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Ta có : \(M=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left[\left(x+1\right)\left(x+4\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]+1=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(t=x^2+5x+5\) \(\Rightarrow M=\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)
Vậy \(M=\left(x^2+5x+5\right)^2\)
\(a^2-6a+5=\left(a^2-5a\right)-\left(a-5\right)=a\left(a-5\right)-\left(a-5\right)=\left(a-1\right)\left(a-5\right)\)
\(a^2-7a+12=\left(a^2-3a\right)-\left(4a-12\right)=a\left(a-3\right)-4\left(a-3\right)=\left(a-4\right)\left(a-3\right)\)
\(4a^2+4a-3=4a^2-2a+\left(6a-3\right)=2a\left(2a-1\right)+3\left(2a-1\right)=\left(2a+3\right)\left(2a-1\right)\)
X2 - 6x + 5
= x2 - 6x + 5 + 4 - 4
= x2 - 6x + 9 - 22
= ( x - 3 )2 - 22
= ( x - 3 - 2 ) ( x - 3 + 2 )
a) x2 + 2x - 3 = x2 - x + 3x - 3 = x( x - 1 ) + 3( x - 1 ) = ( x - 1 )( x + 3 )
b) x2 - 2x - 15 = x2 + 3x - 5x - 15 = x( x + 3 ) - 5( x + 3 ) = ( x + 3 )( x - 5 )
c) x2 - 2x - 48 = x2 + 6x - 8x - 48 = x( x + 6 ) - 8( x + 6 ) = ( x + 6 )( x - 8 )
d) 4x2 + 4x - 15 = ( 4x2 + 4x + 1 ) - 16 = ( 2x + 1 )2 - 42 = ( 2x + 1 - 4 )( 2x + 1 + 4 ) = ( 2x - 3 )( 2x + 5 )
e) 3x2 - 7x + 2 = 3x2 - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )
f) 4x2 - 5x + 1 = 4x2 - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )
(x2+x)2+4x2+4x+12 = (x2+x)2+4(x2+x)+12
= (x2+x) (x2 + x +4)+12
1.xy(14x-21y+28xy)
2. a)\(x^2-4\ne0\Rightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
b)\(\frac{x^2-2x-2x+4}{x^2-4}=\frac{x\left(x-2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\) với đk (a)=> \(b=\frac{x-2}{x+2}=1-\frac{4}{x+2}\)
c) \(C=\frac{-3-2}{-3+2}=-\frac{5}{-1}=5\)
1. \(14x^2y-21xy^2+28x^2y^2\)
\(=7xy\left(2x-3y+4xy\right)\)
2.a)Để phân thức được xác định thì \(x^2-4\ne0\Leftrightarrow x^2\ne4\Leftrightarrow\orbr{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
b) \(\frac{x^2-4x+4}{x^2-4}=\frac{x^2-2.x.2+2^2}{x^2-2^2}\)
\(=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)
c)Thay x=-3 ta có:
\(\frac{-3-2}{-3+2}=\frac{-5}{-1}=5\)