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7 tháng 10 2015

\(5\left(x-y\right)^3-5\left(y-x\right)=5\left(x-y\right)^3+5\left(x-y\right)=5\left(x-y\right)\left[\left(x-y\right)^2+1\right]\)

30 tháng 8 2017

Câu 1:
\(\left(x+y\right)^2+3\left(x+y\right)-10\)
\(=\left(x+y\right)^2+3\left(x+y\right)+2,25-12,25\)
\(=\left(x+y+1,5\right)^2-3,5^2\)
\(=\left(x+y+1,5+3,5\right)\left(x+y+1,5-3,5\right)\)
\(=\left(x+y+5\right)\left(x+y-2\right)\)

Câu 2:
\(2x^2-y^2+xy\)
\(=2x^2-y^2+2xy-xy\)
\(=\left(2x^2+2xy\right)-\left(xy+y^2\right)\)
\(=2x\left(x+y\right)-y\left(x+y\right)\)
\(=\left(2x-y\right)\left(x+y\right)\)

Câu 3:
\(x^{64}+x^{32}+1\)
\(=x^{64}+2x^{32}+1-x^{32}\)
\(=\left(x^{32}+1\right)^2-\left(x^{16}\right)^2\)
\(=\left(x^{32}+1+x^{16}\right)\left(x^{32}+1-x^{16}\right)\)
\(=\left(x^{32}+x^{16}+1\right)\left(x^{32}-x^{16}+1\right)\)

Câu 4:
\(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2\)
\(=x^2+25y^2-10xy+6cd-\left(3cd\right)^2-1\)
\(=\left(x^2+25y^2-10xy\right)-\left(\left(3cd\right)^2+1-6cd\right)\)
\(=\left(x+5y\right)^2-\left(3cd-1\right)^2\)
\(=\left(\left(x+5y\right)+\left(3cd-1\right)\right)\cdot\left(\left(x+5y\right)-\left(3cd-1\right)\right)\)
\(=\left(x+5y+3cd-1\right)\left(x+5y-3cd+1\right)\)

= [\(\left(x-y\right)^3\)\(-1\)] - \(\left[3\left(x-y\right)\left(x-y-1\right)\right]\) 

\(\left\{\left(x-y-1\right)\left[\left(x-y\right)^2+2\left(x-y\right)+1\right]\right\}\)-  \(\left[3\left(x-y\right)\left(x-y-1\right)\right]\)

\(^{\left(x-y-1\right)\left[\left(x-y\right)^2+2\left(x-y\right)+1-3\left(x-y\right)\right]}\)

\(\left(x-y-1\right)\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]\)

\(\left(x-y-1\right)\left(x-y-1\right)^2\)

\(^{\left(x-y-1\right)^3}\)

T*ck mình nha. Suy nghĩ bài này cực lắm đó!

7 tháng 9 2022

= (x-y-1) [(x-y)^2 + (x-y) + 1] - 3(x-y)(x-y-1)

= (x-y-1) [(x-y)^2 + (x-y) + 1 - 3(x-y)]

= (x-y-1) [(x-y)^2 - 2(x-y)+1]

= (x-y-1)(x-y-1)^2

=(x-y-1)^3

14 tháng 11 2016

a) x2+2xy+y2-4= (x+y)2-22 => hiệu hai bình phương

=(x+y+2)(x+y-2)

Bài 2: 

a: \(=\left(x+y\right)^2-4=\left(x+y+2\right)\left(x+y-2\right)\)

b: \(=4x^2-\left(y+2\right)^2\)

\(=\left(2x-y-2\right)\left(2x+y+2\right)\)

c: \(=25a^4-\left(x-2y\right)^2\)

\(=\left(5a^2-x+2y\right)\left(5a^2+x-2y\right)\)

a)(x+y)(x+y)=x^2+2xy+y^2

b)(x-y)(x-y)=x^2-2xy+y^2

c)(x-y)(x-1)=x^2-x-xy+y

d)(x+5)(x-1)+x^2+4x-5