Bài 1. tìm x, biết :
\(\dfrac{X}{2}+\dfrac{X}{14}+\dfrac{X}{35}+\dfrac{X}{65}+\dfrac{X}{104}+\dfrac{X}{209}=1\dfrac{3}{11}\)
Tìm x biết: \(\dfrac{13}{14}+\dfrac{13}{35}+\dfrac{13}{65}+\dfrac{13}{104}+...+\dfrac{13}{1274}-x+2\dfrac{1}{5}=0\)
MỌI NGƯỜI GIÚP EM VỚI
Bài 1: tìm x
a)\(\left|3x-5\right|=4\)
b)\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
c)\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
Bài 2: Tính
a)\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
b)\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
c)\(\dfrac{\left(\dfrac{3}{10}-\dfrac{4}{15}-\dfrac{7}{20}\right).\dfrac{5}{19}}{\left(\dfrac{1}{14}+\dfrac{1}{7}-\dfrac{-3}{35}\right).\dfrac{-4}{3}}\)
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))
Bài 2:
a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)
b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)
\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)
\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)
Bài 1:
a) \(\left|3x-5\right|=4\) (1)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
Tìm x:
a) 2n+3.5n+3=209:29
b)3n+5-3n+4=1458
c)5n+3+5n+2=3750
d)\(\dfrac{2}{7}x\) + \(\dfrac{3}{14}x\) = \(\dfrac{1}{2}\)
e)\(\dfrac{x+2}{-3}\) = \(\dfrac{-2}{x+3}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(2^{n+3}\cdot5^{n+3}=20^9\div2^9\)
`=>`\(\left(2\cdot5\right)^{n+3}=\left(20\div2\right)^9\)
`=>`\(10^{n+3}=10^9\)
`=>`\(n+3=9\)
`=> n = 9 - 3`
`=> n= 6`
Vậy, `n=6`
`b)`
\(3^{n+5}-3^{n+4}=1458\)
`=> 3^n*3^5 - 3^n*3^4 = 1458`
`=> 3^n*(3^5 - 3^4) = 1458`
`=> 3^n*162 = 1458`
`=> 3^n = 1458 \div 162`
`=> 3^n = 9`
`=> 3^n = 3^2`
`=> n=2`
Vậy, `n=2.`
`c)`
\(5^{n+3}+5^{n+2}=3750\)
`=> 5^n*5^3 + 5^n*5^2 = 3750`
`=> 5^n*(5^3+5^2) = 3750`
`=> 5^n*150 = 3750`
`=> 5^n = 3750 \div 150`
`=> 5^n =25`
`=> 5^n = 5^2`
`=> n=2`
Vậy, `n=2.`
`d)`
\(\dfrac{2}{7}x+\dfrac{3}{14}x=\dfrac{1}{2}\)
`=> 1/2x = 1/2`
`=> x = 1/2 \div 1/2`
`=> x=1`
Vậy, `x=1`
`e)`
\(\dfrac{x+2}{-3}=\dfrac{-2}{x+3}\)
`=> (x+2)(x+3) = -3*(-2)`
`=> (x+2)(x+3) = -6`
`=> x(x+3) + 2(x+3) = -6`
`=> x^2 + 3x + 2x + 6 = -6`
`=> x^2 + 5x + 6 - 6 = 0`
`=> x^2 + 5x = 0`
`=> x(x+5) = 0`
`=>`\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy, `x \in {0; -5}`
`@` `\text {Kaizuu lv u}`
Tìm x:
3\(\dfrac{1}{14}\) - 5\(\dfrac{1}{3}\) - \(\dfrac{4}{7}\) + \(\dfrac{11}{21}\) = x -\(\dfrac{1}{2}\)
Làm ra bài giải giúp mk vs
=>x-1/2=3+1/14-5-1/3-4/7+11/21
=>x-1/2=-97/42
=>x-21/42=-97/42
=>x=-97/42+21/42=-76/42=-38/21
\(A.\dfrac{-15}{28}x\dfrac{7}{25}\\ B.\dfrac{-5}{14}x\dfrac{7}{-3}\\ C.\dfrac{-1}{5}-\dfrac{7}{15}x\dfrac{9}{35}\\ D.\dfrac{-3}{4}-(\dfrac{-1}{2})^2\\ E.\dfrac{-4}{5}-\dfrac{-4}{5}x\dfrac{15}{16}\\F.(\dfrac{3}{4}+\dfrac{-7}{2})x(\dfrac{2}{11}+\dfrac{12}{22})\)
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
Tính giá trị của biểu thức : A = \(\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}+\dfrac{1}{209}+\dfrac{1}{275}\)
