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a) \(x^2-2x-15\)
\(\Leftrightarrow x^2-2x+1-16\)
\(\Leftrightarrow\left(x-1\right)^2-4^2\)
\(\Leftrightarrow\left(x-5\right)\left(x-3\right)\)
\(a,x^2-2x-15=\left(x^2-2x+1\right)-16.\)
\(=\left(x-1\right)^2-4^2\)
\(=\left(x-5\right)\left(x+3\right)\)

b) 2x^2 + 7x - 15
2x^2 + 10x - 3x -15
2x(x+5) - 3(x+5)
(x+5)(2x-3)
a, 5x^3y - 10x^2y^2 + 5xy^3 = 5xy. ( x^2 - 2xy + y^2) = 5xy.( x-y)^2
b, 2x^2 + 7x -15 = 2x^2 + 10X - 3x -15
= 2x( x+5) - 3( x+5)
= ( 2x-3) (x+5)

\(x^3+5x^2+3x-9\)
\(=x^3-x^2+6x^2-6x+9x-9\)
\(=x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)
\(x^{16}+x^8-2\)
\(=\left(x^{16}-1\right)+\left(x^8-1\right)\)
\(=\left(x^8-1\right)\left(x^8+1\right)+\left(x^8-1\right)\)
\(=\left(x^8-1\right)\left(x^8+2\right)\)
\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+2\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+2\right)\)
\(c,x^3+5x^2+3x-9\)
\(=x^3+6x^2+9-x^2-6x-9\)
\(=x\left(x^2+6x^2+9\right)-\left(x^2+6x^2+9\right)\)
\(=x.\left(x+3\right)^2-\left(x+3\right)^2\)
\(=\left(x+3\right)^2\left(x-1\right)\)
\(d,x^{16}+x^8-2\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1+x^4\right)\left(x^4+1-x^4\right)\)

x3+5x2+3x-9=x3+6x2+9x-x2-6x-9
=x(x2+6x+9)-(x2+6x+9)
=x.(x+3)2-(x+3)2
=(x+3)2.(x-1)

a) \(x^2+6x+9\)
\(=\left(x+3\right)^2\)
\(=\left(x+3\right)\left(x+3\right)\)
b) \(10x-25-x^2\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
\(=-\left(x-5\right)\left(x-5\right)\)
c) \(8x^3-\frac{1}{8}\)
\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)
\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
d) \(\frac{1}{25}x^2-64y^2\)
\(=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2\)
\(=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)
a) \(x^2+6x+9=x^2+2.3.x+3^2\)\(=\left(x+3\right)^2\)
b)\(10x-25-x^2=-\left(x^2-10x+25\right)\)\(=-\left(x^2-2.5.x+5^2\right)=-\left(x+5\right)^2\)
c)\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)\(=\left(2x-\frac{1}{2}\right)\left(4x+x+\frac{1}{4}\right)\)
d)\(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}\right)^2-\left(8y\right)^2\)\(=\left(\frac{1}{5}-8y\right)\left(\frac{1}{5}+8y\right)\)

a/ \(=5x^2\left(x-y\right)-10x\left(x-y\right)=\left(x-y\right)\left(5x^2-10x\right)=5x\left(x-2\right)\left(x-y\right)\)
b/ \(=2x^2-7x+2x-7=x\left(2x-7\right)+\left(2x-7\right)=\left(x+1\right)\left(2x-7\right)\)
thôi mk ko cần nữa