125a³+75a²-15+1

= 5a³+3.5a².1+3.5a.1+1³

= (5a+1)³

\(12xy-4x^2y+8xy^2\)

\(=4xy\left(3-x+2y\right)\)

\(-125a^3+75a^2-15a+1\)

\(=-\left(125a^3-75a^2+15a-1\right)\)

\(=-\left[\left(5a\right)^3-3.\left(5a\right)^2.1+3.5a.1^3-1^3\right]\)

\(=-\left(5a-1\right)^3\)

a) Ta có: \(-9x^2+12xy-4y^2\)

\(=-\left(9x^2-12xy+4y^2\right)\)

\(=-\left[\left(3x\right)^2-2\cdot3x\cdot2y+\left(2y\right)^2\right]\)

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

b) Ta có: \(-125a^3+75a^2-15a+1\)

\(=\left(-5a\right)^3+3\cdot\left(-5a\right)^2\cdot1+3\cdot\left(-5a\right)\cdot1^2+1^3\)

\(=\left(-5a+1\right)^3\)

\(=\left(1-5a\right)^3\)

c) Ta có: \(64-96a+48a^2-8a^3\)

\(=4^3-3\cdot4^2\cdot2a+3\cdot4\cdot\left(2a\right)^2-\left(2a\right)^3\)

\(=\left(4-2a\right)^3\)

\(=\left[2\cdot\left(2-a\right)\right]^3\)

\(=8\left(2-a\right)^3\)

d) Ta có: \(-\frac{1}{8}m^3n^6-\frac{1}{27}\)

\(=-\left(\frac{1}{8}m^3n^6+\frac{1}{27}\right)\)

\(=-\left[\left(\frac{1}{2}mn^2\right)^3+\left(\frac{1}{3}\right)^3\right]\)

\(=-\left(\frac{1}{2}mn^2+\frac{1}{3}\right)\left(\frac{1}{4}m^2n^4-\frac{1}{6}mn^2+\frac{1}{9}\right)\)

ĐK : \(a\ne\pm2;a\ne\frac{-13}{6}\)

Ta có :

\(\frac{3b}{a^2-4}=\frac{1-125a-3b}{6x+13}=1-125a\)

\(\frac{3b}{a^2-4}=\frac{1-125a-3b}{6a+13}=\frac{1-125a}{1}=\frac{1-125a}{a^2+6a+9}\)

\(\Rightarrow a^2+6a+8=0\) ( \(\left(a\ne\frac{1}{125}\right)\)

Nếu \(a=-2\) ( loại) 

\(a=-4\) ( t/mãn)

Vậy \(a=-4\)là giá trị a cần tìm 

Thay vào biểu thức tìm b .Bạn tự làm nhé !!

a,\(3x-3y+6xy-6y^2=3x\left(1+2y\right)-3y\left(1+2y\right)=\left(3x-3y\right)\left(1+2y\right)\)

b,

\(4x^2-8xy-4z^2+4y^2=\left(2x-2y\right)^2-\left(2z\right)^2=\left(2x-2y-2z^{ }\right)\left(2x-2y+2z\right)\)

c,\(125a^3-8=\left(5a^{ }\right)^3-2^3=\left(5a-2\right)\left(5a^2+10a+4\right)\)

d, \(x^2-y^2-x-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)

a) \(3x-3y+6xy+6y^2=3\left(x-y+2xy+2y^2\right)\)

b) \(4x^2-8xy-4z^2+4y^2=4\left(x^2-2xy+y^2-z^2\right)=4\left(\left(x-y\right)^2-z^2\right)=4\left(x-y-z\right)\left(x-y+z\right)\)

c) \(125a^3-8=\left(5a-2\right)\left(25a^2+10a+4\right)\)

d) \(x^2-y^2-x-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)

e) \(a^2b+ab^2+a^3+b^3=ab\left(a+b\right)+\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left(a^2+b^2\right)\)

\(a.3x-3y+6xy-6y^2=3\left(x-y\right)+6y\left(x-y\right)=3\left(x-y\right)\left(2y+1\right)\) \(b.4x^2-8xy-4z^2+4y^2=\left(2x-2y\right)^2-4z^2=\left(2x-2y-2z\right)\left(2x-2y+2z\right)=4\left(x-y-z\right)\left(x-y+z\right)\) \(c.125a^3-8=\left(5a-2\right)\left(25a^2+10a+4\right)\)

\(d.x^2-y^2-x-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\) \(e.a^2b+ab^2a^3+b^3=b\left(a^2+a^4b+b^2\right)\)