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do các phân số ở hàng số thứ 2 đã tối giản nên x=0=>7x=0 =>tổng các phân số sau đều tối giản
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\frac{1}{7}x-\frac{1}{2}x+\frac{5}{7}x=-\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{7}-\frac{1}{2}+\frac{5}{7}\right)x=-\frac{1}{2}\)
\(\Rightarrow\left(\frac{2}{14}-\frac{7}{14}+\frac{10}{14}\right)x=-\frac{1}{2}\)
\(\Rightarrow\frac{5}{14}x=-\frac{1}{2}\)
\(\Rightarrow x=-\frac{1}{2}:\frac{5}{14}\)
\(\Rightarrow x=-\frac{1}{2}.\frac{14}{5}\)
\(\Rightarrow x=-\frac{7}{5}\)
\(b)(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{49.51})x=-\frac{1}{3}\)
\(\Rightarrow\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{49}-\frac{1}{51}\right)x=-\frac{1}{3}\)
\(\Rightarrow\left(\frac{1}{11}-\frac{1}{51}\right)x=-\frac{1}{3}\)
\(\Rightarrow\left(\frac{51}{561}-\frac{11}{561}\right)x=-\frac{1}{3}\)
\(\Rightarrow\frac{40}{561}x=-\frac{1}{3}\)
\(\Rightarrow x=-\frac{1}{3}:\frac{40}{561}\)
\(\Rightarrow x=-\frac{1}{3}.\frac{561}{40}\)
\(\Rightarrow x=-\frac{187}{40}\)
Chúc bạn học tốt !!!
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Ta có :
\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\)
\(=>\left(7x+2\right)\left(5x+1\right)=\left(7x-1\right)\left(5x+7\right)\)
\(=>35x^2+7x+10x+2=35x^2+49x-5x-7\)
\(=>35x^2+17x+2=35x^2+44x-7\)
\(=>17x+2=44x-7\)
\(=>44x-17x=2+7\)
\(=>27x=9\)
\(=>x=\frac{9}{27}=\frac{1}{3}\)
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\(1,\Rightarrow3^{x-3}=\left(3^2\right)^8:\left(3^3\right)^5=3^{16}:3^{15}=3^1\\ \Rightarrow x-3=1\\ \Rightarrow x=4\\ 2,\Rightarrow7^x\left(1+7^2\right)=350\\ \Rightarrow7^x=\dfrac{350}{50}=7=7^1\\ \Rightarrow x=1\)
\(3,\Rightarrow2^{2+2x+2}-2^{2x}=240\\ \Rightarrow2^{2x}\left(2^4-1\right)=240\\ \Rightarrow2^{2x}=\dfrac{240}{15}=16=2^4\\ \Rightarrow2x=4\Rightarrow x=2\)
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7x.(2+x)-7x.(x+3)=14
7x.(2+x-x-3)=14
7x.(-1)=14
7x=14:(-1)
7x=-14
x=-14:7
x=2
Vậy x=2
tk nha!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(7\frac{x}{2.5}+7\frac{x}{5.8}+.....+7.\frac{x}{17.20}=\frac{21}{10}\)
\(7\left(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}\right)=\frac{21}{10}\)
\(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}=\frac{21}{70}\)
\(\frac{x.3}{2.5.3}+\frac{x.3}{5.8.3}+...+\frac{x.3}{17.20.3}=\frac{21}{70}\)
\(x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\right)=\frac{21}{70}\)
\(x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{21}{70}\)
\(x.\frac{1}{3}.\frac{9}{20}=\frac{21}{70}\)
=> \(x=2\)
\(x=\frac{7x}{2}\)\(-\frac{7x}{5}+\)\(\frac{7x}{5}\)\(-\frac{7x}{8}\)\(+\frac{7x}{8}\)\(-\frac{7x}{11}\)\(+\frac{7x}{11}\)\(-\frac{7x}{14}\)\(+\frac{7x}{14}\)\(-\frac{7x}{17}+\)\(\frac{7x}{17}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x}{2}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.10}{20}\)\(+\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.10+7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.\left(10+2\right)}{20.2}\)\(=\frac{7x.12}{40}\)\(=\frac{21}{10}\)
\(=>\frac{7x.12:4}{40:4}=\)\(\frac{21}{10}\)
\(=>x=1\)
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1.
2|x-6|+7x-2=|x-6|+7x
2|x-6| - |x-6|=7x-(7x-2)
|x-6| = 2
=>x-6 = +2
*x-6=2 *x-6 = -2
x =2+6 x = (-2)+6
x =8 x = 4
2.
|x-5|-7(x+4)=5-7x
|x-5|-7x-28 =5-7x
|x-5|-28 =5-7x+7x
|x-5|-28 = 5
|x-5| = 5+28
|x-5| = 33
=>x-5 = +33
*x-5=33 *x-5=-33
x =38 x = -28
3.
3|x+4|-2(x-1)=7-2x
3|x+4|-2x+2 =7-2x
3|x+4|-2 =7-2x+2x
3|x+4|-2 =7
3|x+4| =7+2
3|x+4| = 9
|x+4| =9:3
|x+4| = 3
=>x+4 = +3
*x+4=3 *x+4=-3
x =-1 x = -7
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<=>3x+x+1+x+2<1053
<=> 33x+3<1053
Ta thấy 36=729; 37=2187
=> 33x+3=36
=> 3x+3=6
<=> 3x=3
<=> x=1