a) \(4-x=2\left(x-4\right)^2\)

\(\Rightarrow4-x=2\left(x^2-8x+16\right)\)

\(\Rightarrow4-x=2x^2+16x-32=0\)

\(\Rightarrow-2x^2+15x-28=0\)

\(\Rightarrow-2x\left(x-4\right)+7\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(-2x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\-2x+7=0\end{cases}}\Rightarrow\orbr{\begin{cases}-2x=-7\\x=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=4\end{cases}}\)

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

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

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

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

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

\(\Rightarrow\orbr{\begin{cases}x^2+3=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=-3\text{(Không thoả mãn)}\\x=2\end{cases}}\)

ta có :

ta có :

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

\(=x^3+8-x^3-2x=8-2x\)

\(\left|x-1\right|=3\Leftrightarrow x-1=\pm3\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\)

Với \(x=4\): \(A=8-2.4=0\).

Với \(x=-2\): \(A=8-2.\left(-2\right)=12\).

a) \(A=\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)\\ A=x^3+8-x^3-2x\\ A=-2x+8\)

b)\(\left|x-1\right|=3\\ TH1\\ x-1=3\\ x=4\left(TM\right)\\ TH2\\ x-1=-3\\ x=-2\left(TM\right)\)

Thay x = 4 vào A

A = - 8 + 8 = 0

Thay x = -2 vào A

A = 4 + 8 = 12

HT  

a/ A = x³ + 8 - x³ - 2x = 8 - 2x

b/ |x-1| = 3 <=> x = 4 || x = -2
+ Với x = 4: A = 8 - 4 * 2 = 0

+ Với x = -2: A = 8 - (-2) * 2 = 12

\(x^4y^4+16+\left(xy+2\right)^4=t^4+16+\left(t+2\right)^4\)(\(t=xy\))

\(=t^4+16+\left(t^2+4t+4\right)^2=t^4+16+\left(t^2+2t+4\right)^2+4t\left(t^2+2t+4\right)+4t^2\)

\(=\left(t^2+2t+4\right)^2+t^4+4t^3+12t^2+16t+16\)

\(=2\left(t^2+2t+4\right)^2\)

\(=2\left(x^2y^2+2xy+4\right)^2\)

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