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\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)ĐKXĐ: \(x\ne-\frac{1}{5};x\ne-\frac{7}{5}\)
\(\Rightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow15x^2+13x+2=15x^2+16x-7\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\)
\(\frac{3x+2}{5x+7}\)= \(\frac{3x-1}{5x+1}ĐKXĐ:x#\)- \(\frac{1}{5};x#-\frac{1}{5};x#-\frac{7}{5}\)
< = > (\(\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
< = > \(3x=9\)
\(x=3\)
số phải tìm :3
![](https://rs.olm.vn/images/avt/0.png?1311)
áp dụng t/c của dãy tỉ số bằng nhau ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\frac{3}{6}=\frac{1}{2}\)
=> \(\frac{3x+2}{5x+7}=\frac{1}{2}\)=> 2(3x+2) = 5x + 7 => 6x + 4 = 5x + 7 => 6x - 5x = 7 - 4 => x = 3
Vậy x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\frac{3x+2}{5x+7}=\frac{3x-1}{5x-1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x-1\right)}=\frac{3}{8};\frac{3x+2}{5x+7}=\frac{3}{8}\Leftrightarrow24x+16=15x+21\Leftrightarrow9x=5\Leftrightarrow x=\frac{5}{9}\) \(b,\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow37.7-7x=3x+39\Leftrightarrow259-7x=3x+39\Leftrightarrow220-7x=3x\Leftrightarrow10x=220\Leftrightarrow x=22\) \(c,\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}=\frac{x+4}{2x+6}=\frac{\left(x+4\right)-\left(x+1\right)}{2x+6-\left(2x+1\right)}=\frac{3}{5};\frac{x+1}{2x+1}=\frac{3}{5}\Leftrightarrow5x+5=6x+3\Leftrightarrow x=2\) \(d,\frac{x-2}{x+2}=\frac{x+3}{x-4}=\frac{\left(x+3\right)-\left(x-2\right)}{\left(x-4\right)-\left(x+2\right)}=\frac{5}{-6};\frac{x-2}{x+2}=\frac{5}{-6}\Leftrightarrow6\left(2-x\right)=5x+10\Leftrightarrow2-6x=5x\Leftrightarrow x=\frac{2}{11}\) \(f,\frac{3x-5}{x}=\frac{9x}{3x+2}=\frac{9x-15}{3x}=\frac{9x-\left(9x-15\right)}{\left(3x+2\right)-3x}=\frac{15}{2};\frac{9x}{3x+2}=\frac{15}{2}\Leftrightarrow18x=45x+30\Leftrightarrow27x+30=0\Leftrightarrow x=\frac{-10}{9}\) \(e,\frac{x+2}{6}=\frac{5x-1}{5}\Leftrightarrow5\left(x+2\right)=6\left(5x-1\right)\Leftrightarrow5x+10=30x-6\Leftrightarrow10=25x-6\Leftrightarrow25x=16\Leftrightarrow x=\frac{16}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x-3}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x-3\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(5x-3\right)-\left(5x+7\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left(15x^2-9x+10x-6\right)-\left(15x^2-5x+21x-7\right)=0\)
\(\Leftrightarrow15x^2-9x+10x-6-15x^2+5x-21x+7=0\)
\(\Leftrightarrow-15x+1=0\)
\(\Leftrightarrow-15x=-1\)
\(\Leftrightarrow x=\dfrac{-1}{-15}=\dfrac{1}{15}\)
Vậy \(x=\dfrac{1}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(|5x-3|-x=7\)
\(|5x-3|=7+x\)
\(\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)
\(\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)
\(\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)
\(\orbr{\begin{cases}x=2,5\\x=\frac{-2}{3}\end{cases}}\)
Vậy x = 2,5 hoặc x = -2/3
Hi Hi!
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(3\left(2x-1\right)-3x\left(-x+2\right)=5x-\left(1-3x\right)\cdot x\\ 6x-3+3x^2-6x=5x-x+3x^2\\ 3x^2-3=4x+3x^2\\ 3x^2-3x^2=4x+3\\ 4x+3=0\\ 4x=-3\\ x=\frac{-3}{4}\)
Vậy \(x=\frac{-3}{4}\)
b, \(x-\frac{x-3}{4}=3-\frac{x-3}{12}\\ \frac{4x-x-3}{4}=\frac{36-x-3}{12}\\ \frac{3x-3}{4}=\frac{33-x}{12}\\ \Rightarrow12\left(3x-3\right)=4\left(33-x\right)\\ 36x-36=132-4x\\ 36x+4x=132+36\\ 40x=168\\ x=\frac{168}{40}=\frac{21}{5}\)
Vậy \(x=\frac{21}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>(3x+2)(5x+1)=(5x+7)(3x-1)
(3x)(5x+1)+2(5x+1)=(5x)(3x-1)+7(3x-1)
15x2+3x+10x+2=15x2-5x+21x-7
(15x2-15x2)+(3x+10x+5x-21x)=-7-2
0-3x=-9
-3x=-9
x=(-9)/(-3)
x=3
\(\frac{3x+2}{5x+7}\)=\(\frac{3x-1}{5x+1}\)
\(\Rightarrow\)\(\frac{3x+2}{3x-1}\)= \(\frac{5x+7}{5x+1}\)\(\Rightarrow\)1+\(\frac{3}{3x-1}\)=1+\(\frac{6}{5x+1}\)
\(\Rightarrow\)\(\frac{3}{3x-1}\)= \(\frac{6}{5x+1}\)\(\Rightarrow\)3.(5x+1) = 6.(3x-1)
\(\Rightarrow\)15x+3 = 18x -6
\(\Rightarrow\)3x = 6 +3
\(\Rightarrow\)x =3
Ta có :\(\frac{\text{3x + 2}}{\text{5x + 7}}=\frac{\text{3x -1}}{\text{5x +1}}\)
=> ( 5x + 1 ) . ( 3x + 2 ) = ( 3x - 1 ) . ( 5x + 7 )
=> 5x(3x + 2 ) + ( 3x + 2 ) = 3x(5x + 7 ) - ( 5x + 7 )
=> ( 15x2 + 10x ) + ( 3x + 2 ) = ( 15x2 + 21x ) - ( 5x + 7 )
=> ( 3x + 2 ) + ( 5x + 7 ) = ( 15x2 + 21x ) - ( 15x2 + 10x )
=> ( 3x + 5x ) + ( 2 + 7 ) = ( 15x2 - 15x2 ) + ( 21x - 10x )
=> 8x + 9 = 11x
=> 9 = 11x - 8x
=> 9 = 3x
=> x = 3
Vậy x = 3
~~Học tốt~~
Ta có \(\frac{3x+2}{5x+7}\)=\(\frac{3x-1}{5x+1}\)=\(\frac{3x+2-3x+1}{5x+7-5x-1}\)=\(\frac{1}{2}\)
=> \(\frac{3x+2}{5x+7}\)=\(\frac{1}{2}\)=> ( 3x + 2 ) . 2 = 5x + 7 . 1 => 6x + 4 = 5x + 7 => x=3