![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Rightarrow\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)
\(\Rightarrow3x\left(5x-34\right)-\left(5x-34\right)=40\left(25-3x\right)-5x\left(25-3x\right)\)
\(\Rightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Rightarrow15x^2-97x+34=1000-245x+15x^2\)
\(\Rightarrow15x^2=1000-34-245x+97x+15^2\)
\(\Rightarrow15x^2=966-148x+15^x\)
\(\Rightarrow0=966-148x\)
\(\Rightarrow x=\frac{996}{148}=\frac{249}{37}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)
\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)
\(\Leftrightarrow138x=966\)
\(\Leftrightarrow x=966:138\)
\(\Leftrightarrow x=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\ne8;x\ne\frac{34}{5}\)
\(\Rightarrow\left(3x-1\right)\left(5x-34\right)=\left(25-3x\right)\left(40-5x\right)\)
\(\Rightarrow15x^2-102x-5x+34=1000-125x-120x+15x^2\)
\(\Rightarrow-102x-5x+125x+120x=1000-34\)
\(\Rightarrow138x=966\Rightarrow x=7\)
Vậy x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
=> (3x-1).(5x-34)=(40-5x).(25-3x)
=> 3x.(5x-34)-(5x-34)= 40.(25-3x)-5x.(25-3x)
=> 15x2-102x-5x+34= 1000-120x-125x+15x2
=> 15x2- 107x+34 = 1000-245x+15x2
=> 107x+34 =1000-245x
=> 138x=966
=> x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có;
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}=\frac{3x-1+25-3x}{40-5x+5x-34}=\frac{24}{6}=4\)
=> 25 - 3x = 4(5x - 34)
=> 25 - 3x = 20x - 136
=> 25 + 136 = 20x + 3x
=> 161 = 23x
=> x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
áp dụng tính chất của dãy tỉ số bằng nhau ta có;
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}=\frac{3x-1+25-3x}{40-5x+5x-34}=\frac{24}{6}=4\)
suy ra:
\(\frac{3x-1}{40x-5}=4\Rightarrow3x-1=4.\left(40-5x\right)\)
3x-1=160-20x
3x+20x=160+1
23x=161
x=161:23
x=7
vậy x=7
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Leftrightarrow\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)
<=>