A chia hết cho B

=>\(49x^2+ax+b⋮7x-1\)

=>\(49x^2-7x+\left(a+7\right)x-\dfrac{1}{7}\left(a+7\right)+b+\dfrac{1}{7}\left(a+7\right)⋮7x-1\)

=>\(7x\left(7x-1\right)+\dfrac{1}{7}\left(a+7\right)\left(7x-1\right)+b+\dfrac{1}{7}\left(a+7\right)=0\)

b+1/7(a+7)=0

=>(a+7)+7b=0

=>a=-7b-7

Vậy: Với a,b là các số nguyên sao cho a=-7b-7 thì A chia hết cho B

a) Ta có  (1 + 2y)² + (1 - 2y)² + 2(1 + 2y)(1 - 2y) 

= (1 + 2y + 1 - 2y)² = 2² = 4

b) Ta có (7x + 2y)² + (7x - 2y)² - 2(49x² - 4y²) 

= (7x + 2y)² + (7x - 2y)² - 2[(7x)² - (2y)²]

= (7x + 2y)² + (7x - 2y)² - 2(7x - 2y)(7x + 2y)

= (7x + 2y - 7x + 2y)² 

= (4y)² = 16y²

a) \(x^4+3x^3-7x^2-27x-18\)

\(=\left(x^4+3x^3+2x^2\right)-\left(9x^2-27x-18\right)\)

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

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

b) \(x^4+5x^3-7x^2-41x-30\)

\(=\left(x^4+2x^3-15x^2\right)+\left(3x^3+6x^2-45x\right)+\left(2x^2+4x-30\right)\)

\(=x^2\left(x^2+2x-15\right)+3x\left(x^2+2x-15\right)+2\left(x^2+2x-15\right)\)

\(=\left(x^2+2x-15\right)\left(x^2+3x+2\right)=\left(x^2+5x-3x-15\right)\left(x^2+x+2x+2\right)\)

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

c) \(x^6-14x^4+49x^2-36\)

\(=\left(x^6-9x^4\right)+\left(-5x^4+45x^2\right)+\left(4x^2-36\right)\)

\(=x^4\left(x^2-9\right)-5x^2\left(x^2-9\right)+4\left(x^2-9\right)\)

\(=\left(x^2-9\right)\left(x^4-5x^2+4\right)=\left(x^2-9\right)\left(x^4-4x^2-x^2+4\right)\)

\(=\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x-3\right)\left(x+3\right)\)

a) \(14\left(x-y\right)^2+21\left(y-x\right)\)

\(=14\left(x-y\right)^2-21\left(x-y\right)\)

\(=7\left(x-y\right)\left[2\left(x-y\right)-3\right]\)

\(=7\left(x-y\right)\left(2x-2y-3\right)\)

b) \(7x^5\left(y-3\right)-49x^4\left(3-y\right)^3\)

\(=7x^4\left(y-3\right)\left[x+7\left(y-3\right)^2\right]\)

\(=7x^4\left(y-3\right)\left(x+7y^2-42y+63\right)\)

c) \(\left(x^2-9\right)^2-x^2\left(x-3\right)^2\)

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

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

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

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

d) \(\left(4x^2-1\right)^2-9\left(2x-1\right)^2\)

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

\(=\left(2x-1\right)^2\left[\left(2x+1\right)^2-9\right]\)

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

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

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

a) Ta có: \(\left(2x+3\right)^2-\left(5+x\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(2x+3+5+x\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(3x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-3\\3x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{2}\\x=\frac{-8}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-3}{2};\frac{-8}{3}\right\}\)

b) Ta có: \(\left(2x+5\right)^2-\left(2x-5\right)^2=6x+8\)

\(\Leftrightarrow\left(2x+5+2x-5\right)\left(2x+5-2x+5\right)-6x-8=0\)

\(\Leftrightarrow40x-6x-8=0\)

\(\Leftrightarrow34x=8\)

\(\Leftrightarrow x=\frac{8}{34}=\frac{4}{17}\)

Vậy: \(x=\frac{4}{17}\)

c) Ta có: \(\left(4x+3\right)^2=4\left(x-1\right)^2\)

\(\Leftrightarrow16x^2+24x+9=4\left(x^2-2x+1\right)\)

\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)

\(\Leftrightarrow12x^2+32x+5=0\)

\(\Leftrightarrow12x^2+2x+30x+5=0\)

\(\Leftrightarrow2x\left(6x+1\right)+5\left(6x+1\right)=0\)

\(\Leftrightarrow\left(6x+1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}6x+1=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-1\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)

d) Ta có: \(\left(7x-1\right)\left(3x-2\right)-49x^2+14x=1\)

\(\Leftrightarrow\left(7x-1\right)\left(3x-2\right)-\left(49x^2-14x+1\right)=0\)

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

\(\Leftrightarrow\left(7x-1\right)\left[3x-2-\left(7x-1\right)\right]=0\)

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

\(\Leftrightarrow\left(7x-1\right)\left(-4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-1=0\\-4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=1\\-4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{7}\\x=\frac{-1}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{7};\frac{-1}{4}\right\}\)