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\(x^2-xy\left(a+b\right)+aby^2=x^2-xya-xyb+aby^2=x\left(x-ya\right)-yb\left(x-ya\right)=\left(x-ya\right)\left(x-yb\right)\)
\(x^2-xy\left(a+b\right)+aby^2\)
\(=x^2-axy-bxy+aby^2\)
\(=x\left(x-ay\right)-by\left(x-ay\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)
a/ x^3z+xyz-x^3-xyz^2
=x3z-xyz2-x3+xyz
=xz.(x2-xyz)-x(x2-xyz)
=(x2-xyz)(xz-x)
=x(x-yz)x(z-1)
=x2(x-yz)(z-1)
b/ x^2-axy-bxy+aby^2
=x(x-ay)-by(x-ay)
=(x-ay)(x-by)
c/ abx^2+a^2xy+aby^2+b^2xy
=ax(bx+ay)+by(ay+bx)
=(ay+bx)(ax+by)
Bạn chuyển tất cả hạng tử từ vế phải sang vế trái ta được
\(^{x^2+5\text{x}^3+x^2y=5\text{x}^3+x^2y}\)
\(x^2+5\text{x}^3+x^2y-5\text{x}^3-x^2y=0\)
Rút gọn ta được
\(x^2=0\)
\(=>x=0\)
tick cho mình nha
\(=x^2-2xy-5xy+10y^2=x\left(x-2y\right)-5y\left(x-2y\right)=\left(x-2y\right)\left(x-5y\right)\)
\(\left(xy+1\right)^2-\left(x+y\right)^2=\left(xy+1-x-y\right)\left(xy+1+x+y\right)=\left[x\left(y-1\right)-\left(y-1\right)\right]\left[x\left(y+1\right)+\left(y+1\right)\right]=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(=\left(xy-x-y+1\right)\left(xy+1+x+y\right)\)
\(=\left(y-1\right)\left(x-1\right)\left(y+1\right)\left(x+1\right)\)
\(\left(ab-1\right)^2+\left(a+b\right)^2=a^2b^2-2ab+1+a^2+2ab+b^2=a^2+b^2+a^2b^2+1=a^2\left(b^2+1\right)+\left(b^2+1\right)=\left(a^2+1\right)\left(b^2+1\right)\)
\(9x^2+6x-4y^2+4y=\left(9x^2+6x+1\right)-\left(4y^2-4y+1\right)=\left(3x+1\right)^2-\left(2y-1\right)^2=\left(3x+1-2y+1\right)\left(3x+1+2y-1\right)\)
x2 - axy - bxy + aby2
= ( x2 - axy) - ( bxy - aby2)
= x( x-ay) - by( x - ay)
= ( x-ay)( x - by)