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Bài 6:
c: \(9x^2+6x+1=\left(3x+1\right)^2\)
d: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)
e: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left(2x+3-2x-5\right)^2\)
=4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^4+8x+63\)
\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)
\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)
\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
c) \(\left(x^2+2x+7\right)+\left(x^2-2x+4\right)\left(x^2+2x+3\right)\left(1\right)\)
Ta có : \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Rightarrow x^2+2x+4=\dfrac{x^3-8}{x-2}\)
\(\left(1\right)\Rightarrow\left[\left(\dfrac{x^3-8}{x-2}+3\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-8}{x-2}-1\right)\right]\)
\(=\left[\left(\dfrac{x^3-3x-14}{x-2}\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-2x-5}{x-2}\right)\right]\)
\(=\dfrac{1}{x-2}\left[x^3-3x-14+\left(x^2-2x+4\right)\left(x^3-2x-5\right)\right]\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x2 + 2x – 3
= x2 + 2x + 1 – 4
= (x + 1)2 – 22
= (x + 1 + 2)(x + 1 – 2)
= (x + 3)(x – 1)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : x2 + 2x - 3
= x2 - x + 3x - 3
= x(x - 1) + 3(x - 1)
= (x + 3)(x - 1)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-3xy+2x-6y\)
= \(x\left(x-3y\right)+2\left(x-3y\right)\)
= \(\left(x+2\right)\left(x-3y\right)\)
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= ( x2 - 4y2 ) - ( 2x + 4y )
= ( x - 2y ) ( x + 2y ) - 2 ( x - 2y )
= ( x - 2y ) ( x + 2y - 2 )
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+2x+1-16=\left(x+1\right)^2-4^2=\left(x+1-4\right).\left(x+1+4\right)=\left(x-3\right).\left(x+5\right)\)
\(x^2+2x+1-16=\left(x^2+2x+1\right)-4^2=\left(x+1\right)^2-4^2=\left(x+1-4\right)\left(x+1+4\right)=\left(x-3\right)\left(x+5\right)\)
Đặt \(A=\left(x^2-2x+3\right)\left(x^2-2x+5\right)-8\). Rút gọn A,ta được:
\(A=x^4-4x^3+12x^2-16x+7\)
\(=x^4-2x^3+x^2-2x^3+4x^2-2x+7x^2-14x+7\)
\(=x^2\left(x^2-2x+1\right)-2x\left(x^2-2x+1\right)+7\left(x^2-2x+1\right)\)
\(=\left(x^2-2x+1\right)\left(x^2-2x+7\right)\)
\(=\left(x-1\right)^2\left(x^2-2x+7\right)\)
Ok chứ?