\(a,VT=\left(a^2-1\right)^2+4a^2\\ =a^4-2a^2+1+4a^2\\ =a^4+2a^2+1\\ =\left(a^2+1\right)^2 =VP\\ b,VT=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)\\ =x^2-2xy+y^2+x^2+y^2+2xy+2x^2-2y^2\\ =4x^2=VP\)

Bài 12:

a) \(\left(\dfrac{1}{2}x+4\right)^2\)

\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)

\(=\dfrac{1}{4}x^2+4x+16\)

b) \(\left(7x-5y\right)^2\)

\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)

\(=49x^2-70xy+25y^2\)

c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)

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

\(=y^4-36x^4\)

d) \(\left(x+2y\right)^2\)

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)

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

e) \(\left(x-3y\right)\left(x+3y\right)\)

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

\(=x^2-9y^2\)

f) \(\left(5-x\right)^2\)

\(=5^2-2\cdot5\cdot x+x^2\)

\(=25-10x+x^2\)

\(11,\)

\(a,\left(7x+4\right)^2-\left(7x+4\right)\left(7x-4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=\left(7x+4\right).8=56x+32\)

\(b,\left(x+2y\right)^2-6xy\left(x+2y\right)\)

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

6x² - 8xy +y² + 1=6x² - 8xy +y² + 1²=x(6x-8y)+(x+1)(x-1)

Cái này hình như k phân thích đc

Đáp án B.

1=1=3464535656567765432345676543234567

 a) 3x2 – 7x + 2

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

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

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

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

 b) a(x2 + 1) – x(a2 + 1)

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

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

.......?

a) Ta có: \(3x^2-7x+2\)

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

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

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

b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)

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

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

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

a) Ta có: \(\dfrac{P}{x+2}=\dfrac{x^2+5x+6}{x^2+4x+4}\)

\(\Leftrightarrow\dfrac{P}{x+2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)

hay P=x+3