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21 tháng 5 2022
a: x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
A=1/7-x-(x+3/5)+4/5
=1/7-2x-3/5+4/5
=-2x+12/35
b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)
\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)
-3/5<x<1/7
nên x-1/7<0; x+3/5>0
\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)
c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)
\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)
Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0
\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)
21 tháng 5 2022
a: x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
A=1/7-x-(x+3/5)+4/5
=1/7-2x-3/5+4/5
=-2x+12/35
b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)
\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)
-3/5<x<1/7
nên x-1/7<0; x+3/5>0
\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)
c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)
\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)
Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0
\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)
D
11 tháng 6 2018
1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
26 tháng 3 2020
a) Ta có: \(\frac{7}{8}+\frac{15}{23}+\frac{1}{8}+\left(-\frac{15}{23}\right)+\frac{1}{3}\)
\(=\left(\frac{7}{8}+\frac{1}{8}\right)+\left(\frac{15}{23}-\frac{15}{23}\right)+\frac{1}{3}\)
\(=1+\frac{1}{3}=\frac{4}{3}\)
b) Ta có: \(\frac{3-\frac{2}{3}}{\frac{1}{3}-3}\)
\(=\left(3-\frac{2}{3}\right)\cdot\frac{1}{\frac{1}{3}-3}=\left(3-\frac{2}{3}\right)\cdot\frac{1}{-\frac{8}{3}}\)
\(=\left(3-\frac{2}{3}\right)\cdot\frac{-3}{8}\)
\(=-\frac{9}{8}+\frac{1}{4}=\frac{-9}{8}+\frac{2}{8}=-\frac{7}{8}\)
c) Ta có: \(8\cdot\left(-\frac{1}{2}\right)^3+3\cdot\left(-\frac{1}{2}\right)^2-0,75\)
\(=8\cdot\frac{-1}{8}+3\cdot\frac{1}{4}-\frac{3}{4}\)
\(=-1\)
NT
0
\(x-\dfrac{1}{2}=\dfrac{3}{2}+\dfrac{1}{7}\)
\(x-\dfrac{1}{2}=\dfrac{21}{14}+\dfrac{2}{14}\)
\(x-\dfrac{1}{2}=\dfrac{23}{14}\)
\(x=\dfrac{23}{14}+\dfrac{1}{2}\)
\(x=\dfrac{23}{14}+\dfrac{7}{14}\)
\(x=\dfrac{30}{14}=\dfrac{15}{7}.\)
x-1/2=3/2+1/7
x-1/2=23/14
x=23/14+1/2
x=15/7