X x \(\frac{11}{5}\)- X x \(\frac{4}{7}\) = \(\frac{5}{77}\)
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T
24 tháng 11 2016
Xx11/5-Xx4/7=5/77
Xx(11/5-4/7)=5/77
Xx57/35=5/77
x=5/77:57/35=25/627
9 tháng 4 2019
1, Tính tổng:
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)=\frac{5}{7}\cdot\frac{-7}{11}=\frac{-5}{11}\)
2, Tìm x:
\(x+\frac{5}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\Rightarrow x+\frac{9}{45}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{8}{45}=\frac{-37}{45}\Rightarrow x=\frac{-37}{45}-\frac{8}{45}=\frac{-45}{45}=-1\)
- Các bài tìm x còn lại bạn cứ theo trình tự thực hiện phép tính mà làm nhé!
BA
9 tháng 4 2019
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)
\(=\frac{5}{7}\cdot-\frac{7}{11}\)
\(=-\frac{5}{11}\)
VM
3 tháng 2 2020
g) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2+98}{98}\right)+\left(\frac{x+4+96}{96}\right)=\left(\frac{x+6+94}{94}\right)+\left(\frac{x+8+92}{92}\right)\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0.\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=0-100\)
\(\Leftrightarrow x=-100.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-100\right\}.\)
h) \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\Leftrightarrow\left(\frac{x-12}{77}-1\right)+\left(\frac{x-11}{78}-1\right)=\left(\frac{x-74}{15}-1\right)+\left(\frac{x-73}{16}-1\right)\)
\(\Leftrightarrow\left(\frac{x-12-77}{77}\right)+\left(\frac{x-11-78}{78}\right)=\left(\frac{x-74-15}{15}\right)+\left(\frac{x-73-16}{16}\right)\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right).\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
Vì \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\ne0.\)
\(\Leftrightarrow x-89=0\)
\(\Leftrightarrow x=0+89\)
\(\Leftrightarrow x=89.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{89\right\}.\)
Chúc bạn học tốt!
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
\(x\times\frac{11}{5}-x\times\frac{4}{7}=\frac{5}{77}\)
\(x\times\left(\frac{11}{5}-\frac{4}{7}\right)=\frac{5}{77}\)
\(x\times\frac{57}{35}=\frac{5}{77}\)
\(x=\frac{5}{77}:\frac{57}{35}\)
\(x=\frac{25}{627}\)
\(x\times\frac{11}{5}-x\times\frac{4}{7}=\frac{5}{77}\)
\(x\times\left(\frac{11}{5}-\frac{4}{7}\right)=\frac{5}{77}\)
\(x\times\frac{57}{35}=\frac{5}{77}\)
\(x=\frac{5}{77}:\frac{57}{35}\)
\(x=\frac{25}{627}\)