a) 27a2b2 – 18ab + 3 = 3(9a2b2 – 6ab + 1) = 3(3ab – 1)2

giải bài toán: cho tam giác MNP, NTlà phân giác của góc N biết MN=4cm, NT=10cm, MP=8cm:TínhTM, TP? 

a: \(3ab-6a^2b\)

\(=3ab\cdot1-3ab\cdot2a\)

=3ab(1-2a)

b: \(x^3-6x\)

\(=x\cdot x^2-x\cdot6\)

\(=x\left(x^2-6\right)\)

c: \(x^2-y^2-9x+9y\)

\(=\left(x^2-y^2\right)-\left(9x-9y\right)\)

\(=\left(x-y\right)\left(x+y\right)-9\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-9\right)\)

d: \(5x^2+10xy+5y^2\)

\(=5\left(x^2+2xy+y^2\right)\)

\(=5\left(x+y\right)^2\)

a) 10x³-90x⁵=10x³(1-9x²)=10x³(1-3x)(1+3x)

b)3a²-6a²b+5a-10ab=(3a²-6a²b)+(5a-10ab)=3a²(1-2b)+5a(1-b)=(3a²+5a)(1-2b)=a(3a+5)(1-2b)

c) 7a³-28a²+28a=7a(a²-4a+4)

d) 45a³-30a²+5a-500=5(9a³-6a²+a-100)

\(2a\left(x^2-9\right)=2a\left(x-3\right)\left(x+3\right)\)

Trả lời:

\(27a^2b^2-18ab+3\)

\(=3\left(9a^2b^2-6ab+1\right)\)

\(=3\left[\left(3ab\right)^2-2.3ab.1+1^2\right]\)

\(=3\left(3ab-1\right)^2\)

a: \(27a^2b^2+18ab+3\)

\(=3\left(9a^2b^2+6ab+1\right)\)

\(=3\left\lbrack\left(3ab\right)^2+2\cdot3ab\cdot1+1^2\right\rbrack\)

\(=3\left(3ab+1\right)^2\)

b: \(5x^2-y+5xy-x\)

=5x(x+y)-(x+y)

=(x+y)(5x-1)

c: \(2x^3y^2-8x^3-12x^2y-6xy^2-y^3+x^2y^3\)

\(=x^2y^2\left(2x+y\right)-\left(2x+y\right)^3\)

\(=\left(2x+y\right)\left\lbrack x^2y^2-\left(2x+y\right)^2\right\rbrack\)

=(2x+y)(xy-2x-y)(xy+2x+y)

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)