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a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
\(a,\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{x-2}{7}=0\Rightarrow x-2=0\Leftrightarrow x=2\)
TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow\frac{-x+3}{5}=0\Rightarrow-x+3=0\Leftrightarrow x=3\)
TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{x+4}{3}=0\Rightarrow x+4=0\Leftrightarrow x=-4\)
\(\Rightarrow x\in\left\{2;3;-4\right\}\)
\(b,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)
\(\Rightarrow\frac{5x+3x-8x}{30}+1=0\)
\(\Rightarrow1=0\)( vô lý )\(\Rightarrow x\in\varnothing\)
a, | x + 1/5 | - 4 = - 2
| x + 1/5 | = - 2 + 4
| x + 1/5 | = 2
=> x + 1/5 = 2 hoặc x + 1/5 = -2
=> x = 9/5 hoặc x = -11/5
\(a,\left|x+\frac{1}{5}\right|-4=-2\)
\(\left|x+\frac{1}{5}\right|=-2+4\)
\(\left|x+\frac{1}{5}\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{5}=2\\x+\frac{1}{5}=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{5}\\x=-\frac{11}{5}\end{cases}}}\)
\(b,-\frac{15}{12}x+\frac{3}{7}=\frac{6}{5}x-\frac{1}{2}\)
\(\frac{6}{5}x+\frac{5}{4}x=\frac{3}{7}+\frac{1}{2}\)
\(\left(\frac{6}{5}+\frac{5}{4}\right)x=\frac{13}{14}\)
\(\frac{49}{20}x=\frac{13}{14}\)
\(x=\frac{130}{343}\)
\(S=\frac{\left(9\frac{3}{8}:5,2+3,4.2\frac{7}{34}\right):1\frac{9}{16}}{0,31.8\frac{2}{2}-5,61:27\frac{1}{3}}\)\(\Rightarrow S=\frac{\left(\frac{75}{8}.\frac{5}{26}+\frac{17}{5}.\frac{75}{34}\right):\frac{25}{16}}{\frac{31}{100}.9-\frac{561}{100}.\frac{3}{82}}\)\(\Rightarrow S=\frac{\left(\frac{75.5}{8.26}-\frac{17.75}{5.34}\right).\frac{16}{25}}{\frac{31.9}{100}-\frac{561.3}{100.82}}\)
\(\Rightarrow S=\frac{\left(\frac{375}{208}-\frac{15}{2}\right).\frac{16}{25}}{\frac{279}{100}-\frac{1682}{8200}}\)\(\Rightarrow S=\frac{\frac{-1185}{208}.\frac{16}{25}}{\frac{21196}{8200}}\)\(\Rightarrow S=\frac{-237}{65}:\frac{21196}{8200}\)\(\Rightarrow S=\frac{-194340}{137774}\)
\(\Rightarrow x=\frac{2}{3}S\Rightarrow x=\frac{2}{3}.\frac{-194340}{137774}\Rightarrow x=\frac{-388680}{413322}\)
\(M=\frac{23\frac{11}{15}-26\frac{13}{20}}{12^2+5^2}:\frac{1-\frac{1}{3}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2}-\frac{19}{37}\)\(\Rightarrow M=\frac{\frac{356}{15}-\frac{533}{20}}{12^2+5^2}:\frac{\frac{5}{8}}{3^2.13.2}-\frac{19}{37}\)
\(\Rightarrow M=\frac{\frac{-35}{12}}{12^2+5^2}.\frac{3^2.13.2}{\frac{5}{8}}-\frac{19}{37}\)\(\Rightarrow M=\frac{-84}{13}-\frac{19}{37}\Rightarrow M=\frac{-3355}{481}\Rightarrow15\%M=\frac{-3355}{481}.15\%\Rightarrow15\%M=\frac{-2013}{1924}\)
1) \(\frac{x+4}{7+y}=\frac{4}{7}\)\(\Rightarrow7\left(x+4\right)=4\left(7+y\right)\)
\(\Rightarrow7x+28=28+4y\)
\(\Rightarrow7x=4y\)
\(\Rightarrow\frac{x}{4}=\frac{y}{7}\)
áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{4}=\frac{y}{7}=\frac{x+y}{4+7}=\frac{22}{11}=2\)
x/4 = 2 => x = 4 x 2 = 8
y/7 = 2 => y = 2 x 7 = 14
a)
\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)
Vậy \(x = \frac{{ - 3}}{2}\).
b)
\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)
Vậy \(x = \frac{{ - 5}}{6}\).
c)
\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)
Vậy \(x = \frac{4}{5}\)
d)
\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)
Vậy \(x = \frac{{ - 2}}{5}\).
Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.
a)
\(\Rightarrow A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(\Rightarrow A=\frac{1}{5}+\frac{2}{7}\)
\(\Rightarrow A=\frac{17}{35}\)
b)
\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)
\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(\Rightarrow B=5.\frac{50}{671}=\frac{250}{671}\)
c)
\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow C=1-1-\frac{1}{25}\)
\(\Rightarrow C=\frac{1}{25}\)