\(\frac{2x+1}{3}\)=\(\frac{x-5}{2}\)
=>(2x+1)x2 = (x-5)x3
=>4x+2 = 3x-15
=> 4x -3x =-15 -2
=> x = -17
Vậy x= -17
Mỹ Châu
\(\frac{2x+1}{3}\)=\(\frac{x-5}{2}\)
=>(2x+1)x2 = (x-5)x3
=>4x+2 = 3x-15
=> 4x -3x =-15 -2
=> x = -17
Vậy x= -17
Mỹ Châu
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
Bài 1: Tìm các số x; y; z biết rằng \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)và 2x + 3y - z = 124.
Bài 2: Tìm các số x; y; z biết rằng \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)
tìm x, biết:
\(\frac{7^{x-2}+7^{x-10}+7^x}{57}=\frac{5^{2x}+5^{2x-1}+5^{2x+3}}{131}\)
Tìm x, biết:
a)\(-\frac{2}{3}\left(x-\frac{1}{4}\right)=\frac{1}{3}\left(2x-1\right)\)
b) \(\frac{1}{5}.2^x+\frac{1}{3}.2^{x+1}=\frac{1}{5}.2^7+\frac{1}{3}.2^8\)
Bài 2: Tìm x biết
a, \(\frac{3}{5}+x=\frac{5}{6}\)
b, \(\left(3\frac{1}{2}+2x\right).2\frac{2}{5}=5\frac{1}{3}\)
c, \(2\frac{2}{3}.x-8\frac{2}{3}=3\frac{1}{3}\)
d, \(\frac{5}{13}+2x=\frac{3}{13}\)
Tìm x biết:
\(x+\frac{1}{3}=\frac{1}{6}x\)
\(\left(\frac{1}{2}+2x\right).\frac{2}{3}=\frac{5}{6}\)
tìm x biết
1)\(-\frac{2}{3}\cdot\left(x-\frac{1}{4}\right)=\frac{1}{3}\cdot\left(2x-1\right)\)
2)\(\frac{1}{5}\cdot2^x+\frac{1}{5}\cdot2^{x+1}=\frac{1}{5}\cdot2^7+\frac{1}{3}\cdot2^8\)
Tìm x, biết :
a, \(60\%x+0,4x+x:3=2\)
b, \(\left|2x-5\right|-7=\left(\frac{1}{49}-\frac{1}{3^2}\right)\left(\frac{1}{49}-\frac{1}{4^2}\right)...\left(\frac{1}{49}-\frac{1}{2015^2}\right)\)
c, \(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+...+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+...+\frac{40}{39}\)