right)\right]\)

\(=\left(x+1\right)\left(x^6+x^4+x^2+1\right)\)

\(=\left(x+1\right)\left(x+1\right)\left(x^4+1\right)\)

\(=\left(2^4+1\right)\left(2+1\right)^2=17\cdot9=153\)

a) Ta có: \(36x^3-4x=0\)

\(\Leftrightarrow4x\left(9x^2-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)

b) Ta có: \(3x\left(x-2\right)+x-2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)

d) Ta có: \(x^2-6x-16=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

e) Ta có: \(x^4-6x^2-7=0\)

\(\Leftrightarrow\left(x^2-7\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x\in\left\{\sqrt{7};-\sqrt{7}\right\}\)

a. \(x^4-16=0\\ \Leftrightarrow\left(x^2-4\right)\left(x^2+4\right)=0\)

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

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

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

a. x⁴ - 16 = 0
=> x⁴ = 16
=> x⁴ = 2⁴
=> x = 2
b. x² - 9x + 8 = 0
=> x² - 8x - x + 8 = 0
=> ( x² - x ) - ( 8x - 8 ) = 0
=> x(x-1) - 8(x-1)=0
=> (x-1)(x-8)=0
=>TH1: x-1=0       TH2 : x-8=0
=> x=1                       => x=8
 

A. x^4 - 16 = 0

=>x^4        = 16

=>x           = 2

B.x^2 - 9x + 8 = 0

=>x(x - 9)        = -8

=>x(x + 9)       = 8

=>x = -1

Câu b chưa chắc mình làm đúng đâu nha.

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

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

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

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

\(\Rightarrow x^4+2x^3-4x-4=0\\ \Rightarrow x^4-2x^2+2x^3-4x+2x^2-4=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+2x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=2\\\left(x+1\right)^2+1=0\left(vô.lí\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)