Tim x:
a) \(\dfrac{13+x}{20}=\dfrac{3}{4}\)
b) \(\dfrac{23-x}{25}=\dfrac{4}{5}\)
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e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
\(a,x-\dfrac{5}{7}=\dfrac{3}{5}\times\dfrac{1}{2}\)
\(x-\dfrac{5}{7}=\dfrac{3}{10}\)
\(x=\dfrac{3}{10}+\dfrac{5}{7}\)
\(x=\dfrac{21}{70}+\dfrac{50}{70}\)
\(x=\dfrac{71}{70}\)
\(Vayx=\dfrac{71}{70}\)
\(b,x\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{3}{4}\)
\(x\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{6}{8}\)
\(x\times\dfrac{2}{3}=\dfrac{11}{8}\)
\(x=\dfrac{11}{8}:\dfrac{2}{3}\)
\(x=\dfrac{11}{8}\times\dfrac{3}{2}\)
\(x=\dfrac{33}{16}\)
\(Vayx=\dfrac{33}{16}\)
a) \(x+\dfrac{4}{15}=\dfrac{4}{12}\)
\(x=\dfrac{4}{12}-\dfrac{4}{15}\)
\(x=\dfrac{20}{60}-\dfrac{16}{60}\)
\(x=\dfrac{1}{15}\)
b) \(x-\dfrac{5}{8}=1\dfrac{2}{3}\)
\(x-\dfrac{5}{8}=\dfrac{5}{3}\)
\(x=\dfrac{5}{3}+\dfrac{5}{8}\)
\(x=\dfrac{40}{24}+\dfrac{15}{24}\)
\(x=\dfrac{55}{24}\)
a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))
= \(\dfrac{1}{13}\) \(\times\) \(\dfrac{317}{73}\)
= \(\dfrac{317}{949}\)
b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)
= \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)
= \(\dfrac{199}{45}\)
c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)
= \(\dfrac{20696}{315}\)
d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)
= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)
= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))
= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)
= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)
e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))
= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= -1 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) - \(\dfrac{8}{33}\)
= - \(\dfrac{47}{33}\)
\(a,x-\dfrac{1}{4}=\dfrac{7}{2}.\dfrac{-3}{5}\\ \Rightarrow x-\dfrac{1}{4}=\dfrac{-21}{10}\\ \Rightarrow x=\dfrac{-21}{10}+\dfrac{1}{4}\\ \Rightarrow x=\dfrac{-37}{20}\\ b,\dfrac{x}{134}=\dfrac{9}{7}.\dfrac{5}{-11}\\ \Rightarrow\dfrac{x}{134}=\dfrac{-45}{77}\\ \Rightarrow x=\dfrac{-45}{77}.134\\ \Rightarrow x=\dfrac{-6030}{77}\)
\(x-\dfrac{1}{4}=\dfrac{7}{2}\cdot\dfrac{-3}{5}\)
\(x-\dfrac{1}{4}=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}+\dfrac{1}{4}\)
\(x=\dfrac{-37}{20}\)
b ) \(\dfrac{x}{134}=\dfrac{9}{7}\cdot\dfrac{5}{-11}\)
\(\dfrac{x}{134}=\dfrac{9}{7}\cdot\dfrac{-5}{11}\)
\(\dfrac{x}{134}=\dfrac{-45}{77}\)
\(x=\dfrac{-45}{77}\cdot134\)
\(x=-\dfrac{6034}{77}\)
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
`(13+x)/20 = 3/4`
`(13+x) xx4=3xx20`
`(13+x)xx4=60`
`13+x=60:4`
`13+x=15`
`x=15-13`
`x=2`
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`(23-x)/25 =4/5`
`(23-x)xx5=4xx25`
`(23-x)xx5=100`
`23-x=100:5`
`23-x=20`
`x=23-20`
`x=3`
Giỏi v