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a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
Bài 2:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
a: x=4/27-2/3=4/27-18/27=-14/27
b: =>3/4x-1/4x=1/6+7/3
=>1/2x=1/6+14/6=5/2
hay x=5
c: =>13/10x=7/2+5/2=6
=>x=13/10:6=13/60
d: (3x+2)(-2/5x-7)=0
=>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-35/2
a, 2/5 + 3/4 : x = -1/2
3/4 : x = -1/2 - 2/5
3/4 : x = -9/10
x = 3/4 : -9/10
x = -5/6
b, 5/7 - 2/3 . x = 4/5
2/3 . x = 4/5 + 5/7
2/3 . x = 53/35
x = 53/35 : 2/3
x = 159/70
a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27
b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4
tick cho mik nhé
a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)
\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(x=\dfrac{-3}{4}\)
b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)
\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(x=-\dfrac{2}{3}\)
c, \(\dfrac{-1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\dfrac{-1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)
\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)
\(x=\dfrac{1}{5}\)
d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)
\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)
\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)
\(\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(x=\dfrac{2}{3}:\dfrac{5}{8}\)
\(x=\dfrac{16}{15}\)
#YVA6
\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)
\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)
\(\Leftrightarrow x=-\dfrac{3}{4}\)
\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)
\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c,-\dfrac{1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)
\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)
\(\Leftrightarrow x=\dfrac{1}{5}\)
\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)
\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{16}{15}\)
a, \(\left(2x-3\right)\left(\dfrac{3}{4}x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=3\\\dfrac{3}{4}x=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-4}{3}\end{matrix}\right.\)
Vậy......
b, \(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy......
c, \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)
\(\Rightarrow\dfrac{1}{7}:x=\dfrac{-3}{14}\)
\(\Rightarrow x=\dfrac{1}{7}:\dfrac{-3}{14}=\dfrac{-2}{3}\)
Vậy.....
Chúc bạn học tốt!!!
a,(2x-3)(\(\dfrac{3}{4}\)x+1)=0
*2x-3=0\(\rightarrow\)2x=3\(\rightarrow\)x=\(\dfrac{3}{2}\)
*\(\dfrac{3}{4}\)x+1=0\(\rightarrow\)\(\dfrac{3}{4}\)x=\(-1\)\(\rightarrow\)x=\(\dfrac{-3}{4}\)
Vậy x\(\in\){\(\dfrac{-3}{4};\dfrac{3}{2}\)}
b,(5x-1)(2x-1/3)=0
*5x-1=0\(\rightarrow\)5x=1\(\rightarrow\)x=1/5
*2x-1/3=0\(\rightarrow\)2x=1/3\(\rightarrow\)x=1/6
Vậy x\(\in\){1/5;1/6}
c,3/7+1/7:x=3/14
1/7:x=-3/14
x=-14/21 Vậy x=-14/21
Lời giải:
$\frac{x}{3}-\frac{1}{4}=\frac{-5}{6}$
$\frac{x}{3}=\frac{1}{4}-\frac{5}{6}=\frac{-7}{12}$
$x=\frac{-7}{12}.3=\frac{-7}{4}$
-------------------
$\frac{2x}{3}=\frac{6}{x}$ ($x\neq 0$)
$\Rightarrow 2x^2=18$
$x^2=18:2=9=(-3)^2=3^2$
$\Rightarrow x=\pm 3$
---------------------
$\frac{x-1}{14}=\frac{4}{x}$ ($x\neq 0$)
$\Rightarrow x(x-1)=14.4=56$
$x^2-x-56=0$
$(x+7)(x-8)=0$
$\Rightarrow x+7=0$ hoặc $x-8=0$
$\Leftrightarrow x=-7$ hoặc $x=8$
---------------------------
$\frac{-x}{8}=\frac{-50}{x}$ ($x\neq 0$)
$\Rightarrow -x^2=8(-50)$
$x^2=400=20^2=(-20)^2$
$\Rightarrow x=\pm 20$
\(a,\dfrac{8}{14}=\dfrac{x+1}{7}\\ \Leftrightarrow\dfrac{x+1}{7}=\dfrac{4}{7}\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\\ b,\dfrac{-4}{3}=\dfrac{5x+1}{-27}\\ \Leftrightarrow\dfrac{36}{-27}=\dfrac{5x+1}{-27}\\ \Leftrightarrow5x+1=36\\ \Leftrightarrow5x=35\\ \Leftrightarrow x=7\)