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\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
a) \(\frac{2}{5}x-x=\frac{\left(-2018\right)^0}{5^2}\\ x\left(\frac{2}{5}-1\right)=\frac{1}{25}\\ x\left(\frac{2}{5}-\frac{5}{5}\right)=\frac{1}{25}\\ x\cdot\frac{-3}{5}=\frac{1}{25}\\ x=\frac{1}{25}:\frac{-3}{5}\\ x=\frac{1}{25}\cdot\frac{-5}{3}\\ x=\frac{-1}{15}\)Vậy \(x=\frac{-1}{15}\)
b) \(\left|-1\frac{1}{2}x+2x\right|-\frac{7}{4}=0,5\\ \left|x\left(-1\frac{1}{2}+2\right)\right|-\frac{7}{4}=\frac{1}{2}\\ \left|x\cdot\frac{1}{2}\right|=\frac{1}{2}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{2}{4}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x\cdot\frac{1}{2}=\frac{9}{4}\\x\cdot\frac{1}{2}=\frac{-9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}:\frac{1}{2}\\x=\frac{-9}{4}:\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}\cdot2\\x=\frac{-9}{4}\cdot2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=\frac{-9}{2}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{9}{2};\frac{-9}{2}\right\}\)
c) \(x+\left(x+\frac{2}{7}\right)+\frac{-5}{11}=\frac{4}{11}\\ x+x+\frac{2}{7}=\frac{4}{11}-\frac{-5}{11}\\ 2x+\frac{2}{7}=\frac{4}{11}+\frac{5}{11}\\ 2x+\frac{2}{7}=\frac{9}{11}\\ 2x=\frac{9}{11}-\frac{2}{7}\\ 2x=\frac{63}{77}-\frac{22}{77}\\ 2x=\frac{41}{77}\\ x=\frac{41}{77}:2\\ x=\frac{41}{77\cdot2}\\ x=\frac{41}{154}\)Vậy \(x=\frac{41}{154}\)
d) \(\left|0,25x-20\%\right|+\frac{3}{8}=1\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\frac{3}{8}-\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\\ \Rightarrow\left[{}\begin{matrix}\frac{1}{4}x-\frac{1}{5}=1\\\frac{1}{4}x-\frac{1}{5}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=1+\frac{1}{5}\\\frac{1}{4}x=\left(-1\right)+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{5}{5}+\frac{1}{5}\\\frac{1}{4}x=\frac{-5}{5}+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{6}{5}\\\frac{1}{4}x=\frac{-4}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}:\frac{1}{4}\\x=\frac{-4}{5}:\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}\cdot4\\x=\frac{-4}{5}\cdot4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{24}{5}\\x=\frac{-16}{5}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{24}{5};\frac{-16}{5}\right\}\)
mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn
a)\(\frac{5}{6}-x=-\frac{7}{12}+\frac{2}{3}\)
\(\frac{5}{6}-x=\frac{1}{12}\)
\(x=\frac{5}{6}-\frac{1}{12}\)
\(\Rightarrow x=\frac{3}{4}\)
b)\(\left(2,4x-36\right):1\frac{5}{7}=-14\)
\(\left(2,4x-36\right)=-24\)
\(2,4x=12\)
\(\Rightarrow x=5\)
c)\(\left(3\frac{1}{2}+2x\right).3\frac{2}{3}=5\frac{1}{3}\)
\(3\frac{1}{2}+2x=\frac{16}{11}\)
\(2x=-\frac{45}{22}\)
\(x=-\frac{45}{44}\)
d)\(\frac{5}{6}-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{3}{8}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{11}{24}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{11}{24}\\\frac{1}{2}x-\frac{1}{3}=-\frac{11}{24}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{19}{12}\\x=-\frac{1}{4}\end{cases}}\)
e)\(\left|\frac{1}{4}-2x\right|-\frac{3}{4}=0\)
\(\left|\frac{1}{4}-2x\right|=\frac{3}{4}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{4}-2x=\frac{3}{4}\\\frac{1}{4}-2x=-\frac{3}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{2}\end{cases}}\)
a) \(\frac{x}{5}=\frac{2}{3}\)
\(\Rightarrow\)\(x=\frac{2.5}{3}=\frac{10}{3}\)
Vậy....
