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4x5x6/3x10x8=120/240=120:120/240:120=1/2 nha
~HT~
k cho mình nha
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1,
a, \(x\)x 124= 4829-365
\(x\)x 124 = 4464
\(x\)= 4464 :124
\(x\)= 36.
b,\(x\)x 137 = 25 x 3
\(x\)x 137 = 75
\(x\)= \(\frac{75}{137}\)
2,
\(\frac{4}{5}\)x\(\frac{3}{7}+\frac{4}{5}\)x\(\frac{6}{7}-\frac{4}{5}\)x\(\frac{4}{10}\)
=\(\frac{4}{5}\)x\(\left(\frac{3}{7}+\frac{6}{7}-\frac{4}{10}\right)\)=\(\frac{4}{5}\)x\(\frac{31}{35}\)\(\frac{124}{175}\)
\(\text{Bài }1\text{ : Tìm }x\text{ : }\)
\(a,\text{ }x\text{ x }124=4829-365\)
\(x\text{ x }124=4464\)
\(x=4464\text{ : }124\)
\(x=36\)
\(b,\text{ }x\text{ x }137=25\text{ x }3\)
\(x\text{ x }137=75\)
\(x=75\text{ : }137\)
\(x=\frac{75}{137}\)
\(\text{Bài }2\text{ : Tính bằng cách thuận tiện nhất : }\)
\(\frac{4}{5}\text{ x }\frac{3}{7}+\frac{4}{5}\text{ x }\frac{6}{7}-\frac{4}{5}\text{ x }\frac{4}{10}\)
\(=\frac{4}{5}\text{ x }\left(\frac{3}{7}+\frac{6}{7}-\frac{4}{10}\right)\)
\(=\frac{4}{5}\text{ x }\left(\frac{9}{7}-\frac{4}{10}\right)\)
\(=\frac{4}{5}\text{ x }\frac{31}{35}\)
\(=\frac{124}{175}\)
a) \(x+\frac{4}{5}=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}-\frac{4}{5}\)
\(\Rightarrow x=\frac{7}{10}\)
Vậy \(x=\frac{7}{10}.\)
b) \(x-\frac{3}{2}=\frac{11}{4}\)
\(\Rightarrow x=\frac{11}{4}+\frac{3}{2}\)
\(\Rightarrow x=\frac{17}{4}\)
Vậy \(x=\frac{17}{4}.\)
c) \(\frac{25}{3-x}=\frac{5}{6}\)
\(\Rightarrow\left(3-x\right)5=25.6\)
\(\Rightarrow15-5x=150\)
\(\Rightarrow5x=15-150\)
\(\Rightarrow5x=-135\)
\(\Rightarrow x=-135:5\)
\(\Rightarrow x=-27\)
Vậy \(x=-27.\)
\(\frac{23}{6}:x-\frac{2}{5}=\frac{1}{4}\)
\(\frac{23}{6}:x=\frac{1}{4}+\frac{2}{5}\)
\(\frac{23}{6}:x=\frac{13}{20}\)
\(x=\frac{23}{6}:\frac{13}{20}\)
\(x=\frac{230}{39}\)
Vậy \(x=\frac{230}{39}\)
\(\frac{23}{6}\) : X - \(\frac{2}{5}\) = \(\frac{1}{4}\)
\(\frac{23}{6}\) : X = \(\frac{1}{4}\) + \(\frac{2}{5}\)
\(\frac{23}{6}\) : X = \(\frac{13}{20}\)
X = \(\frac{23}{6}\) : \(\frac{13}{20}\)
X = \(\frac{23}{6}\) x \(\frac{20}{13}\)
X = \(\frac{230}{39}\)
~ Hok T ~
\(x+\dfrac{3}{4}=\dfrac{28}{6}+\dfrac{1}{2}\\ x+\dfrac{3}{4}=\dfrac{31}{6}\\ x=\dfrac{31}{6}-\dfrac{3}{4}\\ x=\dfrac{53}{12}\)
\(x+\dfrac{3}{4}=\dfrac{28+3}{6}=\dfrac{31}{6}\Leftrightarrow x=\dfrac{31}{6}-\dfrac{3}{4}=\dfrac{62-9}{12}=\dfrac{53}{12}\)