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\(x^2-xy\left(a+b\right)+aby^2=x^2-xya-xyb+aby^2=x\left(x-ya\right)-yb\left(x-ya\right)=\left(x-ya\right)\left(x-yb\right)\)
\(x^2-xy\left(a+b\right)+aby^2\)
\(=x^2-axy-bxy+aby^2\)
\(=x\left(x-ay\right)-by\left(x-ay\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)
\(A=-x-z\left(x-y\right)+y=-x-xz+zy+y=-x\left(1+z\right)+y\left(1+z\right)=\left(1+z\right)\left(y-x\right)\)
x^5+x^4+1
=x5+x4+x3+x2+x+1-x3-x2-x
=x3.(x2+x+1)+(x2+x+1)-x.(x2+x+1)
tự xử tiếp
\(x^2+2\left(a+b\right)x+4ab\)
\(=x^2+2\left(a+b\right)x+a^2+2ab+b^2-a^2-b^2+2ab\)
\(=\left(x+a+b\right)^2-\left(a-b\right)^2\)
\(=\left(x+2a\right)\left(x+2b\right)\)
\(a,2x^2-xy-y^2=2x^2-2xy+xy-y^2.\)
\(=2x\left(x-y\right)+y\left(x-y\right)=\left(x-y\right)\left(2x+y\right)\)
\(b,x^8+4=\left(x^4\right)^2+4x^4+4-4x^4\)
\(=\left(x^4+2\right)^2-\left(2x^2\right)^2\)
\(=\left(x^4-2x^2+2\right)\left(x^4+2x^2+2\right)\)
\(c,x^2+2\left(a+b\right)x+4ab\)
\(=x^2+2\left(a+b\right)x+\left(a+b\right)^2+4ab-\left(a+b\right)^2\)
\(=\left(x+a+b\right)^2+4ab-a^2-2ab-b^2\)
\(=\left(x+a+b\right)^2-a^2+2ab-b^2\)
\(=\left(x+a+b\right)-\left(a-b\right)^2\)
\(=\left(x+a+b+a-b\right)\left(x+a+b-a+b\right)\)
\(=\left(x+2a\right)\left(x+2b\right)\)
\(\left(x-5\right)\left(x-1\right)\left(x+3\right)\left(x+7\right)+60\)
\(=\left(x^2+2x-35\right)\left(x^2+2x-3\right)+60\)
\(=\left(x^2+2x\right)^2-38\left(x^2+2x\right)+105+60\)
\(=\left(x^2+2x\right)^2-3\left(x^2+2x\right)-35\left(x^2+2x\right)+165\)
\(=\left(x^2+2x-3\right)\left(x^2+2x-35\right)\)
\(=\left(x+3\right)\left(x-1\right)\left(x+7\right)\left(x-5\right)\)
\(2x^2+x-6\)
\(=2x^2-3x+4x-6\)
\(=x\left(2x-3\right)+2\left(2x-3\right)\)
\(=\left(2x-3\right)\left(x+2\right)\)
\(x^4-x^3-x+1=\left(x^4-x^3\right)-\left(x-1\right)=x^3\left(x-1\right)-\left(x-1\right)=\left(x^3-1\right)\left(x-1\right)=\left(x-1\right)^2.\left(x^2+x+1\right)\)
x4 - x3 - x + 1
= (x4 - x3) - (x - 1)
= x3(x - 1) - (x - 1)
= (x3 - 1)(x - 1)
\(x^2-x-2020.2021=x^2+2020x-2021x-2020.2021=x\left(x+2020\right)-2021\left(x+2020\right)=\left(x+2020\right)\left(x-2021\right)\)
\(x^2-x-2020\cdot2021\)
\(=\left(x-2021\right)\left(x+2020\right)\)
\(3x^2+x-4=3x^2-3x+4x-4=3x\left(x-1\right)+4\left(x-1\right)=\left(3x+4\right)\left(x-1\right)\)
\(=4ab\left(ab+ax+bx+x^2\right)=4a^2b^2+4a^2bx+4ab^2x+4abx^2\)
Nhân tử mà ??!!