\(\Leftrightarrow A=\frac{1}{2}+\frac{1}{2\cdot7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\frac{1}{13}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}\)
\(\Rightarrow A=-\frac{1}{19}\)
Tìm x, biết:
a) \(\dfrac{3}{7}\)x - \(\dfrac{2}{3}\)x = \(\dfrac{10}{21}\)
b) \(\dfrac{7}{35}\) : (x - \(\dfrac{1}{3}\)) = \(-\dfrac{2}{25}\)
c) 3.(x - \(\dfrac{1}{2}\)) - 5. (x + \(\dfrac{3}{5}\)) = -x + \(\dfrac{1}{5}\)
a, \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\) = \(\dfrac{10}{21}\)
(\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\) = \(\dfrac{10}{21}\)
- \(\dfrac{5}{21}\) \(\times\) \(x\) = \(\dfrac{10}{21}\)
\(x\) = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))
\(x\) = -2
b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)
\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))
\(x\) - \(\dfrac{1}{3}\) = - \(\dfrac{5}{2}\)
\(x\) = - \(\dfrac{5}{2}\) + \(\dfrac{1}{3}\)
\(x\) = - \(\dfrac{13}{6}\)
c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)
3\(x\) - \(\dfrac{3}{2}\) - 5\(x\) - 3 = - \(x\) + \(\dfrac{1}{5}\)
- \(x\) + 5\(x\) - 3\(x\) = - \(\dfrac{3}{2}\) - 3 - \(\dfrac{1}{5}\)
\(x\) = - \(\dfrac{47}{10}\)
\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)
\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)
a,73x−32x=2110⇒x(73−32)=2110⇒x.−215=2110⇒x=−2b,357:(x−31)=−252⇒51:(x−31)=−252⇒x−31=−25⇒x=−613c,3.(x−21)−5.(x+53)=−x+51⇒3x−23−5x+5=−x+51
Bài 2. Tìm x biết
\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
a) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\)
\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11-8}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{5}{20}-\dfrac{8}{20}\)
\(\Leftrightarrow x=\dfrac{-3}{20}\)
Vậy x= \(\dfrac{-3}{20}\)
b) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{8-15}{20}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Leftrightarrow x=\dfrac{1}{4}.\dfrac{-20}{7}\)
\(\Leftrightarrow x=\dfrac{-5}{7}\)
Vậy x= \(\dfrac{-5}{7}\)
a) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow x+\dfrac{2}{5}=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{11}{12}-\dfrac{8}{12}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{5}{20}-\dfrac{8}{20}=\dfrac{-3}{20}\)
b) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{-7}{20}\)
hay \(x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{1}{4}\cdot\dfrac{-20}{7}=\dfrac{-5}{7}\)
Bài 1: Tìm số nguyên x, biết:
a)\(\dfrac{6}{x-3}\) = \(\dfrac{2}{3}\)
b) \(\dfrac{14}{13}\) = \(\dfrac{-28}{10-x}\)
c) \(\dfrac{1}{5}\) = \(\dfrac{x:4-1}{10}\)
d) \(\dfrac{x}{4}\)= \(\dfrac{1}{x}\)
e) \(\dfrac{x-2}{50}\) = \(\dfrac{2}{x-2}\)
giúp ưm
a: =>x-3=9
=>x=12
b: =>10-x=-26
=>x=36
c: =>x:4-1=2
=>x:4=3
=>x=12
d: =>x^2=4
=>x=2 hoặc x=-2
e: =>(x-2)^2=100
=>x-2=10 hoặc x-2=-10
=>x=12 hoặc x=-8