a: |x|=4
=>x=4(nhận) hoặc x=-4(loại)
b: |-x|=1
=>|x|=1
=>x=1(nhận) hoặc x=-1(loại)
c: |x|=7
=>x=7(loại) hoặc x=-7(nhận)
d: |-x|=|-2|
=>|x|=2
=>x=2 hoặc x=-2
Bài 1: Tìm x:
a) x - \(\frac{4}{5}=\frac{7}{10}-\frac{3}{4}\)
b)\(2\frac{1}{3}-x=\frac{-5}{9}+2x\)
c) \(\frac{x+3}{2016}+\frac{x+2}{2017}=\frac{x+1}{2018}+\frac{x}{2019}\)
d) \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
a\(\left(4x-5\right)\left(3x+2\right)=0\)
b\(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)
Tìm x biết
a)\(\dfrac{2}{\left(x+2\right)\times\left(x+4\right)}+\dfrac{4}{\left(x+4\right)\times\left(x+8\right)}+\dfrac{6}{\left(x+8\right)\times\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\times\left(x+14\right)}\)
Tìm x để các biểu thức sau là số hữu tỉ âm :
A = \(\dfrac{\left(x+3\right).\left(4x+5\right)}{\left(20-4x\right).\left(2.x\right)}\)
B= \(\dfrac{\left(x+1\right).\left(7-2x\right)}{\left(5x-4\right).\left(10-2x\right)}\)
a, \(x\left(x+y+z\right)=13;y\left(x+y+z\right)=7;z\left(x+y+z\right)=-4\)
giup minh nhe minh dang can gap
e) \(\dfrac{3}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)
f) \(\dfrac{1}{x-1}+\dfrac{-2}{3}\left(\dfrac{3}{4}-\dfrac{6}{5}\right)=\dfrac{5}{2-2x}\)
tim x y z
\(x\left(x+y+z\right)=13;y\left(x+y+z\right)=7;z\left(x+y+z\right)=-4\)
giup minh nhe minh dang can gap
1.Tìm x
a) \(1\dfrac{4}{5}=-0,15-x\)
b) \(-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\)
c) \(\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)
tim x, y,z
\(x\left(x+y+z\right)=13\) \(y\left(x+y+z\right)=7\) \(z\left(x+y+z\right)=-4\)
