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a, \(\left(a-b\right)\left(b-a\right)y+a-b\)
\(=\left(a-b\right)\left(by-ay+1\right)\)
b, \(\left(x-y+z\right)\left(a+y-x-z\right)b-x+y-z\)
\(=\left(x-y+z\right)\left(a+y-x-z\right)b-\left(x-y+z\right)\)
\(=\left(x-y+z\right)\left(ab+yb-xb-xz-1\right)\)
c, \(\left(2a+3\right)x-\left(2a+3\right)y+2a+3\)
\(=\left(2a+3\right)\left(x-y\right)+\left(2a+3\right)\)
\(=\left(2a+3\right)\left(x-y+1\right)\)
d, \(\left(a-b\right)x+\left(b-a\right)y-a+b\)
\(=\left(a-b\right)x-\left(a-b\right)y-\left(a-b\right)\)
\(=\left(a-b\right)\left(x-y-1\right)\)
a. (a-b)(b-a)y+a-b=(a-b)[(b-a)y+1]
b. (x-y+z)(a+y-x-z)b-x+y-z=(x-y+z)(a-y-x-z)b-(x-y+z)=(x-y+z)[(a-y-x-z)b-1]
c. (2a+3)x-(2a+3)y+2a+3=(2a+3)(x-y+1)
d. (a-b)x+(b-a)y-a+b=(a-b)x-(a-b)y-(a-b)=(a-b)(x-y-1)
a: \(=-y\left(a-b\right)^2+\left(a-b\right)\)
\(=\left(a-b\right)\left(-ya+yb+1\right)\)
b: \(=\left(x-y+z\right)\left(a+y-x-z\right)\cdot b-\left(x-y+z\right)\)
\(=\left(x-y+z\right)\left(ab-yb-xb-zb-1\right)\)
c: \(\left(2a+3\right)\cdot x-\left(2a+3\right)\cdot y+2a+3\)
\(=\left(2a+3\right)\left(x-y+1\right)\)
d: \(=\left(a-b\right)\cdot x-\left(a-b\right)\cdot y-\left(a-b\right)\)
\(=\left(a-b\right)\left(x-y-1\right)\)
a,ta có : x^3+y^3-xy(x+y)=(x+y)(x^2+y^2-xy) -xy(x+y)=(x+y)(x^2+y^2-2xy=(x+y)(x-y)^2
(đpcm)
b)x^3-y^3+xy(x-y)=(x-y)(x^2+y^2+xy)+xy(x-y)=(x-y)(x^2+y^2+2xy)=(x-y)(x+y)^2 (đpcm)
Bài 1:
a) \(\left(x+y\right)^2-y^2=x^2+2xy+y^2-y^2=x^2+2xy=x\left(x+2y\right)\)
b) Sửa đề: \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)^2\left(x+y\right)^2\)
c) \(x\left(x-3y\right)^2+y\left(y-3x\right)^2=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\)
\(=x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y\)
\(=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3\)
Bài 2:
a) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)
\(=2a\left(a^2+3b^2\right)\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(b^2+3a^2\right)\)
1, \(2\left(x+y\right)-5a\left(x+y\right)=\left(x+y\right)\left(2-5a\right)\)
2, \(a^2\left(x-5\right)-3\left(x-5\right)=\left(a^2-3\right)\left(x-5\right)\)
3, \(4x\left(a-b\right)+6xy\left(b-a\right)=\left(4x-6xy\right)\left(a-b\right)=2x\left(2-3y\right)\left(a-b\right)\)
4, \(y\left(a-b\right)-x\left(b-a\right)=\left(x+y\right)\left(a-b\right)\)
5, \(6x\left(x-y\right)+8y\left(y-x\right)=\left(x-y\right)\left(6x-8y\right)=2\left(3x-4y\right)\left(x-y\right)\)
6, \(4\left(x-3\right)^2-2x\left(x-3\right)=\left(x-3\right)\left[4\left(x-3\right)-2x\right]=2\left(x-3\right)\left(x-6\right)\)
Em làm thử nếu sai thì thôi ạ (vì mới học lớp 6)
a)
Ta có:
\(\left(a+b\right)^2-\left(a-b\right)^2=a^2.b^2-a^2:b^2\)
\(=a^2.b^2-a^2.\frac{1}{b^2}=a^2.\left(b^2-\frac{1}{b^2}\right)\)
Chắc thế ạ, em chỉ làm 1 phần vì sợ sai