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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: \(=-y\left(a-b\right)^2+\left(a-b\right)\)
\(=\left(a-b\right)\left(-ya+yb+1\right)\)
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)x+2a.(x-y)-y=(x-y)+2a(x-y)
=(x-y)(1+2a)
b)x^2-(a+b)x+ab=[x^2-(a+b)x]+a
=x(x-a-b)+a
a,\(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-2b\right)\)
\(=\left(a-b\right)2\left(a-b\right)\)
\(=2\left(a-b\right)^2\)
b,\(\left(x+y\right)\left(2x-y\right)+\left(2x-y\right)\left(3x-y\right)-\left(y-2x\right)\)
\(=\left(x+y\right)\left(2x-y\right)+\left(2x-y\right)\left(3x-y\right)+\left(2x-y\right)\)
\(=\left(2x-y\right)\left(x+y+3x-y+1\right)\)
\(=\left(2x-y\right)\left(4x+1\right)\)
c,\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z+y^2z-y^2x+z^2\left(x-y\right)\)
\(=x^2y-y^2x-x^2z+y^2z+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x^2-y^2\right)+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
a x+ 2a(x-y)-y
=x-y+2a(x-y)
=(x-y)(1+2a)
b x^2-(a+b)x+ab
=x^2-(xa+xb)+ab
=x^2-xa-xb+ab
=x(x-a)-b(x-a)
=(x-a)(x-b)
f x(y^2-z^2)+y(z^2-x^2)+z(x^2-y^2)
=xy^2-xz^2+yz^2-yx^2+zx^2-zy^2
=-xzy^2-xyz^2-zyx^2
=-xyz(y+z+x)
2
a
\(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow\left(x+y\right)^3=\left(-z\right)^3\)
\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)
\(\Rightarrow x^3+y^3+z^3=3xy\left(x+y\right)=3xyz\)
b
Đặt \(a-b=x;b-c=y;c-a=z\Rightarrow x+y+z=0\)
Ta có bài toán mới:Cho \(x+y+z=0\).Phân tích đa thức thành nhân tử:\(x^3+y^3+z^3\)
Áp dụng kết quả câu a ta được:
\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(\left(2a+3\right)\left(2a+3\right)y+\left(2a+3\right)\)
\(=\left(2a+3\right)[y\left(2a+3\right)+1]\)
\(=\left(2a+3\right)\left(2ay+3y+1\right)\)
\(\left(a-b\right)x+\left(b-a\right)y-\left(a-b\right)\) (Sửa đề)
\(=\left(a-b\right)x-\left(a-b\right)y-\left(a-b\right)\)
\(=\left(a-b\right)\left(x-y-1\right)\)
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)