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Câu 1:
a) Ta có: \(\left(x-\frac{5}{8}\right)\cdot\frac{5}{18}=\frac{15}{36}\)
\(\Leftrightarrow x-\frac{5}{8}=\frac{15}{36}:\frac{5}{8}=\frac{15}{36}\cdot\frac{8}{5}=\frac{120}{180}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}+\frac{5}{8}=\frac{16}{24}+\frac{15}{24}=\frac{41}{24}\)
Vậy: \(x=\frac{41}{24}\)
b) Ta có: \(\left|x-\frac{1}{3}\right|=\frac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{5}{6}\\x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{6}+\frac{1}{3}=\frac{5}{6}+\frac{2}{6}=\frac{7}{6}\\x=\frac{-5}{6}+\frac{1}{3}=\frac{-5}{6}+\frac{2}{6}=\frac{-3}{6}=\frac{-1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{7}{6};\frac{-1}{2}\right\}\)
a, (10/3:x).(-5/4)=-10/3
10/3:x=-10/3:(-5/4)
10/3:x=8/3
x=10/3:8/3
x=5/4
b,(-6/5+x):(-18/5)=-1/4
-6/5+x=-1/4.(-18/5)
-6/5+x=9/10
x=9/10-(-6/5)=9/10+6/5
x=21/10
c,-22/15.x+1/3=-2/5
-22/15.x=-2/5-1/3=-11/15
x=-11/15:(-22/15)
x=11/21
d,(0,25-30%x).1/3=-31/6+1/4=-59/12
1/4-3/10x=-59/12:1/3=-59/4
3/10x=1/4-(-59/4)=1/4+59/4=15
x=15:3/10
x=50
e,(0,5x-3/7):1/2=8/7
1/2x=8/7.1/2=4/7
x=4/7:1/2
x=8/7
\(a,\left(x-\dfrac{5}{8}\right).\dfrac{5}{8}=-\dfrac{15}{36}\)
\(\left(x-\dfrac{5}{8}\right)=-\dfrac{15}{36}\div\dfrac{5}{8}\)
\(x-\dfrac{5}{8}=-\dfrac{2}{3}\)
\(x=-\dfrac{2}{3}+\dfrac{5}{8}\)
\(x=-\dfrac{1}{24}\)
\(b,\left(x-\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x-\dfrac{1}{3}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{1}{3}\)
\(x=\dfrac{7}{6}\)
\(a,\left(x-\dfrac{5}{8}\right)\cdot\dfrac{8}{18}=-\dfrac{15}{16}\\ x-\dfrac{5}{8}=-\dfrac{15}{36}:\dfrac{8}{18}\\ x-\dfrac{5}{8}=-\dfrac{15}{16}\\ x=-\dfrac{15}{16}+\dfrac{5}{8}\\ x=-\dfrac{15}{16}+\dfrac{10}{16}\\ x=-\dfrac{5}{16}\\ b,x-\dfrac{1}{3}=\dfrac{5}{6}\\ x=\dfrac{5}{6}+\dfrac{1}{3}\\ x=\dfrac{5}{6}+\dfrac{2}{6}\\ x=\dfrac{7}{6}\)
A/ X= \(-\frac{7}{8}\)
B/ X= \(\frac{7}{6}\)