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b) \(x^2-8x-4y^2+16=\left(x^2-8x+16\right)-4y^2=\left(x-4\right)^2-4y^2=\left(x-4-2y\right)\left(x-4+2y\right)\)

a) Ta có: \(x^3+6x-7\)

\(=x^3-x+7x-7\)

\(=x\left(x-1\right)\left(x+1\right)+7\left(x-1\right)\)

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

b) Ta có: \(4x^2+8x-5\)

\(=4x^2+10x-2x-5\)

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

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

c) Ta có: \(9x^2-4y^2+6x-4y\)

\(=9x^2+6x+1-\left(4y^2+4y+1\right)\)

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

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

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

1) \(4x^5y^2-8x^4y^2+4x^3y^2\)

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

\(=4x^3y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

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

2) \(5x^4y^2-10x^3y^2+5x^2y^2\)

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

\(=5x^2y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

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

3) \(12x^2-12xy+3y^2\)

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

\(=3\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

4) \(8x^3-8x^2y+2xy^2\)

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

\(=2x\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

5) \(20x^4y^2-20x^3y^3+5x^2y^4\)

\(=5x^2y^2\left(4x^2-4xy+y^2\right)\)

\(=5x^2y^2\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

1: 4x^5y^2-8x^4y^2+4x^3y^2

=4x^3y^2(x^2-2x+1)

=4x^3y^2(x-1)^2

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

3: \(=3\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)^2\)

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

5: \(=5x^2y^2\left(4x^2-4xy+y^2\right)=5x^2y^2\left(2x-y\right)^2\)

Lời giải:

ĐKĐB $\Leftrightarrow (x^2+4y^2-4xy)+8x=5$

$\Leftrightarrow (x-2y)^2+8x=5$.

Đặt $x-2y=a; x=b$ thì bài toán trở thành:

Cho $a,b$ thực thỏa mãn $a^2+8b=5$. Tìm max của $B=-2a+8b$

Áp dụng BĐT AM-GM:

$a^2+1\geq 2\sqrt{a^2}=2|a|\geq -2a$

$\Rightarrow a^2+1\geq -2a$

$\Rightarrow a^2+8b+1\geq -2a+8b$

$\Leftrightarrow 6\geq B$. Vậy $B_{\max}=6$

a: \(=\left(-\dfrac{6}{2}\right)\cdot\dfrac{x^3}{x}\cdot\dfrac{y^2}{y^2}=-3x^2\)

b: \(=\left(-\dfrac{1}{4}:\dfrac{1}{2}\right)\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^3}{y^2}=-\dfrac{1}{2}xy\)

c: \(=\dfrac{8}{4}\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^5}{y^4}=2xy\)

\(a,-6x^3y^2:2xy^2=-3x^2\)

\(b,-\dfrac{1}{4}x^4y^3:\dfrac{1}{2}x^3y^2=-\dfrac{1}{2}xy\)

\(c,8x^4y^5:4x^3y^4=2xy\)

#Urushi

a: \(\left(2x+7y\right)^2=4x^2+28xy+49y^2\)

b: \(\left(8x-\dfrac{5}{4}y\right)^2=64x^2-20xy+\dfrac{25}{16}y^2\)

a, Đề sai bạn ơi phải là cộng 16 chứ không phải cộng 4

b,B= (x-2y+1)^2

thế còn c với d