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Bài làm:
a) \(\frac{x}{3}=\frac{2}{3}+\left(-\frac{1}{7}\right)\)
<=>\(\frac{x}{3}=\frac{2.7-1.3}{3.7}\)
<=>\(\frac{x}{3}=\frac{11}{21}\)
<=> \(21.x=33\)
<=> \(x=\frac{11}{7}\)
b) \(x-\frac{1}{4}=\frac{2}{13}\)
<=> \(x=\frac{1}{4}+\frac{2}{13}\)
<=> \(x=\frac{13.1+2.4}{4.13}\)
<=> \(x=\frac{21}{52}\)
c)\(\frac{4}{7}.x=\frac{9}{8}-\frac{125}{1000}\)
<=> \(\frac{4}{7}.x=\frac{9}{8}-\frac{1}{8}\)
<=> \(\frac{4}{7}.x=\frac{8}{8}\)
<=> \(x=1:\frac{4}{7}\)
<=> \(x=\frac{7}{4}\)
Học tốt!!!!
7 + 67 + 56 - 76 + 368 - 93 + 45 -88 =
=? ? ? ? ? ? ?
= ? ? ? ? ?
= ?
Bài 1: Tìm \(x\)
a; \(x-2\) + 7 = 1.3.(-9)
\(x\) - 2 + 7 = 3.(-9)
\(x\) - 2 + 7 = - 27
\(x\) = - 27 - 7 + 2
\(x\) = - 34 + 2
\(x\) = - 32
Vậy \(x=-32\)
Bài 1
c; - 2\(x\) + 5 = 7
- 2\(x\) = 7 - 5
- 2\(x\) = - 2
\(x\) = -2 : (-2)
\(x\) = - 1
Vậy \(x\) = - 1
a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
a) x=3/7
b)x=8/7
c)x=6/7
a)\(x=\dfrac{2}{7}:\dfrac{2}{3}=\dfrac{3}{7}\)
b)\(x=\dfrac{13}{7}\times\dfrac{8}{13}=\dfrac{8}{7}\)
c)\(x=\dfrac{3}{2}:\dfrac{7}{4}=\dfrac{6}{7}\)