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) \(16x^2-\left(x^2+4\right)^2=\left(4x\right)^2-\left(x^2+4\right)^2\)

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

\(=\left(x+2\right)^2\left\{-\left(x^2-4x+4\right)\right\}=\left(x+2\right)^2\left\{-\left(x-2\right)^2\right\}\)

Ở đây mình không đổi \(-\left(x-2\right)^2=\left(2-x\right)^2\)được vì vốn dĩ \(\left(x-2\right)^2=\left(2-x\right)^2\)

b) \(\left(x^2+9\right)^2-36=\left(x^2+9\right)^2-6^2\)

\(=\left(x^2+9+6\right)\left(x^2+9-6\right)=\left(x^2+15\right)\left(x^2-3\right)\)

c) \(x^2-2xy+y^2-z^2+2zt-t^2\)

\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t\right)^2\)

\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)

f) \(x^4y^4-z^4=\left\{\left(x^2y^2\right)^2-\left(z^2\right)^2\right\}\)

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

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

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 !!

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405

c

C

C.3/10