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\(=\left(x+4\right)^2-1=\left(x+4-1\right)\left(x+4+1\right)=\left(x+3\right)\left(x+5\right)\)
\(x^2+8x+15=x\left(x+3\right)+5\left(x+3\right)=\left(x+3\right)\left(x+5\right)\)
\(\frac{-x^6}{125}-\frac{y^3}{64}\)
\(=\frac{-\left(x^2\right)^3}{5^3}-\frac{y^3}{4^3}\)
\(=\left(\frac{-x^2}{5}\right)^3-\left(\frac{y}{4}\right)^3\)
\(=\left(\frac{-x^2}{5}-\frac{y}{4}\right)\cdot\left(\frac{x^4}{25}-\frac{x^2y}{20}+\frac{y^2}{16}\right)\)
Tham khảo nhé~
\(x^3+27y^3=x^3+\left(3y\right)^3=\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)
\(a^6-8b^3=\left(a^2\right)^3-\left(2b\right)^3=\left(a^2-2b\right)\left(a^4+2a^2b+4b^2\right)\)
\(8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)
\(8z^3+x^3=\left(2z\right)^3+x^3=\left(2z+x\right)\left(4z^2-2xz+x^2\right)\)
a/ \(\left(a^2-b^2+1\right)\left(a^2-ab+b^2\right)\left(a^2+ab+b^2\right)\)
b/ \(\left(x+y-1\right)\left(y^2-xy+y+x^2+x+1\right)\)
a) \(a^2+2a+1-b^2\)
\(=\left(a+1\right)^2-b^2\)
\(=\left(a+1-b\right)\left(a+1+b\right)\)
b) \(4a^2+4a+1-9b^2\)
\(=\left(2a\right)^2+4a+1-\left(3b\right)^2\)
\(=\left(2a+1\right)^2-\left(3b\right)^2\)
\(=\left(2a+1-3b\right)\left(2a+1+3b\right)\)
mày không phân tích được ak,tự làm đi
\(x^2-x-6=x^2+2x-3x-6=x\left(x+2\right)-3\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
\(x^3-19x-30=x^3+6x-25x-30=x\left(x^2-25\right)+6x-30=x\left(x^2-25\right)+6\left(x-5\right)\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)=\left(x-5\right)\left[\left(x\right)\left(x+5\right)+6\right]\)
\(\left(a^6-b^3\right)=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
\(=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)