dấu ^ là mũ nha mn

3)(9a)²-(5a-3b)²

= (9a-5a+3b)(9a+5a-3b)

= (4a+3b)(14a-3b)

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

16x^4_y2-25a²b²

1) \(x^6+1\)

\(=x^6+x^4-x^4+x^2-x^2+1\)

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

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

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

2) \(x^6-y^6\)

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

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

3) \(5x^2y^4-10x^4y^2+5x^2y^2\)

\(=5x^2y^2\left(y^2-2x^2+1\right)\)

4) \(25a^2-49b^2\)

\(=\left(5a+7b\right)\left(5a-7b\right)\)

5) \(36\left(x-y\right)^2-25\left(2x-1\right)^2\)

\(=\left[6\left(x-y\right)+5\left(2x-1\right)\right]\left[6\left(x-y\right)-5\left(2x-1\right)\right]\)

\(=\left[6x-6y+10x-5\right]\left[6x-6y-10x+5\right]\)

\(=\left(16x-6y-5\right)\left(-4x-6y+5\right)\)

a) = (xyz+xy) +(z+1) +(yz+zx)+(x+y)

 = xy(z+1) +(z+1)+z(x+y)+(x+y)

= (z+1)(xy+1)+(x+y)(Z+1)

=(z+1)(xy+1+x+y)

dễ mà bn

mk bk lm mà lm biếng gõ qá

a) \(\left(2x+5\right)^2\)\(-\left(x-9\right)^2\)

=\(\left(2x+5+x-9\right).\left(2x+5-x+9\right)\)

=\(\left(3x-4\right).\left(x+14\right)\)

Lời giải:
$x^4y^4-z^4=(x^2y^2)^2-(z^2)^2=(x^2y^2-z^2)(x^2y^2+z^2)$

$=(xy-z)(xy+z)(x^2y^2+z^2)$

$(x+y+z)^2-4z^2=(x+y+z)^2-(2z)^2=(x+y+z-2z)(x+y+z+2z)$
$=(x+y-z)(x+y+3z)$

$\frac{-1}{9}x^2+\frac{1}{3}xy-\frac{1}{4}y^2=\frac{-4x^2+12xy-9y^2}{36}$

$=-\frac{4x^2-12xy+9y^2}{36}=-\frac{(2x-3y)^2}{36}=-\left(\frac{2x-3y}{6}\right)^2$

Câu trả lời của cô quá đúng luôn đấy

a) Ta có: \(x^4y^4-z^4\)

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

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

b) Ta có: \(\left(x+y+z\right)^2-4z^2\)

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

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