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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 \(x\in\left\{\frac{24}{5};\frac{-16}{5}\right\}\)
a) \(1\frac{5}{8}:x-1\frac{1}{4}=2\)
=> \(\frac{13}{8}:x-\frac{5}{4}=2\)
=> \(\frac{13}{8}:x=\frac{13}{4}\)
=> \(x=\frac{13}{8}:\frac{13}{4}=\frac{13}{8}\cdot\frac{4}{13}=\frac{1}{2}\)
b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{2}=0\)
=> \(\left(x-\frac{1}{3}\right)^2=\frac{1}{2}\)
=> x không thỏa mãn
c) 2(x - 1) = 3x + 1
=> 2x - 2 = 3x + 1
=> 2x - 2 - 3x - 1 = 0
=> 2x - 3x - 2 - 1 = 0
=> -x = 3
=> x = -3
d) \(\frac{-2}{x}=\frac{x}{-8}\)=> x2 = 16 => x = \(\pm\)4
e) |7 - x| + 2x = 11
=> |7 - x| = 11 - 2x
=> 7 - x = 11 - 2x
=> 7 - x - 11 + 2x = 0
=> 7 - 11 - x + 2x = 0
=> -4 + x = 0
=> -4 = -x
=> x = 4
a)=> 13/8 : x-5/4 =2
<=> 13/8 : x= 13/4
<=> x=1/2
b)=>x2 - (1/3)2 = 1/2
=>x2-1/9=1/2
=>x2=11/18
=>x=0.78 (bằng xấp xỉ thôi nhé bạn :33)
c)=>2x-2=3x-1
=>2x-3x=2-1
=>-x=1
=>x=-1
còn 2 câu bạn làm nốt nhé :33
k đúng cho mk nhé!!!!
\(\frac{1}{2}+\frac{5}{6}-\frac{3}{8}\)
\(=\frac{12}{24}+\frac{20}{24}-\frac{9}{24}\)
\(=\frac{23}{24}\)
\(\frac{10}{15}\cdot\frac{7}{4}+\frac{7}{4}\cdot\frac{9}{15}-\frac{4}{15}\cdot\frac{7}{4}\)
\(=\frac{7}{4}\cdot\left(\frac{10}{15}+\frac{9}{15}-\frac{4}{15}\right)\)
\(=\frac{7}{4}\cdot1=\frac{7}{4}\)
\(\left(\frac{4}{5}+\frac{1}{2}\right)\left(\frac{3}{13}-\frac{8}{13}\right)\)
\(=\left(\frac{8}{10}+\frac{5}{10}\right)\cdot\left(-\frac{5}{13}\right)\)
\(=\frac{13}{10}\cdot\left(-\frac{5}{13}\right)=-\frac{1}{2}\)
a, 3x-12 = 30
=> 3x = 30 + 12
=> 3x = 42
=> x = 42 : 3 = 14
Vậy x = 14
b, \(\frac{2}{3}x+\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{7}{12}-\frac{1}{4}\)
\(\Rightarrow\frac{2}{3}x=\frac{7}{12}-\frac{3}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}\div\frac{2}{3}\Rightarrow\frac{1}{3}\cdot\frac{3}{2}=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
c, 2x2 = 32
=> x2 = 32 : 2
=> x2 = 16
=> x2 = 42
=> x = 4
Vậy x = 4
câu 1 :
\(A=\frac{-7}{12}:\frac{49}{11}\cdot\frac{5}{121}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{8}{7}:8-3:\frac{3}{4}\cdot-2^3\)
\(A=\frac{-11}{84}\cdot\frac{5}{121}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{8}{7}:\frac{8}{1}-\frac{3}{1}:\frac{3}{4}\cdot\left(-2^3\right)\)
\(A=\frac{-5}{924}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{1}{7}-\left(-32\right)\)
\(A=\frac{-361}{308}\) \(B=\frac{-1}{56}-\left(-32\right)\)
\(B=\frac{1791}{56}\)
Câu 2 :
a)\(\frac{22}{7}:x=\frac{11}{7}\) b)\(\left(1-3x\right)\cdot\frac{4}{3}=-2^3\)
\(x=\frac{22}{7}:\frac{11}{7}\) \(\left(1-3x\right)\cdot\frac{4}{3}=-8\)
\(x=2\) \(\left(1-3x\right)=-8:\frac{4}{3}\)
\(\left(1-3x\right)=-6\)
\(3x=-6-1=7\)
\(3x=7:3=\frac{7}{3}\)
c ) bằng \(\frac{27}{5}\)nhé