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11 tháng 5 2022
Bài 3:
\(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{1\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}=\dfrac{2}{3}+\dfrac{1}{3}=1\)
PT
28 tháng 9 2017
a) \(\dfrac{x+5}{5}+\dfrac{x+5}{7}+\dfrac{x+5}{9}=\dfrac{x+5}{11}+\dfrac{x+5}{13}\)
\(\Rightarrow\left(x+5\right)\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)=\left(x+5\right)\left(\dfrac{1}{11}+\dfrac{1}{13}\right)\)
\(\Rightarrow\dfrac{143}{315}\left(x+5\right)=\dfrac{24}{143}\left(x+5\right)\)
\(\Rightarrow\dfrac{143}{315}\left(x+5\right)-\dfrac{24}{143}\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(\dfrac{143}{315}-\dfrac{24}{143}\right)=0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
b) \(\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(3+\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=3+\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(1+\dfrac{x+2}{100}+1+\dfrac{x+3}{99}+1+\dfrac{x+4}{98}=1+\dfrac{x+5}{97}+1+\dfrac{x+6}{96}+1+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(\dfrac{100}{100}+\dfrac{x+2}{100}+\dfrac{99}{99}+\dfrac{x+3}{99}+\dfrac{98}{98}+\dfrac{x+4}{98}=\dfrac{97}{97}+\dfrac{x+5}{97}+\dfrac{96}{96}+\dfrac{x+6}{96}+\dfrac{95}{95}+\dfrac{x+7}{95}\)\(\Rightarrow\)\(\dfrac{x+102}{100}+\dfrac{x+102}{99}+\dfrac{x+102}{98}=\dfrac{x+102}{97}+\dfrac{x+102}{96}+\dfrac{x+102}{95}\)
\(\Rightarrow\)\(\left(x+102\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{98}\right)=\left(x+102\right)\left(\dfrac{1}{97}+\dfrac{1}{96}+\dfrac{1}{95}\right)\)
\(\Rightarrow\)\(x+102=0\)
\(\Rightarrow x=-102\)
PT
28 tháng 9 2017
c) \(\left(x+2\right)-\left(x+3\right)>0\)
\(\Rightarrow x+2-x-3>0\Rightarrow-1>0\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)\left(x+\dfrac{7}{3}\right)\ge0\)
TH1: \(\left\{{}\begin{matrix}x-5\ge0\\x+\dfrac{7}{3}\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\ge\dfrac{-7}{3}\end{matrix}\right.\)
\(\Rightarrow x\ge\dfrac{-7}{3}\)
TH2: \(\left\{{}\begin{matrix}x-5\le0\\x+\dfrac{7}{3}\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le5\\x\le\dfrac{-7}{3}\end{matrix}\right.\)
\(\Rightarrow x\le5\)
TH3: \(\left[{}\begin{matrix}x-5=0\\x+\dfrac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)
28 tháng 9 2017
a) Ta có : \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\left(\frac{x+5}{11}+\frac{x+5}{13}\right)=0\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\frac{x+5}{11}-\frac{x+5}{13}=0\)
\(\Rightarrow\left(x+5\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\right)=0\)
Do \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\ne0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
Vậy x = -5
b) Ta có : \(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)
\(\Rightarrow\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}+3=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}+3\)
\(\Rightarrow\frac{x+2}{100}+1+\frac{x+3}{99}+1+\frac{x+4}{98}+1=\frac{x+5}{97}+1+\frac{x+6}{96}+1+\frac{x+7}{95}+1\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}=\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\left(\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\right)=0\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\frac{x+102}{97}-\frac{x+102}{96}-\frac{x+102}{95}\)
\(\Rightarrow\left(x+102\right)\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Do \(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
\(\Rightarrow x+102=0\Rightarrow x=-102\)
Vậy x = -102
c) Ta có : (x + 2) - (x + 3) = x + 2 - x - 3
= x - x + 2 - 3
= -1
mà (x + 2) - (x + 3) > 0 => không tồn tại x sao cho (x + 2) - (x + 3) > 0
d) Ta có : \(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)
\(\Rightarrow\orbr{\begin{cases}x\ge5\\x\ge\frac{-7}{3}\end{cases}}\)
\(\Rightarrow x\ge\frac{-7}{3}\)
Vậy \(x\ge\frac{-7}{3}\)
6 tháng 2 2017
X=3/7:5/7-3/11:5/11+3/13:5/13
+
1/2:5/4-1/3+1/4:5/6
=3/5-3/5+3/5 + 2/5-1/3+3/10
=3/5 + 11/10
=17/10
TH
22 tháng 9 2018
a) \(\left(\frac{1}{7}x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\)TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\) TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\) TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\)
\(\frac{1}{7}x=\frac{2}{7}\) \(-\frac{1}{5}x=\frac{3}{5}\) \(\frac{1}{3}x=\frac{4}{3}\)
\(x=\frac{2}{7}\cdot7\) \(x=\frac{3}{5}\cdot-5\) \(x=\frac{4}{3}\cdot3\)
\(x=2\) \(x=-3\) \(x=4\)
Vậy x = 2 hoặc x = -3 hoặc x = 4
b) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(x\cdot\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{5}\right)=1\)
\(x\cdot\frac{5+3-24}{30}=1\)
\(x\cdot\frac{-8}{15}=1\)
\(x=1\cdot\frac{-15}{8}=\frac{-15}{8}\)
Vậy x = \(\frac{-15}{8}\)
23 tháng 7 2017
\(3x.\left(x-\frac{2}{3}\right)=0\)
\(\Leftrightarrow3x=0\)hoặc \(x-\frac{2}{3}=0\)
\(3x=0\Rightarrow x=0\)
\(x-\frac{2}{3}=0\Rightarrow x=0+\frac{2}{3}=\frac{2}{3}\)
Vậy..
NN
4
21 tháng 6 2016
\(\frac{10}{3}:\frac{5}{2}=\frac{10}{3}.\frac{2}{5}=\frac{20}{15}=\frac{3}{4}\)
x = 2,(142857) = 15/7
xin tiick
x=15/7