pt <=> \(\frac{2}{\left|x-2\right|+2}=\frac{3}{3\left|2-x\right|+1}\)
<=> \(6\left|2-x\right|+2=3\left|x-2\right|+6\)
<=> \(3\left|x-2\right|=4\)( vì | x - 2 | = | 2 - x | )
<=> \(\left|x-2\right|=\frac{4}{3}\)
TH1: \(x-2=\frac{4}{3}\)
<=> \(x=\frac{10}{3}\)
TH2: \(x-2=-\frac{4}{3}\)
<=> \(x=\frac{2}{3}\)
Vậy x = 10/3 hoặc x = 2/3
Tìm x, biết:
a) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)\div\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
b) \(-\frac{3x}{4}.\left(\frac{1}{x}+\frac{2}{7}\right)=0\)
c) \(3-\frac{1-\frac{1}{2}}{1+\frac{1}{x}}=2\frac{2}{3}\)
Help me, cần gấp lắm. Nhanh lên ạ!
Tìm x, y,z biết:
a) \(\frac{x}{y+z+1}=\frac{y}{x+z+2}=\frac{y}{x+y-3}\)
b) \(6\left(x-\frac{1}{y}\right)=3\left(y-\frac{1}{2}\right)=2\left(z-\frac{1}{x}\right)=xyz-\frac{1}{xyz}\)
Help me ! mik hứa sẽ tk
Tìm x , biết : \(\frac{2}{\left|x-2\right|+2}=\frac{3}{\left|6-3x\right|+1}\)
Tìm x biết : \(\frac{2}{\left|x-2\right|+2}=\frac{3}{\left|6-3x\right|+1}\)
Tìm nguyên x,y : \(\left|3x+1\right|+\left|3x-5\right|=\frac{12}{\left(y+3\right)^2+2}\)
Mọi người ơi !! Giúp mình với :-0 Help me
2/ Tìm x biết:
a,\(\frac{3x}{5}-\frac{7}{10}x-\frac{x-3}{2}=0\)
b,\(\left(4x+3\right)^2=\frac{2}{3}:6\)
c,\(\left|3x+\frac{1}{2}\right|-2=\sqrt{\left(-\frac{3}{2}\right)^2}\)
Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1.Tìm x, biết:(help me)
a)\(|x-1|—\left(-2\right)^3=9\times\left(-1\right)^{100}\)
b)\(\frac{x-2}{-4}\)=\(\frac{-9}{x-2}\)
c)\(\frac{x-5}{3}=\frac{-12}{5-x}\)
d)\(8x-|4x+\frac{3}{4}|=x+2\)
e) anh \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2019}{2020}\)
Tìm x biết:
\(\frac{-1}{2}\times\left(3x-1\right)+\frac{3}{4}\left(3-2x\right)=-3\times\left(\frac{x}{2}-1\right)-\left(\frac{4}{5}\right)^{-1}\)
