a) \(=\left(x-2\right)^2\)

b) \(=\left(2x+1\right)^2\)

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

d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)

e) \(=\left(2x-3x+1\right)\left(2x+3x-1\right)=\left(1-x\right)\left(5x-1\right)\)

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

g) \(=\left(x+3\right)\left(x^2-3x+9\right)\)

h) \(=\left(x+2\right)^3\)

i) \(=\left(1-x\right)^3\)

a/ $=(x-2)^2$

b/ $=(2x+1)^2$

c/ $=(4x-3y)(4x+3y)$

d/ $=(1-x)(x+7)$

e/ $=(-x+1)(5x-1)$

f/ $=(x-y)(x^2+xy+y^2)$

g/ $=(3+x)(9-3x+x^2)$

h/ $=(x+2)^3$

i/ $=(1-x)^3$

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(4x^2+4x+1=\left(2x+1\right)^2\)

g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

1) \(\left(3x+7\right)^2-\left(2x-3\right)^2=0\)

\(\Leftrightarrow\left(3x+7-2x+3\right)\left(3x+7+2x-3\right)=0\)

\(\Leftrightarrow\left(x+10\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+10=0\\5x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-10\\x=\frac{-4}{5}\end{cases}}\)

Vạy ...

phần 2 tương tự áp dụng \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

\((4x-1)^2-(5-3x)^2=0\)

\(\Leftrightarrow(4x-1-5-3x)(4x+1+5-3x)=0\)

\(\Leftrightarrow(x-6)(x+6)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

Vậy : ...

Sai rồi nhé bỏ qua , mk xin lỗi

a: =4x^2+8x-3x-6

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

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

b: =3(3x^2-2x-1)

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

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

c: =2x^2-4x+x-2

=2x(x-2)+(x-2)

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

d: =3x^2+3x-2x-2

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

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

e: =3x^2+9x+x+3

=3x(x+3)+(x+3)

=(x+3)(3x+1)

a) \(4x^2+5x-6\)

\(=4x^2+8x-3x-6\)

\(=\left(4x^2+8x\right)-\left(3x+6\right)\)

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

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

b) \(9x^2-6x-3\)

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

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

\(=3\left[3x\left(x-1\right)+\left(x-1\right)\right]\)

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

c) \(2x^2-3x-2\)

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

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

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

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

d) \(3x^2+x-2\)

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

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

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

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

e) \(3x^2+10x+3\)

\(=3x^2+9x+x+3\)

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

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

:) bóc lột !

Câu 1: 

a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x

b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)

\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)

\(=2x^2+6x+17\)

c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)

Câu 2: 

\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x^2-2.x.3+3^2\right)=2x\left(x-3\right)^2\)

a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)

b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)

c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)

d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)

tik nhé

​

a) Ta có: \(a^3y^3+125\)

\(=\left(ay+5\right)\left(a^2y^2-5ay+25\right)\)

b) Ta có: \(8x^3-y^3-6xy\cdot\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy-6xy+y^2\right)\)

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

2) 9x²+ 12x+ 4

<=>(3x)²+ 2.3x.2+ 2² <=>(3x+ 2)²

3) 4x⁴+ 20x²+ 25

<=>(2x²)²+ 2.2x².5+ 5² <=>(2x²+5)²

4) 25x²- 20xy+ 4y²

<=> (5x)²- 2.5x.2y+ (2y)²<=> (5x-2y)²

5) 9x⁴- 12x²y+ 4y²

<=> (3x²)²- 2.3x².2.y+ (2y)²<=> (3x²- 2y)²

6) 4x⁴- 16x²y³+ 16y⁶

<=> (2x²)²- 2.2x².4y³+ (4y³)²<=> (2x²- 4y³)²

7) 9x⁴- 12x⁵+ 4x⁶

<=> (3x²)²- 2.3x².2x³+ (2x³)²<=> (3x²- 2x³)²