Tìm x
\(\frac{2\cdot x}{3}=\frac{10}{15}\)
K
Khách
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JN
30 tháng 7 2018
\(x.\left(\frac{1}{6}.\frac{72}{10}+\frac{13}{10}+\frac{1}{2}\right)+15=19.75\)
\(\Leftrightarrow x.\left(\frac{6}{5}+\frac{13}{10}+\frac{1}{2}\right)=4,75\)
\(\Leftrightarrow x.3=4,75\) \(\Rightarrow x=1,583\)
Ủa mà có bài thì tự đi mà làm bài này có khó lắm đâu
NV
1
TL
15 tháng 3 2019
\(x\times\frac{6}{25}=\frac{15}{-13}\)
x=\(\frac{15}{-13}\div\frac{6}{25}\)
x=\(-\frac{125}{26}\)
các câu còn lại làm tương tự nha!!!
KN
15 tháng 3 2019
\(1.x.\frac{6}{25}=\frac{15}{-13}\\ x=\frac{15}{-13}:\frac{6}{25}\\ x=-\frac{125}{26}\)
\(2.x:\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\\ x:\frac{4}{10}=\frac{11}{45}\\ x=\frac{11}{45}.\frac{4}{10}\\ x=\frac{22}{225}\)
\(3.\frac{3}{8}-\frac{1}{6}.x=\frac{1}{4}\\ \frac{1}{6}.x=\frac{3}{8}-\frac{1}{4}\\ \frac{1}{6}.x=\frac{1}{8}\\ x=\frac{1}{8}:\frac{1}{6}\\ x=\frac{3}{4}\)
\(4.\frac{1}{3}+\frac{1}{2}:x=-4\\ \frac{1}{2}:x=-4-\frac{1}{3}=-\frac{13}{3}\\ x=\frac{1}{2}:\left(-\frac{13}{3}\right)=-\frac{3}{26}\)
\(5.x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}=\frac{5}{6}\\ x=\frac{5}{6}-\frac{7}{12}\\ x=\frac{1}{4}\)
21 tháng 9 2017
Theo đề ta có :
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}+\frac{\left(x+10\right)-\left(x+5\right)}{\left(x+5\right)\left(x+10\right)}+\frac{\left(x+17\right)-\left(x+10\right)}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{\left(x+17\right)-\left(x+2\right)}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\left(x+17\right)-\left(x+2\right)=x\)
\(\Rightarrow x=15\)
20 tháng 5 2018
a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)
Do đó \(x\in\left\{0;1;2\right\}\)
25 tháng 7 2018
b)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
KN
5 tháng 8 2020
\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right):\left(1+8+15+...+281+288\right)=\frac{1}{3}\)
\(\Rightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right):\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)
\(\Rightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right):6069=\frac{1}{3}\)
\(\Rightarrow\frac{1488x-279+279-476,5x}{10}=2023\)
\(\Rightarrow1488x-476,5x=20230\)
\(\Rightarrow1011,5x=20230\)
\(\Rightarrow x=20\)
B
16 tháng 9 2020
Bài làm :
Ta có :
\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right)\div\left(1+8+15+...+281+288\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right)\div\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right)\div6069=\frac{1}{3}\)
\(\Leftrightarrow\frac{1488x-279+279-476,5x}{10}=2023\)
\(\Leftrightarrow1488x-476,5x=20230\)
\(\Leftrightarrow1011,5x=20230\)
\(\Leftrightarrow x=20\)
Vậy x=20
\(\text{Tìm }x\text{ }:\)
\(\frac{2\text{ x }x}{3}=\frac{10}{15}\)
\(\frac{2\text{ x }x}{3}=\frac{2}{3}\)
\(\frac{2\text{ x }x}{3}=\frac{2}{3}\text{ }\Leftrightarrow\text{ }x=1\)
\(\text{Chúc bạn học tốt !}\)
\(\frac{2x}{3}=\frac{10}{15}\)
\(\Rightarrow2\cdot x\cdot15=3\cdot10\)
\(\Rightarrow30x=30\)
\(\Rightarrow x=1\)
Vậy x=1