Tìm x biết : x+ \(\frac{8}{5}\) = \(\frac{7}{25}\)
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2/5.x:7/3.1/8=25/15
2/5.x:7/24=5/3
2/5.x=5/3.7/24
2/5.x=6/34
x=6/34:2/5
x=15/34
\(\frac{2}{5}x:\frac{7}{3}.\frac{1}{8}=\frac{25}{15}\Leftrightarrow\frac{2}{5}x:\frac{7}{3}.\frac{1}{8}=\frac{5}{3}\Leftrightarrow\frac{2}{5}x:\frac{7}{3}=\frac{5}{3}:\frac{1}{8}\Leftrightarrow\frac{2}{5}x:\frac{7}{3}=\frac{5}{3}.8\Leftrightarrow\frac{2}{5}x:\frac{7}{3}=\frac{40}{3}\)
\(\Leftrightarrow\frac{2}{5}x=\frac{40}{3}.\frac{7}{3}\Leftrightarrow\frac{2}{5}x=\frac{280}{3}\Leftrightarrow x=\frac{280}{3}:\frac{2}{5}\Leftrightarrow x=\frac{280}{3}.\frac{5}{2}\Leftrightarrow x=\frac{700}{3}\)
Vậy .........
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{\frac{3}{3}+\frac{3}{5}-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{3.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}{8.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}=\frac{3.1}{8.1}=\frac{3}{8}\)
\(\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}=\frac{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}+\frac{7}{16}}=\frac{1.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}{7.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}=\frac{1.1}{7.1}=\frac{1}{7}\)
=>\(\frac{3}{8}-\frac{1}{7}=\frac{13}{56}\)
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a,
x.2/7.3/4=5/21
x.3/14=5/21
x=5/21:3/14
x=10/9
b,
x.1/2=1/3
x=1/3:1/2
x=2/3
c,
x:4/5=25/8:5/4
x:4/5=5/2
x=5/2.4/5=2
A, x= 5/25 : 3/4 : 2/7 = 14/15
B, x=1/3 : 1/2 = 2/3
C, x=(25/8 : 5/4)x4/5 = 5/2 x 4/5 = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a/\(\frac{13}{11}.\frac{22}{26}-x^2=\frac{7}{16}\)
\(\Rightarrow1-x^2=\frac{7}{16}\)
\(\Rightarrow x^2=\frac{9}{16}\)
\(\Rightarrow x=\orbr{\begin{cases}\frac{3}{4}\\-\frac{3}{4}\end{cases}}\)
\(a,\frac{13}{11}.\frac{22}{26}-x^2=\frac{7}{16}\)
\(1-x^2=\frac{7}{16}\)
\(x^2=1-\frac{7}{16}=\frac{9}{16}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\)
\(b,x^2+-\frac{9}{25}=\frac{2}{5}.\frac{8}{5}\)
\(x^2+-\frac{9}{25}=\frac{16}{25}\)
\(x^2=\frac{16}{25}--\frac{9}{25}=1\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
học tốt ~~~
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1. \(x=\frac{61}{42}\)
2. \(x=\frac{-36}{5}\)
3. \(x=\frac{13}{11}\)
4. \(x=\frac{1}{12}\)
5.\(x=\frac{-5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\begin{array}{l}a)x - \left( {\dfrac{5}{4} - \dfrac{7}{5}} \right) = \dfrac{9}{{20}}\\x = \dfrac{9}{{20}} + \left( {\dfrac{5}{4} - \dfrac{7}{5}} \right)\\x = \dfrac{9}{{20}} + \dfrac{{25}}{{20}} - \dfrac{{28}}{{20}}\\x = \dfrac{{6}}{{20}}\\x = \dfrac{{ 3}}{{10}}\end{array}\)
Vậy \(x = \dfrac{{ 3}}{{10}}\)
\(\begin{array}{*{20}{l}}{b)9 - x = \dfrac{8}{7} - \left( { - \dfrac{7}{8}} \right)}\\\begin{array}{l}9 - x = \dfrac{8}{7} + \dfrac{7}{8}\\9 - x = \dfrac{{64}}{{56}} + \dfrac{{49}}{{56}}\\9 - x = \dfrac{{113}}{{56}}\end{array}\\{x = 9 - \dfrac{{113}}{{56}}}\\{x = \dfrac{{504}}{{56}} - \dfrac{{113}}{{56}}}\\{x = \dfrac{{391}}{{56}}}\end{array}\)
Vậy \(x = \dfrac{{391}}{{56}}\)
\(\frac{-33}{25}\)
\(x+\frac{8}{5}=\frac{7}{25}\)
\(x=\frac{7}{25}-\frac{8}{5}\)
\(x=\frac{-33}{25}\)