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a) \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
\(\Leftrightarrow3-2x=-1\)
\(\Leftrightarrow-2x=-1-3\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
b) \(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{5}{3}x+\dfrac{1}{5}=x-\dfrac{7}{8}\)
\(\Leftrightarrow200x+24=120x-105\)
\(\Leftrightarrow80x=-129\)
\(\Leftrightarrow x=-\dfrac{129}{80}\)
Vậy \(x=-\dfrac{129}{80}\)
c) \(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)
\(\Leftrightarrow-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow-\left|x+0,5\right|=-\dfrac{11}{20}\)
\(\Leftrightarrow\left|x+0,5\right|=\dfrac{11}{20}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+0,5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{21}{20};x_2=\dfrac{1}{20}\)
d) \(\left(x+0,2\right)^2+0,75=1\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2+\dfrac{3}{4}=1\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=1-\dfrac{3}{4}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow x+\dfrac{1}{5}=\pm\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{1}{2}\\x+\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{7}{10};x_2=\dfrac{3}{10}\)
a, \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
=>\(-\dfrac{1}{2}x=-\dfrac{1}{4}-\dfrac{3}{4}\)
=>\(-\dfrac{1}{2}x=-1\)
=>\(x=-1:(-\dfrac{1}{2})\)
=>\(x=2\)
vậy ...........
b,\(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
=>\(\dfrac{5}{3}x+0,2=x-\dfrac{7}{8}\)
=>\(\dfrac{5}{3}x-x=-\dfrac{7}{8}-0,2\)
=>\(\dfrac{2}{3}x=-\dfrac{43}{40}\)
=>\(x=-\dfrac{43}{40}:\dfrac{2}{3}\)
=>\(x=-\dfrac{192}{80}\)
vậy...................
c,\(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)
=>\(-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)
=>\(-\left|x+0,5\right|=-\dfrac{11}{20}\)
=>\(\left|x+0,5\right|=\dfrac{11}{20}\)
=>\(\left[{}\begin{matrix}x+0.5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)
vậy ....... hoặc.....
d,\((x+0,2)^2+0,75=1\)
=>\(\left(x+0,2\right)^2=1-0,75\)
=>\(\left(x+0,2\right)^2=0,25\)
=>\(\left[{}\begin{matrix}x+0,2=0,5\\x+0,2=-0,5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0,3\\x=-0,7\end{matrix}\right.\)
vậy..........................
HỌC TỐT NHA !!!!!
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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1:
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x^2+5x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x\right)^2+10\left(x^2+5x\right)=0\)
\(\Leftrightarrow x^2+5x=0\)
=>x=0 hoặc x=-5
3: \(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
=>(x+2)(x-1)=0
=>x=-2 hoặc x=1
b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)
\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)
\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)
\(\Leftrightarrow3x\left(x+4\right)=0\)
=>x=0(nhận) hoặc x=-4(loại)
