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\(\begin{array}{l}a)x + 7,25 = 15,75\\x = 15,75 - 7,25\\x = 8,5\end{array}\)
Vậy x = 8,5
\(\begin{array}{l}b)\left( { - \frac{1}{3}} \right) - x = \frac{{17}}{6}\\\left( { - \frac{1}{3}} \right) - \frac{{17}}{6} = x\\\frac{{ - 2}}{6} - \frac{{17}}{6} = x\\\frac{{ - 19}}{6} = x\\x = \frac{{ - 19}}{6}\end{array}\)
Vậy \(x = \frac{{ - 19}}{6}\)
Chú ý: A = B và B = A là tương đương nhau
a,x+1,25=30
x=30-1,25
x=28,75
b,x.4,25=57,8
x=57,8:4,25
x=13,6
\(a,\frac{11}{6}=-\frac{x}{5}\)
\(\Leftrightarrow-x\times6=11\times5\)
\(\Leftrightarrow-6x=55\)
\(\Leftrightarrow x=55:\left(-6\right)\)
\(\Leftrightarrow x=-\frac{55}{6}\)
\(b,4,25:8=-3,5:x\)
\(\Leftrightarrow\frac{17}{32}=--\frac{7}{2}:x\)
\(\Leftrightarrow x=-\frac{7}{2}:\frac{17}{32}\)
\(\Leftrightarrow x=-\frac{112}{17}\)
a) \(\frac{11}{6}=\frac{-x}{5}\)
\(\Rightarrow\frac{55}{30}=\frac{\left(-x\right)\cdot6}{30}\)
\(\Rightarrow55=\left(-x\right)\cdot6\)
\(\Rightarrow-x=\frac{55}{6}\)
\(\Rightarrow x=-\frac{55}{6}\)
b) \(4,25:8=-3,5:x\)
\(\Rightarrow0,53125=-3,5:x\)
\(\Rightarrow x=-3,5:0,53125\)
\(\Rightarrow x=-6,5882...\)
a) \(\frac{x}{4}=\frac{16}{128}\)
\(x.128=4.16\)
\(x.128=64\)
\(x=64:128\)
\(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
b) \(\frac{11}{6}=\frac{-x}{5}\)
\(11.5=\left(-x\right).6\)
\(55=\left(-x\right).6\)
\(-x=55:6\)
\(-x=\frac{55}{6}\)
\(\Rightarrow x=-\frac{55}{6}\)
Vậy \(x=-\frac{55}{6}\)
c) \(4,25:8=-3,5:x\)
\(\frac{17}{32}=-3,5:x\)
\(x=\frac{-35}{10}:\frac{17}{32}\)
\(x=-6\frac{10}{17}\)
Vậy \(x=-6\frac{10}{17}\)
\(a.\)
\(\frac{x}{4}=\frac{16}{128}\)
\(\Rightarrow x=\frac{16.4}{128}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy : \(x=\frac{1}{2}\)
\(b.\)
\(\frac{11}{6}=\frac{-x}{5}\)
\(\Rightarrow x=\frac{11.5}{-6}\)
\(\Rightarrow x=-\frac{55}{6}\)
Vậy : \(x=-\frac{55}{6}\)
\(c.\)
\(\frac{4,25}{8}=\frac{-3,5}{x}\)
\(\Rightarrow x=\frac{-3,5.8}{4,25}\)
\(\Rightarrow x=-6\frac{10}{7}\)
Vậy : \(x=-6\frac{10}{7}\)
a) Ta có: \(x-\frac{2}{3}=0,75\)
\(\Leftrightarrow x-\frac{2}{3}=\frac{3}{4}\)
hay \(x=\frac{3}{4}+\frac{2}{3}=\frac{17}{12}\)
Vậy: \(x=\frac{17}{12}\)
b) Ta có: \(\frac{1}{3}+\frac{2}{3}x=-1\)
\(\Leftrightarrow\frac{2}{3}x=-1-\frac{1}{3}=-\frac{4}{3}\)
hay \(x=\frac{-4}{3}:\frac{2}{3}=\frac{-4}{3}\cdot\frac{3}{2}=-2\)
Vậy: x=-2
c) Ta có: \(\left|2x-3\right|-\frac{3}{4}=4,25\)
\(\Leftrightarrow\left|2x-3\right|=\frac{17}{4}+\frac{3}{4}=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Vậy: x∈{-1;4}
a, \(x-\frac{2}{3}=0.75\)
\(x=0.75-\frac{2}{3}\)
\(x=\frac{1}{12}\)
Vậy...
b, \(\frac{1}{3}+\frac{2}{3}\cdot x=-1\)
\(\frac{2}{3}\cdot x=-1-\frac{1}{3}=-\frac{4}{3}\)
\(x=-\frac{4}{3}:\frac{2}{3}=-2\)
Vậy x = -2
c, \(|2x-3|-\frac{3}{4}=4.25\)
\(|2x-3|=4.25-\frac{3}{4}=\frac{7}{2}\)
=> \(2x-3=\frac{7}{2}hay2x-3=-\frac{7}{2}\)
2x = \(\frac{7}{2}+3\) 2x = \(-\frac{7}{2}+3\)
2x = \(\frac{13}{2}\) 2x = \(-\frac{1}{2}\)
x = \(\frac{12}{2}:2=\frac{13}{4}\) x = \(-\frac{1}{2}:2\) = \(-\frac{1}{4}\)
Vậy...
\(\left(25\%\cdot x-7,25\right)+4,25=10\)
=>\(x\cdot0,25-3=10\)
=>\(x\cdot0,25=13\)
=>\(x=\dfrac{13}{0,25}=52\)