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Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
\(a,x+\dfrac{1}{4}\times\dfrac{1}{5}=\dfrac{4}{5}\)
\(x+\dfrac{1}{20}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{1}{20}\)
\(x=\dfrac{16}{20}-\dfrac{1}{20}\)
\(x=\dfrac{15}{20}=\dfrac{3}{4}\)
\(b,3\dfrac{2}{3}:x=2\dfrac{2}{3}-1\dfrac{1}{2}\)
\(\dfrac{11}{3}:x=\dfrac{8}{3}-\dfrac{3}{2}\)
\(\dfrac{11}{3}:x=\dfrac{7}{6}\)
\(x=\dfrac{11}{3}:\dfrac{7}{6}=\dfrac{22}{7}\)
a: =>x+1/20=4/5
=>x=4/5-1/20=16/20-1/20=15/20=3/4
b: =>11/3:x=8/3-3/2=16/6-9/6=7/6
=>x=11/3:7/6=11/3*6/7=66/21=22/7
a)
`x+1/4xx1/5=4/5`
`x+1/20=4/5`
`x=4/5-1/20`
`x=16/20-1/20`
`x=15/20=3/4`
b)
`3 2/3:x=2 2/3-1 1/2`
`11/3:x=8/3-3/2`
`11/3:x=16/6-9/6`
`11/3:x=7/6`
`x=11/3:7/6`
`x=11/3xx6/7`
`x=22/7`
\(a,x+\dfrac{1}{4}.\dfrac{1}{5}=\dfrac{4}{5}\)
\(\Rightarrow\)\(x+\dfrac{1}{20}=\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{4}{5}-\dfrac{1}{20}=\dfrac{16}{20}-\dfrac{1}{20}=\dfrac{3}{4}\)
\(b,3\dfrac{2}{3}:x=2\dfrac{2}{3}-1\dfrac{1}{2}\)
\(\Rightarrow\dfrac{11}{3}:x=\dfrac{7}{6}\)
\(\Rightarrow x=\dfrac{11}{3}:\dfrac{7}{6}=\dfrac{22}{7}\)
a) \(x-\dfrac{3}{4}=6.\dfrac{3}{8}=\dfrac{9}{4}\)
\(\Rightarrow x=\dfrac{9}{4}+\dfrac{3}{4}=3\)
c) \(x+\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{3}{4}\)
\(\Rightarrow x+\dfrac{1}{6}=\dfrac{3}{4}\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{9}{12}-\dfrac{2}{12}=\dfrac{7}{12}\)
Bạn xem lại đề c
a: 2x(x+1)-135=-200
=>2(x^2+x)=-65
=>2x^2+2x+65=0
=>x^2+x+32,5=0
=>x^2+x+0,25+32,25=0
=>(x+0,5)^2+32,25=0(vô lý)
b: 4x-5(x-1)+15=13
=>4x-5x+5=-2
=>5-x=-2
=>x=5+2=7
c: 2/3x-1/4=3/5-7/8
=>2/3x=3/5-7/8+1/4=24/40-35/40+10/40=-1/40
=>x=-1/40:2/3=-1/40*3/2=-3/80
d: 1/2(2x-3)+105/2=-137/2
=>1/2(2x-3)=-137/2-105/2=-242/2=-121
=>2x-3=-242
=>2x=-239
=>x=-239/2
a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)
hay x=0
Vậy: x=0
b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)
\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)
hay \(x=\dfrac{-1}{9}\)
Vậy: \(x=\dfrac{-1}{9}\)
\(\dfrac{1}{2}-\dfrac{1}{2}:x=\dfrac{3}{4}\\ =>\dfrac{1}{2}:x=\dfrac{1}{2}-\dfrac{3}{4}=-\dfrac{1}{4}\\ =>x=\dfrac{1}{2}:\left(-\dfrac{1}{4}\right)=-2\)
`1/2 - 1/2 : x =3/4`
`1/2 : x = 1/2 -3/4`
`1/2:x=2/4-3/4`
`1/2:x=-1/4`
`x=1/2:(-1/4)`
`x=1/2.(-4)`
`x=-2`