\(\dfrac{x}{5}=\dfrac{-12}{10}\)
Tính:
a) \(\dfrac{11}{10}+\dfrac{3}{5}:\dfrac{2}{3}\)
b) \(\dfrac{4}{3}\) + 5 x \(\dfrac{5}{8}\)
c) \(\left(\dfrac{2}{5}+\dfrac{3}{7}\right)x\dfrac{25}{29}\)
d) \(\dfrac{1}{4}x\dfrac{5}{12}+\dfrac{5}{12}x\dfrac{4}{5}\)
a) \(\dfrac{11}{10}+\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{11}{10}+\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)
b) \(\dfrac{4}{3}+5\times\dfrac{5}{8}=\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)
c) \(\left(\dfrac{2}{5}+\dfrac{3}{7}\right)\times\dfrac{25}{29}=\left(\dfrac{14}{35}+\dfrac{15}{35}\right)\times\dfrac{25}{39}=\dfrac{29}{35}\times\dfrac{25}{39}=\dfrac{145}{274}\)
d) \(\dfrac{1}{4}\times\dfrac{5}{12}+\dfrac{5}{12}\times\dfrac{4}{5}=\dfrac{5}{12}\times\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}\times\dfrac{21}{20}=\dfrac{105}{240}=\dfrac{7}{16}\)
a) \(\dfrac{11}{10}+\dfrac{3}{5}x\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)
b) \(\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)
c) \(\dfrac{29}{35}x\dfrac{25}{29}=\dfrac{5}{7}\)
\(=\dfrac{5}{12}x\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}x\dfrac{21}{20}=\dfrac{7}{16}\)
1. Tính : \(\dfrac{12}{4-\sqrt{10}}\)-6\(\sqrt{\dfrac{5}{2}}\)+\(\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}\)
2,Rút gọn:A=(\(\dfrac{\sqrt{x}}{\sqrt{x}-5}\)-\(\dfrac{5}{\sqrt{x}+5}\)+\(\dfrac{10\sqrt{x}}{25-x}\)):\(\dfrac{3}{\sqrt{x}+5}\)
1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)
\(\dfrac{-29}{12}+1+\dfrac{19}{-12}\)\(\le\)x\(\le\dfrac{-1}{3}+\dfrac{3}{4}+\dfrac{7}{12}\)
\(\dfrac{-5}{3}+1+\dfrac{1}{-3}\le x\le\dfrac{8}{10}+\dfrac{1}{5}+2\)
Tìm x, biết:
a) \(\dfrac{-1}{10}\) + \(\dfrac{2}{5}\)x + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)
b) \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\) : x= \(-\dfrac{1}{5}\)
c) \(-\dfrac{2}{3}\) : x + \(\dfrac{5}{8}\) = \(-\dfrac{7}{12}\)
a, - \(\dfrac{1}{10}\) + \(\dfrac{2}{5}\)\(x\) + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\)\(x\) = \(\dfrac{1}{10}\) - \(\dfrac{7}{20}\) + \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\) \(x\) = - \(\dfrac{3}{20}\)
\(x\) = - \(\dfrac{3}{20}\): \(\dfrac{2}{5}\)
\(x\) = - \(\dfrac{3}{8}\)
b, \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{8}{15}\)
\(x\) = \(\dfrac{1}{2}\): (- \(\dfrac{8}{15}\))
\(x\) = - \(\dfrac{15}{16}\)
c, - \(\dfrac{2}{3}\): \(x\) + \(\dfrac{5}{8}\) = - \(\dfrac{7}{12}\)
\(\dfrac{2}{3}\): \(x\) = \(\dfrac{7}{12}\) + \(\dfrac{5}{8}\)
\(\dfrac{2}{3}\) : \(x\) = \(\dfrac{29}{24}\)
\(x\) = \(\dfrac{2}{3}\) : \(\dfrac{29}{24}\)
\(x\) = \(\dfrac{16}{29}\)
1,Tính \(\dfrac{12}{4-\sqrt{10}}-6\sqrt{\dfrac{5}{2}}+\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}\)
2,Rút gọn:A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{5}{\sqrt{x}+5}+\dfrac{10\sqrt{x}}{25-x}\right):\dfrac{3}{\sqrt{x}+5}\)
1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)
Tìm x, y, z, t ∈ Z biết:
a, \(\dfrac{5}{x}=\dfrac{-10}{12}\) b, \(\dfrac{4}{-6}=\dfrac{x+3}{9}\) c, \(\dfrac{x-1}{25}=\dfrac{4}{x-1}\) d, \(\dfrac{x+1}{y}=\dfrac{-3}{5}\)
