Câu 1:

a) \(\left(x^2+y^2-36\right)^2-4x^2y^2\)

\(=\left(x^2+y^2-36\right)^2-\left(2xy\right)^2\)

\(=\left(x^2+y^2+2xy-36\right)\left(x^2+y^2-2xy-36\right)\)

\(=\left[\left(x+y\right)^2-36\right]\left[\left(x-y\right)^2-36\right]\)

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

b) \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)

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

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

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

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

1) a) (x² + y² - 36)² - 4x²y² 

= (x² + y² - 36 - 2xy)(x² + y² - 36 + 2xy)

= [(x - y)² - 36][(x + y)² - 36]

= (x - y - 6)(x - y  + 6)(x + y + 6)(x + y - 6)

b) (x² + x)² - 5(x² + x) + 6

= (x² + x)² - 2(x² + x) - 3(x² + x) + 6

= (x²  + x)(x² + x - 2) - 3(x² + x - 2)

= (x² + x - 3)(x² + 2x - x - 2)

=  (x² + x - 3)(x - 1)(x + 2)

2) Đặt tính là đc

c: \(\Leftrightarrow2x^3-6x^2+4x+x^2-3x+2+a-2⋮x^2-3x+2\)

=>a-2=0

=>a=2

d: \(\dfrac{5x^3+4x^2-6x-a}{5x-1}=\dfrac{5x^3-x^2+5x^2-x-5x+1-a-1}{5x-1}\)

\(=x^2+x-1+\dfrac{-a-1}{5x-1}\)

Để dư bằng -3 thì -a-1=-3

=>a+1=3

=>a=2

3: \(\Leftrightarrow a-15=0\)

hay a=15

\(\frac{x^4+5x^3-2x^2+ax+40}{x^2-3x+b}\)

\(=\frac{x^4-3x^3+bx^2+8x^3-24x^2+8xb+\left(-b+22\right)x^2-3\left(-b+22\right)x+b\left(-b+22\right)}{x^2-3x+b}\) +\(\frac{x\left(a-11b+66\right)+40-b\left(-b+22\right)}{x^2-3x+b}\)

\(=x^2+8x+\left(-b+22\right)+\frac{x\left(a-11b+66\right)+40+b^2-22b}{x^2-3x+b}\)

Để \(\frac{x^4+5x^3-2x^2+ax+40}{x^2-3x+b}\) là phép chia hết thì \(\begin{cases}b^2-22b+40=0\\ a-11b+66=0\end{cases}\Rightarrow\begin{cases}\left(b-20\right)\left(b-2\right)=0\\ a=11b-66\end{cases}\)

TH1: b-20=0

=>b=20

\(a=11b-66=11\cdot20-66=220-66=154\)

TH2: b-2=0

=>b=2

a=11b-66

=22-66

=-44