b) \(\frac{x+3}{15}=\frac{1}{5}\)
\(\Leftrightarrow\)\(5\left(x+3\right)=15\)
\(\Leftrightarrow\)\(x+3=3\)
\(\Leftrightarrow\)\(x=0\)
Vậy....
a,
3x + 3 - [7x+4] = 7 + [4x-1]
=> 3x + 3 - x - 4 = 7 + 4x - 1
=> 2x - 1 = 6 + 4x
=> 2x - 4x = 6 + 1
=> -2x = 7
=> x = -7/2
b,
3x+1 + 3x+3 =810
=> 3x+1[1 + 32] = 810
=> 3x+1 = 810 / 10
=> 3x+1 = 81
=> x = 4
c, \(1\frac{1}{2}:\left[\frac{1}{2}-\frac{1}{3}\right]-x=5\)
\(\Rightarrow\frac{3}{2}:\frac{1}{6}-x=5\Leftrightarrow9-x=5\)
\(\Leftrightarrow x=4\)
d,
\(2,4:\left[25\%+\frac{x}{40}\right]-\frac{12}{15}=3\frac{1}{5}\)
\(\Rightarrow\frac{12}{5}:\left[\frac{1}{4}+\frac{x}{40}\right]-\frac{12}{15}=\frac{16}{5}\)
\(\Leftrightarrow\frac{12}{5}:\left[\frac{10}{40}+\frac{x}{40}\right]=\frac{16}{5}+\frac{12}{15}\Leftrightarrow\frac{12}{5}:\left[\frac{10}{40}+\frac{x}{40}\right]=4\)
\(\Rightarrow\frac{10+x}{40}=\frac{12}{5}:4\Leftrightarrow\frac{10+x}{40}=\frac{3}{5}\)
\(\Rightarrow\frac{10+x}{40}=\frac{24}{40}\Leftrightarrow10+x=24\Rightarrow x=14\)
a) 3x + 3 - ( x + 4 ) = 7 + ( 4x - 1 )
3x + 3 - x - 4 = 7 + 4x - 1
2x - 1 = 6 + 4x
-2x = 7
\(\Rightarrow\)x = \(\frac{-7}{2}\)
b) 3x+1 + 3x+3 = 810
3x . 3 + 3x . 33 = 810
3x . ( 3 + 33 ) = 810
3x . 30 = 810
3x = 810 : 30
3x = 27
3x = 33
\(\Rightarrow\)x = 3
c) \(1\frac{1}{2}:\left(\frac{1}{2}-\frac{1}{3}\right)-x=5\)
\(\frac{3}{2}:\left(\frac{1}{2}-\frac{1}{3}\right)-x=5\)
\(\frac{3}{2}:\frac{1}{6}-x=5\)
\(9-x=5\)
\(\Rightarrow x=9-5\)
\(\Rightarrow x=4\)
d) 2,4 : ( 25% + \(\frac{x}{40}\)) - \(\frac{12}{15}\)= \(3\frac{1}{5}\)
\(\frac{12}{5}\) : ( \(\frac{1}{4}\)+ \(\frac{x}{40}\)) - \(\frac{12}{15}\)= \(\frac{16}{5}\)
\(\frac{12}{5}:\left(\frac{1}{4}+\frac{x}{40}\right)=\frac{16}{5}+\frac{12}{15}\)
\(\frac{12}{5}:\left(\frac{1}{4}+\frac{x}{40}\right)=4\)
\(\frac{1}{4}+\frac{x}{40}=\frac{12}{5}:4\)
\(\frac{1}{4}+\frac{x}{40}=\frac{3}{5}\)
\(\frac{x}{40}=\frac{3}{5}-\frac{1}{4}\)
\(\frac{x}{40}=\frac{7}{20}\)
\(\Rightarrow\frac{x}{40}=\frac{14}{40}\)
\(\Rightarrow x=14\)