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\(\left(x+2\right)\times\left(x+4\right)\times\left(x+6\right)\times\left(x+8\right)+16\)
\(=\left(x+2\right)\times\left(x+8\right)\times\left(x+4\right)\times\left(x+6\right)+16\)
\(=\left(x^2+10x+16\right)\times\left(x^2+10x+24\right)+16\)
Đặt \(t=x^2+10x+16\), ta được :
\(t\times\left(t+8\right)+16\)
\(=t^2+8t+16\)
\(=\left(t+4^2\right)\)
Thay \(t=x^2+10x+16\), ta được :
\(\left(x^2+10x+16\right)^2\)
\(=\left[\left(x+2\right)\times\left(x+8\right)\right]^2\)
\(=\left(x+2\right)^2\times\left(x+8\right)^2\)
\(=\left(x+2\right)^2\left(x+8\right)^2\)
_ Vậy \(\left(x+2\right)\times\left(x+4\right)\times\left(x+6\right)\times\left(x+8\right)+16\)\(=\left(x+2\right)^2\left(x+8\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+x-2\)
\(=x^2-x+2x-2\)
\(=x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
chì làm được 1 cách thôi
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\(x^2-6x+8\)
\(=\left(x^2-6x+9\right)-1\)
\(=\left(x-3\right)^2-1^2\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
(Tíck cho mìk vs nha!)
cách 2:
x2 -6x +8 = x2 -2x -4x+8= x(x-2) -4(x-2)
= (x-2)(x-4)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x-6\right)+32x^2\)
\(=\left(x^2-7x+6\right)\left(x^2+5x+6\right)+32x^2\)
Đặt : \(x^2+6=a\left(a< 0\right)\). Khi đó pt trở thành:
\(\left(a-7x\right)\left(a+5x\right)+32x^2\)
\(=a^2-2ax-3x^2=\left(a+x\right)\left(a-3x\right)\)
\(=\left(x^2+x+6\right)\left(x^2-3x+6\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
( x2 - x + 4 )( x2 - x + 5 ) - 6
Đặt t = x2 - x + 4
Đa thức đã cho trở thành
t( t + 1 ) - 6
= t2 + t - 6
= t2 - 2t + 3t - 6
= t( t - 2 ) + 3( t - 2 )
= ( t - 2 )( t + 3 )
= ( x2 - x + 4 - 2 )( x2 - x + 4 + 3 )
= ( x2 - x + 2 )( x2 - x + 7 )
![](https://rs.olm.vn/images/avt/0.png?1311)
Có : x^2-x-6 = (x^2-3x)+(2x-6) = x.(x-3)+2.(x-3) = (x-3).(x+2)
Tk mk nha
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\(^{x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x\left(x-3\right)+2\left(x-3\right)=\left(x-3\right)\left(x+2\right)}\)
x^2-x-6=x^2-3x+2x-6
=(x^2-3x)+(2x-6)
=x(x-3)+2(x-3)
=(x+2).(x-3)