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Làm bài 1 thôi !! Mấy bài kia tương tự . Tìm nhân tử chung ra .
a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)
b) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2=\left(x^2+x-1+x^2+2x+3\right)\left(x^2+x-1-x^2-2x-3\right)\)
\(=\left(2x^2+3x+2\right)\left(-x-4\right)\)
c) \(-16+\left(x-3\right)^2=\left(x-3+4\right)\left(x-3-4\right)=x\left(x-7\right)\)
d) \(64+16y+y^2=\left(y+8\right)\left(y+8\right)\)
a)x3+3x2+3x+1
=x3+3x2*1+3x*12+13
=(x+1)3
b)(x+y)2-9x2
=y2+2xy+x2-9x2
=y2-2xy+4xy-8x2
=y(y-2x)+4x(y-2x)
=(y-2x)(y+4x)
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)
\(x^3-9x^2+27x-27\)
\(=x^3-3x^2.3+3.x.3^2-3^3\)
\(=\left(x-3\right)^3\)
\(27x^6-y^3\)
\(=\left(3x^2\right)^3-y^3\)
\(=\left(3x^2-y\right)\left[\left(3x^2\right)^2+3x^2y+y^2\right]\)
\(=\left(3x^2-y\right)\left(9x^4+3x^2y+y^2\right)\)
a) \(x^2+2x+1\)
\(=\left(x+1\right)^2\)
b) \(x^2-6x+9\)
\(=\left(x-3\right)^2\)
c) \(x^2+4x+4\)
\(=\left(x+2\right)^2\)
d) \(x^3+9x^2+27x+27\)
\(=\left(x+3\right)^3\)
\(x^3-0.36x\)
\(=x\left(x^2-0.36\right)\)
\(=x\left(x-0.6\right)\left(x+0.6\right)\)
\(x^3-9x=x\left(x^2-9\right)\)
\(=x\left(x-3\right)\left(x+3\right)\)
x − 0.36x
= x x − 0.36
= x x − 0.6 x + 0.6 x − 9x
= x x − 9 = x x − 3 x + 3 3 ( 2 ) ( ) ( ) 3 ( 2 )