a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)

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

b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

c) Đổi đề: \(a^2x+a^2y-7x-7y\)

\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)

d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)

e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)

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

g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)

i) \(=\left(x+1\right)^2-4=\left(x+1-2\right)\left(x+1+2\right)=\left(x-1\right)\left(x+3\right)\)

a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)

b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)

e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)

g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)

i)\(=\left(x-1\right)^2-4=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\)

a,

\(A=4(x-2)(x+1)+(2x-4)^2+(x+1)^2\\=[2(x-2)]^2+2\cdot2(x-2)(x+1)+(x+1)^2\\=[2(x-2)+(x+1)]^2\\=(2x-4+x+1)^2\\=(3x-3)^2\)

Thay $x=\dfrac12$ vào $A$, ta được:

\(A=\Bigg(3\cdot\dfrac12-3\Bigg)^2=\Bigg(\dfrac{-3}{2}\Bigg)^2=\dfrac94\)

Vậy $A=\dfrac94$ khi $x=\dfrac12$.

b,

\(B=x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\\=(x^9-1)-(x^7-x^4)-(x^6-x^3)-(x^5-x^2)\\=[(x^3)^3-1]-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1)-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1-x^4-x^3-x^2)\\=(x^3-1)(x^6-x^4-x^2+1)\)

Thay $x=1$ vào $B$, ta được:

\(B=(1^3-1)(1^6-1^4-1^2+1)=0\)

Vậy $B=0$ khi $x=1$.

$Toru$

a) \(x^4+8x+63\)

\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)

\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)

\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)

c) \(\left(x^2+2x+7\right)+\left(x^2-2x+4\right)\left(x^2+2x+3\right)\left(1\right)\)

Ta có : \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Rightarrow x^2+2x+4=\dfrac{x^3-8}{x-2}\)

\(\left(1\right)\Rightarrow\left[\left(\dfrac{x^3-8}{x-2}+3\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-8}{x-2}-1\right)\right]\)

\(=\left[\left(\dfrac{x^3-3x-14}{x-2}\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-2x-5}{x-2}\right)\right]\)

\(=\dfrac{1}{x-2}\left[x^3-3x-14+\left(x^2-2x+4\right)\left(x^3-2x-5\right)\right]\)

a) x²y+xy+x+1= (x²y+xy)+(x+1)=xy(x+10+(x+1)=(x+1)(xy+1)

b) x²-(a+b)x+ab=x²-ax-bx+ab=(x²-ax)-(bx-ab)=x(x-a)-b(x-a)=(x-a)(x-b)

c) ax²+ay-bx²-by=(ax²+ay)-(bx²+by)=a(x²+y)-b(x²+y)=(a-b)(x²+y)

d) ax-2x-a²+2a=(ax-2x)-(a²-2a)=x(a-2)-a(a-2)=(a-2)(x-a)

e) 2x²+4ax+x+2a=(2x²+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)

f) x³+ax²+x+a=(x³+ax²)+(x+a)=x²(x+a)+(x+a)=(x²+1)(x+a)

g: Ta có: \(x^4+2x^3-4x-4\)

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

\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)

a: Ta có: \(x^2-4y^2-2x-4y\)

\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

c: Ta có: \(x^3+2x^2y-x-2y\)

\(=x^2\left(x+2y\right)-\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)

d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)

\(=3\left(x-y\right)\left(x+y\right)-2\cdot\left(x-y\right)^2\)

\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)

\(=\left(x-y\right)\left(x+5y\right)\)

e: Ta có: \(x^3-4x^2-9x+36\)

\(=x^2\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

f: Ta có: \(x^2-y^2-2x-2y\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-2\right)\)

1: =(x-1-y)(x-1+y)

3: =(x-1)(x+1)(x-2)

1: \(=\left(x-1\right)^2\)

2: \(x\in\left\{0;20\right\}\)

Câu 13:

\(1,=\left(x-1\right)^2\\ 2,\Leftrightarrow x\left(x-20\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=20\end{matrix}\right.\\ 3,\text{Đề lỗi}\)

Câu 14:

\(1,ĐK:x\ne-2\\ 2,=\dfrac{\left(x+2\right)^2}{x+2}=x+2\\ 3,\Leftrightarrow x+2=0\Leftrightarrow x=-2\left(ktm\right)\Leftrightarrow x\in\varnothing\)

Câu 16:

\(A=x^2-4x+4+20=\left(x-2\right)^2+20\ge20\)

Dấu \("="\Leftrightarrow x=2\)

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