e, \(\dfrac{-12}{6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{Z}{-17}=\dfrac{-t}{-9}\)
h, \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{Z^3}{-2}\)
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)
a
ĐK: \(x\ne5\)
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\\ \Leftrightarrow\dfrac{x-5}{3}=\dfrac{12}{x-5}\\ \Leftrightarrow\left(x-5\right)^2=12.3=36\\ \Leftrightarrow\left\{{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=11\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
b
ĐK: \(x\ne0;x\ne-1\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\\ \Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{x}.\dfrac{1}{x+1}\right)=\dfrac{2023}{2024}\\ \Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2023}{2024}\\ \Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2023}{4048}\\ \Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2023}{4048}=\dfrac{1}{4048}\\ \Leftrightarrow4048=x+1\\ \Leftrightarrow x=4047\left(tm\right)\)
a: =>(x-5)/3=12/(x-5)
=>(x-5)^2=36
=>x-5=6 hoặc x-5=-6
=>x=11 hoặc x=-1
b: =>\(2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2023}{2024}\)
=>1/2-1/3+1/3-1/4+...+1/x-1/x+1=2023/4048
=>1/2-1/x+1=2023/4048
=>1/(x+1)=1/4048
=>x+1=4048
=>x=4047
\(\dfrac{4}{5}+\dfrac{1}{10}\)
\(\dfrac{7}{12}\) \(-\) \(\dfrac{1}{4}\)
5 x \(\dfrac{4}{15}\)
\(\dfrac{2}{5}:\dfrac{6}{5}\)
a: 4/5+1/10=8/10+1/10=9/10
b: 7/12-1/4=7/12-3/12=4/12=1/4
c: 5x4/15=20/15=4/3
`4/5+1/10=8/10+1/10=9/10`
`7/12-1/4=7/12-3/12=4/12=1/3`
`5xx4/15=[5xx4]/[3xx5]=4/3`
`2/5:6/5=2/5xx5/6=[2xx5]/[5xx2xx3]=1/3`
\(\dfrac{4}{5}+\dfrac{1}{10}=\dfrac{8}{10}+\dfrac{1}{10}=\dfrac{9}{10}\)
\(\dfrac{7}{12}-\dfrac{1}{4}=\dfrac{7}{12}-\dfrac{3}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)
\(5\times\dfrac{4}{15}=\dfrac{5\times4}{15}=\dfrac{20}{15}=\dfrac{4}{3}\)
\(\dfrac{2}{5}:\dfrac{6}{5}=\dfrac{2}{5}\times\dfrac{5}{6}=\dfrac{10}{30}=\dfrac{1}{3}\)
Tìm X,biết:
a) \(X+X+\dfrac{1}{2}x\dfrac{2}{5}+X+\dfrac{8}{10}=121\) b) \(\dfrac{12+x}{42}=\dfrac{5}{6}\)
`#3107.101107`
a)
\(x+x+\dfrac{1}{2}\times\dfrac{2}{5}+x+\dfrac{8}{10}=121\\3x+\dfrac{1}{5}+\dfrac{4}{5}=121\\ 3x+1=121\\ 3x=121-1\\ 3x=120\\ x=40 \)
Vậy, `x = 40`
b)
\(\dfrac{12+x}{42}=\dfrac{5}{6}\\ \dfrac{12+x}{42}=\dfrac{35}{42}\\ \dfrac{12+x}{42}-\dfrac{35}{42}=0\\ \dfrac{12+x-35}{42}=0\\ \dfrac{x-\left(35-12\right)}{42}=0\\ \dfrac{x-23}{42}=0\\ x-23=0\\ x=23\)
Vậy,` x = 23.`
a: \(x+x+\dfrac{1}{2}\cdot\dfrac{2}{5}+x+\dfrac{8}{10}=121\)
=>\(3x+\dfrac{1}{5}+\dfrac{4}{5}=121\)
=>3x+1=121
=>3x=120
=>x=40
b: \(\dfrac{x+12}{42}=\dfrac{5}{6}\)
=>\(x+12=42\cdot\dfrac{5}{6}=35\)
=>x=35-12=23
Tìm x, biết:
a) \(\dfrac{1}{20}\) - (x - \(\dfrac{8}{5}\)) = \(\dfrac{1}{10}\)
b) \(\dfrac{7}{4}\) - (x + \(\dfrac{5}{3}\)) = \(\dfrac{-12}{5}\)
c) x - [\(\dfrac{17}{2}\) - \(\left(\dfrac{-3}{7}+\dfrac{5}{3}\right)\)] = \(\dfrac{-1}{3}\)
a) 1/20 - (x - 8/5) = 1/10
x - 8/5 = 1/20 - 1/10
x - 8/5 = -1/20
x = -1/20 + 8/5
x = 31/20
b) 7/4 - (x + 5/3) = -12/5
x + 5/3 = 7/4 + 12/5
x + 5/3 = 83/20
x = 83/20 - 5/3
x = 149/60
c) x - [17/2 - (-3/7 + 5/3)] = -1/3
x - (17/2 - 26/21) = -1/3
x - 305/42 = -1/3
x = -1/3 + 305/42
x = 97/14