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Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
Câu 1 :
a, 8.( -5 ).( -4 ).2
= [ 8.2 ].[( -5 ).(-4 ]
= 16.20
= 320
b, \(1\frac{3}{7}+\frac{-1}{3}+2\frac{4}{7}\)
\(=\frac{10}{7}+\frac{-1}{3}+\frac{18}{7}\)
\(=\frac{11}{3}\)
c, \(\frac{8}{5}.\frac{2}{3}+\frac{-5.5}{3.5}\)
\(=\frac{8}{3}+\frac{-5}{3}\)
\(=\frac{3}{3}=1\)
d, \(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{16}.4\)
\(=\frac{55}{56}-\frac{3}{4}\)
\(=\frac{13}{56}\)
Câu 2 :
a, 2x + 10 = 16
2x = 16 + 10
2x = 26
x = 26 : 2
x = 13
b, \(x-\frac{1}{3}=\frac{5}{4}\)
\(x=\frac{5}{4}+\frac{1}{3}\)
\(x=\frac{19}{12}\)
c, \(2x+3\frac{1}{3}=7\frac{1}{3}\)
\(2x+\frac{10}{3}=\frac{22}{3}\)
\(2x=\frac{22}{3}-\frac{10}{3}\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
d, \(\left(\frac{2}{11}+\frac{1}{3}\right)x=\left(\frac{1}{7}-\frac{1}{8}\right).56\)
\(\frac{17}{33}x=1\)
\(x=1-\frac{17}{33}\)
\(x=\frac{16}{33}\)
a) 3/8 . x = 9/8 - 1
3/8 . x = 1/8
x = 1/8 : 3/8
x = 1/3
b) 4/5 . x = 7/5 - 1/5
4/5 . x = 6/5
x = 6/5 : 4/5
x = 3/2
c) 12/7 : x + 2/3 = 7/5
12/7 : x = 7/5 - 2/3
12/7 : x = 11/15
x = 12/7 : 11/15
x = 180/77
d) 3.(x + 7) - 15 = 27
3.(x + 7) = 27 + 15
3.(x + 7) = 42
x + 7 = 42 : 3
x + 7 = 14
x = 14 - 7
x = 7
a) \(\dfrac{3}{8}x=\dfrac{9}{8}-1\)
\(\Rightarrow\dfrac{3}{8}x=\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{1}{8}:\dfrac{3}{8}=\dfrac{1}{3}\)
b) \(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{1}{5}\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{6}{5}:\dfrac{4}{5}=\dfrac{3}{2}\)
c) \(\dfrac{12}{7}:x+\dfrac{2}{3}=\dfrac{7}{5}\)
\(\Rightarrow\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{12}{7}:x=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{12}{7}:\dfrac{11}{15}=\dfrac{180}{77}\)
d) \(3\left(x+7\right)-15=27\)
\(\Leftrightarrow3\left(x+7\right)=42\)
\(\Leftrightarrow x+7=14\Leftrightarrow x=7\)
1) |x + 2| = 4
\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
2) 3 – |2x + 1| = (-5)
\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)
3) 12 + |3 – x| = 9
\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)
=>\(x=\varnothing\)
1) I x+2 I=4
\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)
2) \(3-|2x+1|=-5\)
\(\Leftrightarrow|2x+1|=8\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)
3) \(12+|3-x|=9\)
\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)
a:x=3/5-7/8=24/40-35/40=-11/40
b: =>1/3:(2x-1)=-1/6
=>2x-1=-2
=>2x=-1
=>x=-1/2
c: =>x-3/4=17/2+7/4=34/4+7/4=41/4
=>x=11
d: =>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-7:2/5=-35/2
a. \(\dfrac{8}{7}-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=-1\)
\(-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{15}{7}\)
\(\dfrac{x}{3}-2=\dfrac{-1}{15}\)
\(\dfrac{x}{3}=\dfrac{29}{15}\)
\(x=5,8\)
b. \(\dfrac{5}{8}+\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{4}\)
\(\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{8}\)
\(2x-1=\dfrac{5}{2}\)
\(2x=\dfrac{7}{2}\)
\(x=\dfrac{7}{4}\)