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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) 9x2+30x+25=32x2+2.3.5x+52=(3x+5)2
b)12/5x2y2-9x4-4/25y4=-(9x4-12/5x2y2+4/25y4)=-(3x-2/5y)2
c)a2y2+b2x22axby=(ax-by)2
d)64x2-(8a+b)2=(8x-8a-b)(8x+8a+b)
a/ 9x2-12xy+4y2 = (3x - 2y)2
b/ 25x2-10x+1 = (5x - 1)2
c/ 9x2-12x+4 = (3x - 2)2
d/ 4x2+20x+25 = (2x + 5)2
e/ x4-4x2+4 = (x2 - 2)2
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
\(9x^2-12xy+16y^2\)
\(=\left(3x\right)^2-2.\left(3x\right)\left(4y\right)+\left(4y\right)^2\)
\(=\left(3x-4y\right)^2\)
\(P=\frac{x^2}{4}+x^2+1=\left(\frac{x}{2}\right)^2+2.x^2.\frac{1}{2}+1=\left(\frac{x}{2}+1\right)^2\)
2, a, \(9x^2-12x+9=\left(3x\right)^2-2.3.x.3+3^2=\left(3x-3\right)^2\ge0\)