\(\frac{12x-3}{5x+1}\)=\(\frac{12y+1}{5y+3}\)và 2x+3y=0
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a)\(\frac{12x^5y^2}{8x^3y^5}=\frac{3x^2}{2y^3}\)
b)\(\frac{x^2+2x+1}{5x^2+5x}=\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{x+1}{5x}\)
a/ \(\frac{3x^2}{2y^3}\)
b/ \(\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{x+1}{5x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{12x^5y^2}{8x^3y^5}\)
=\(\frac{3x^2}{2y^3}\)
b)\(\frac{x^2+2x+1}{5x^2+5x}\)
=\(\frac{\left(x+1\right)^2}{5x\left(x+1\right)}\)
=\(\frac{x+1}{5x}\)
t tôi nhé bn
\(a,\frac{12x^5y^2}{8x^3y^5}=\frac{4x^3y^2.3x^2}{4x^3y^2.2y^3}=\frac{3x^2}{2y^3}\)
\(b,\frac{x^2+2x+1}{5x^2+5x}=\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{\left(x+1\right)\left(x+1\right)}{5x\left(x+1\right)}=\frac{x+1}{5x}\)
Xong rồi nhé! Chúc bạn học giỏi.
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{12x^5y^2}{8x^3y^5}=\frac{3x^2}{2y^3}\)
b, \(\frac{x^2+2x+1}{5x^2+5x}=\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{x+1}{5x}\)
a)\(\frac{12x^5y^2}{8x^3y^5}=\frac{4\cdot3\cdot x^3x^2y^2}{4\cdot2\cdot x^3y^2y^3}=\frac{3x}{2y^3}\)
b)\(\frac{x^2+2x+1}{5x^2+5x}=\frac{x^2+2\cdot x\cdot1+1^2}{5x\left(x+1\right)}=\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{x+1}{5x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x−3y/5=5y−2z/3=3z−5x/2=10x-15y/25=15y-6z/9=6z-10x/4=...+..+..../25+9+4=0/31=0
=> 2x=3y; 5y=2z ; 3z=5x => x/3=y/2; y/2=z/5
=> x/3=y/2 =z/5 = 12x/36=5y/10=3z/15= (12x+5y-3z)/31
x/3 = 3y/6=2z/10 = (x-3y+2z)/7
=> (12x+5y-3z)/ (x-3y+2z)=31/7
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu a nhìn là bt mà
Còn câu b chưa học nên ko giúp đc, xin lỗi nhá