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Đặt \(x^2+3x+1=t\)
\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)
\(=t^2+t-6=\left(t^2-2t\right)+\left(3t-6\right)\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1=a\)ta có :
\(a\left(a+1\right)-6\)
\(=a^2+a-6\)
\(=a^2+6a-a-6\)
\(=\left(a^2+6a\right)-\left(a+6\right)\)
\(=a\left(a+6\right)-\left(a+6\right)\)
\(=\left(a+6\right)\left(a-1\right)\)
Thay \(a=x^2+3x+1\)vào A ta có :
\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)
\(=\left(x^2+3x+7\right)\left(x^2+3x\right)\)
\(3x\left(x-2\right)-x+2+5x\left(x-2\right)=\left(x-2\right)\left(8x-1\right)\)
\(3x\left(x-2\right)-x+2+5x\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)+5x\left(x-2\right)=\left(x-2\right)\left(3x=1+5x\right)=\left(x-2\right)\left(8x-1\right)\)
Ta có: \(x^2-3x-28\)
\(=x^2-7x+4x-28\)
\(=x\left(x-7\right)+4\left(x-7\right)\)
\(=\left(x-7\right)\left(x+4\right)\)
\(x^3-3x^2+6x-4\)
\(=x^3-2x^2+4x-x^2+2x-4\)
\(=\left(x^3-2x^2+4x\right)-\left(x^2-2x+4\right)\)
\(=x\left(x^2-2x+4\right)-\left(x^2-2x+4\right)\)
\(=\left(x-1\right)\left(x^2-2x+4\right)\)
x^3 - 3x^2 + 6x - 4
<=> x^3-3x^2+3x-1+3x-3
<=>(x-1)^3+3(x-1)
<=>(x-1)+((x-1)^2+3)
<=>(x-1)+(x^2-2x+4)
a: \(=3\left(x^2-y^2-x+y\right)\)
\(=3\left[\left(x-y\right)\left(x+y\right)-\left(x-y\right)\right]\)
=3(x-y)(x+y-1)
b: =(x-4)(x+1)
c: =x(x-1)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
( x + 2 ) ( x2 - 2x ) - 3x - 6
= ( x + 2 ) ( x2 - 2x ) - ( 3x + 6 )
= ( x + 2 ) ( x2 - 2x ) - 3 ( x + 2 )
= ( x + 2 ) ( x2 - 2x - 3 )
= ( x + 2 ) [ ( x2 + x ) - ( 3x + 3 ) ]
= ( x + 2 ) [ x ( x + 1 ) - 3 ( x + 1 ) ]
= ( x + 2 ) ( x - 3 ) ( x + 1 )
( x + 2 )( x2 - 2x ) - 3x - 6
= ( x + 2 )( x2 - 2x ) - 3( x + 2 )
= ( x + 2 )( x2 - 2x - 3 )
= ( x + 2 )[ ( x2 - 2x + 1 ) - 4 ]
= ( x + 2 )[ ( x - 1 )2 - 22 ]
= ( x + 2 )( x - 1 - 2 )( x - 1 + 2 )
= ( x + 2 )( x - 3 )( x + 1 )
x2 - x - 2x + 2
= (x2 - x) - (2x - 2)
= x(x - 1) - 2(x - 1)
= (x - 2)(x - 1)