\(|x-\frac{3}{2}|+\frac{5}{4}=\frac{x}{2}\)
\(|\frac{4x}{4}-\frac{6}{4}|+\frac{5}{4}=\frac{2x}{4}\)
\(|4x-6|+5=2x\)
\(|4x-6|=2x-5\) ( 1 )
( ĐK \(2x-5\ge0\) )
\(x\ge\frac{5}{2}\)
( 1 ) \(\Leftrightarrow\orbr{\begin{cases}4x-6=2x-5\\4x-6=5-2x\end{cases}}\)
\(\orbr{\begin{cases}2x=1\\6x=11\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{11}{6}\end{cases}}\) ( loại cả hai )
Vậy không có hoặc nói cách khác là x vô nghiệm
tìm x, x\(\in\)Q:
a) \(\frac{\left(x+\frac{3}{4}\right).\frac{7}{2}-\frac{1}{6}}{-\left(\frac{4}{5}+\frac{1}{3}\right).\frac{1}{2}+1}=2\frac{33}{52}\)
b) \(\frac{\left(5-\frac{2}{7}\right).\frac{7}{9}:\frac{3}{5}}{\left(3x-\frac{5}{6}\right):\frac{1}{7}}=5\frac{5}{21}\)
Tìm x\(\in\)Q biết:
a,\(\frac{2}{3x}-\frac{3}{12}=\frac{4}{5}-\left(\frac{7}{x}-2\right)\)
b,\(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
\(\left(\frac{x}{2}\right)^2+\left(\frac{x}{3}\right)^2+\left(\frac{x}{4}\right)^2=\left(\frac{x}{5}\right)^2+\left(\frac{x}{6}\right)^2+\left(\frac{x}{7}\right)^2\)
tìm x
Tìm x
a/\(\frac{x+7}{2003}+\frac{x+4}{2006}=\frac{x-1}{2011}+\frac{x-5}{2015}\)
b/\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
c/\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
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
\(\left(\frac{x}{2}\right)^2+\left(\frac{x}{3}\right)^2+\left(\frac{x}{4}\right)^2+\left(\frac{x}{5}\right)^2+\left(\frac{x}{6}\right)^2+\left(\frac{x}{7}\right)^2\) . Tìm giá trị thỏa mãn của x
TÌM X
a,\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=6\)
b,\(\left(x^2-4\right).\left(2x+\frac{4}{3}\right)=0\)
tìm x
\(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\left(\frac{1}{2}x-\frac{2}{3}\right)\left(\frac{4x^2}{3}+1\right)=\frac{22}{45}x^2\)
Tìm x:
a)\(\left(x-\frac{2}{5}\right)\left(x+\frac{3}{7}\right)\left(x+\frac{3}{4}\right)>0\)
b)\(\left(x-2\right)\left(x+\frac{3}{5}\right)\left(x-\frac{2}{3}\right)< 0\)
