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a: \(\Leftrightarrow\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{5}{3}=\dfrac{1}{6}\\x\cdot\dfrac{5}{3}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}:\dfrac{5}{3}=\dfrac{3}{30}=\dfrac{1}{10}\\x=-\dfrac{1}{10}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x-1\right|=\dfrac{3}{2}:\dfrac{3}{4}=2\)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
c: \(\Leftrightarrow\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)
\(\Leftrightarrow x+\dfrac{3}{5}=\dfrac{7}{3}-x\)
=>2x=44/15
hay x=22/15
\(\dfrac{x-1}{7}+\dfrac{x-2}{3}+\dfrac{x-3}{5}+\dfrac{x-4}{2}=6\\ =>\left(\dfrac{x-1}{7}-1\right)+\left(\dfrac{x-2}{3}-2\right)+\left(\dfrac{x-3}{5}-1\right)+\left(\dfrac{x-4}{2}-2\right)=0\\ =>\dfrac{x-8}{7}+\dfrac{x-8}{3}+\dfrac{x-8}{5}+\dfrac{x-8}{2}=0\\ =>\left(x-8\right)\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{2}\right)=0\\ =>x-8=0\\ =>x=8\)
a: \(\dfrac{x-3}{5-x}=\dfrac{5}{7}\left(x\ne5\right)\)
=>7(x-3)=5(5-x)
=>7x-21=25-5x
=>12x=46
=>x=23/6
b: \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)(ĐKXĐ: \(x\notin\left\{1;-7\right\}\))
=>(x-2)(x+7)=(x+4)(x-1)
=>\(x^2+5x-14=x^2+3x-4\)
=>5x-14=3x-4
=>2x=10
=>x=5(nhận)
a, x + 1/3 = 3/4
- > x = 3/4 - 1/3 = 5/12
b, x - 2/5 = 5/4
-> x = 5/4 + 2/5 = 33/20
c, -x - 2/3 = 6/7
-> -x = 6/7 + 2/3 = 32/21
-> x = -32/21
d, 4/7 - x = 1/3
x = 4/7 - 1/3 = 5/21
a) x+1/3= 3/4
x= 3/4 - 1/3
x= 5/12
b) x-2/5 = 5/4
x= 5/4 + 2/5
x = 33 / 20
c) -x - 2/3 = 6/7
-x = 6/7 + 2/3
-x = 32/21 => x = -32/21
d) 4/7 -x = 1/3
x= 4/7 -1/3
x = 5/21
\(\frac{3}{x+3}>\frac{1}{x}>\frac{4}{x+7}\)
\(\Leftrightarrow\frac{3.4}{4\left(x+3\right)}>\frac{12}{12.x}>\frac{3.4}{\left(x+7\right).3}\)
\(\Leftrightarrow\frac{12}{4x+12}>\frac{12}{12x}>\frac{12}{3x+21}\Leftrightarrow4x+12<12x<3x+21\)
\(\Leftrightarrow\int^{4x+12<12x}_{12x<3x+21}\Leftrightarrow\int^{12<8x}_{8x<21}\)
\(\Leftrightarrow12<8x<21\Leftrightarrow8x\in\left\{16\right\}\Leftrightarrow x=2\)
Vậy x=2