phân tích đa thức thành nhân tử
-7xy+x\(^2\)+10y\(^2\)
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\(\left(xy+1\right)^2-\left(x+y\right)^2=\left(xy+1-x-y\right)\left(xy+1+x+y\right)=\left[x\left(y-1\right)-\left(y-1\right)\right]\left[x\left(y+1\right)+\left(y+1\right)\right]=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(=\left(xy-x-y+1\right)\left(xy+1+x+y\right)\)
\(=\left(y-1\right)\left(x-1\right)\left(y+1\right)\left(x+1\right)\)
\(=x\left[x^2\left(x-y\right)^2-36y^2\right]\\ =x\left[x\left(x-y\right)-6y\right]\left[x\left(x-y\right)+6y\right]\\ =x\left(x^2-xy-6y\right)\left(x^2-xy+6y\right)\)
\(x^2-xy\left(a+b\right)+aby^2=x^2-xya-xyb+aby^2=x\left(x-ya\right)-yb\left(x-ya\right)=\left(x-ya\right)\left(x-yb\right)\)
\(x^2-xy\left(a+b\right)+aby^2\)
\(=x^2-axy-bxy+aby^2\)
\(=x\left(x-ay\right)-by\left(x-ay\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)
\(\left(x+y\right)^2+3\left(x+y\right)-10=\left[\left(x+y\right)^2+2\left(x+y\right).\dfrac{3}{2}+\dfrac{9}{4}\right]-\dfrac{49}{4}\)
\(=\left(x+y+\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\left(x+y+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x+y+\dfrac{3}{2}+\dfrac{7}{2}\right)=\left(x+y-2\right)\left(x+y+5\right)\)
\(\left(x+y\right)^2+3\left(x+y\right)-10\)
\(=\left(x+y\right)^2+5\left(x+y\right)-2\left(x+y\right)-10\)
\(=\left(x+y+5\right)\left(x+y-2\right)\)
\(\left(x^2+x\right)^2+4x^2+4x-12=\left[\left(x^2+x\right)^2+4\left(x^2+x\right)+4\right]-16=\left(x^2+x+2\right)-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
\(\left(x^2+x\right)^2+4x^2+4x-12\\ =\left(x^2+x+2\right)-4\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\)
\(\left(x-2\right)\left(x-1\right)x\left(x+1\right)-24\)
\(=\left(x^2-x-2\right)\left(x^2-x\right)-24\)
\(=\left(x^2-x\right)-2\left(x^2-x\right)-24\)
\(=\left(x^2-x-6\right)\left(x^2-x+4\right)\)
\(=\left(x-3\right)\left(x+2\right)\left(x^2-x+4\right)\)
\(=x^2\left(x^2+2x+1\right)+x+1\)
\(=x^2\left(x+1\right)^2+x+1\)
\(=\left(x+1\right)\left[x^2\left(x+1\right)+1\right]\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
\(x^4+2x^3+x^2+x+1\)
\(=x^2\left(x+1\right)^2+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
\(\left(x+y+z\right)^2+\left(x+y-z\right)^2-4z^2=\left(x+y+z\right)^2+\left(x+y-z-2z\right)\left(x+y-z+2z\right)=\left(x+y+z\right)^2+\left(x+y-3z\right)\left(x+y+z\right)=\left(x+y+z\right)\left(x+y+z+x+y-3z\right)=\left(x+y+z\right)\left(2x+2y-2z\right)=2\left(x+y+z\right)\left(x+y-z\right)\)
Ta có:
(x + y + z)2 + (x + y – z)2 – 4z2
\(=\left(x+y-z\right)^2+\left(x+y-z\right)\left(x+y+3z\right)\)
\(=\left(x+y-z\right)\left(x+y+3z+x+y-z\right)\)
\(=2\left(x+y-z\right)\left(x+y+z\right)\)
\(=x^2-2xy-5xy+10y^2=x\left(x-2y\right)-5y\left(x-2y\right)=\left(x-2y\right)\left(x-5y\right)\)
Đây bạn nhé!