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B
1
13 tháng 3 2018
a) \(\frac{1}{9}+3,25+5\frac{3}{16}+4\frac{1}{3}+2,8+0,5=\frac{1}{9}+\frac{13}{4}+\frac{83}{16}+\frac{13}{3}+\frac{14}{5}+\frac{1}{2}\)
\(=\frac{11651}{720}\)
B) \(2\frac{1}{3}+0,45+4,25+\frac{1}{81}+6\frac{8}{27}=\frac{7}{3}+\frac{9}{20}+\frac{17}{4}+\frac{1}{81}+\frac{170}{27}\)
\(=\frac{10807}{810}\)
C) \(1,25+2\frac{1}{4}+4\frac{2}{5}+0,3+2,14+4\frac{1}{8}=\frac{5}{4}+\frac{9}{4}+\frac{22}{5}+\frac{3}{10}+\frac{107}{50}+\frac{33}{8}\)
\(=\frac{2893}{200}\)
CHÚC BN HỌC TỐT!!!!!
27 tháng 6 2019
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
23 tháng 8 2017
B=\(\frac{1}{3}+\frac{-0,175+\frac{5}{11}}{0,2625-\frac{15}{22}}\)
B=\(\frac{1}{3}+\frac{\frac{123}{440}}{-\frac{369}{880}}\)
B=\(0,2161148416\)
23 tháng 8 2017
Thay x vào ta có:
\(B=\frac{1}{3}+\frac{0,2-0,375+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{22}}\)
\(B=\frac{1}{3}+\frac{\frac{2}{10}-\frac{375}{1000}+\frac{5}{11}}{-\frac{3}{10}+\frac{9}{16}-\frac{15}{22}}\)
\(B=\frac{1}{3}+\frac{\frac{1}{5}-\frac{3}{8}+\frac{5}{11}}{-\frac{3}{10}+\frac{9}{16}-\frac{15}{22}}\)
\(B=\frac{1}{3}+\frac{\frac{123}{440}}{\frac{-369}{880}}\)
\(B=\frac{1}{3}+\frac{123}{440}.\frac{880}{-369}\)
\(B=\frac{1}{3}+\frac{-2}{3}\)
\(B=\frac{-1}{3}\)
Vậy...
30 tháng 3 2020
a) Ta có: \(\left(x-2\right)^3+\frac{8}{27}=0\)
\(\Leftrightarrow\left(x-2\right)^3=\frac{-8}{27}\)
\(\Leftrightarrow\left(x-2\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow x-2=\frac{-2}{3}\)
hay \(x=\frac{-2}{3}+2=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
b) Ta có: \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)
\(\Leftrightarrow\frac{13}{3}\cdot\frac{4}{x}=20\)
\(\Leftrightarrow\frac{4}{x}=20:\frac{13}{3}=20\cdot\frac{3}{13}=\frac{60}{13}\)
hay \(x=\frac{13\cdot4}{60}=\frac{13}{15}\)
Vậy: \(x=\frac{13}{15}\)
c) Ta có: \(\left(0,25-30\%x\right)\cdot\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)
\(\Leftrightarrow\left(\frac{1}{4}-\frac{3x}{10}\right)\cdot\frac{1}{3}=\frac{31}{6}+\frac{1}{4}=\frac{65}{12}\)
\(\Leftrightarrow\frac{1}{4}-\frac{3x}{10}=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}\cdot3=\frac{65}{4}\)
\(\Leftrightarrow\frac{3x}{10}=\frac{1}{4}-\frac{65}{4}=-16\)
\(\Leftrightarrow3x=-160\)
hay \(x=\frac{-160}{3}\)
Vậy: \(x=\frac{-160}{3}\)
d) Ta có: \(\frac{x-2}{-\frac{2}{9}}=\frac{-2}{x-2}\)
\(\Leftrightarrow\left(x-2\right)^2=-2\cdot\left(-\frac{2}{9}\right)=\frac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\frac{2}{3}\\x-2=-\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}+2\\x=\frac{-2}{3}+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{8}{3};\frac{4}{3}\right\}\)
TG
30 tháng 3 2020
a/ (x - 2)3 + \(\frac{8}{27}\) = 0
=> (x - 2)3 = 0 - \(\frac{8}{27}\) = \(\frac{-8}{27}\)
=> x - 2 = \(-\frac{2}{3}\)
=> x = \(-\frac{2}{3}+2=\frac{4}{3}\)
b/ \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)
=> \(4\frac{1}{3}:\frac{x}{4}=6:\frac{3}{10}=6.\frac{10}{3}=20\)
=> \(\frac{x}{4}=4\frac{1}{3}:20=\frac{13}{3}.\frac{1}{20}=\frac{13}{60}\)
=> \(x=\frac{13}{60}.4=\frac{13}{15}\)
c/ \(\left(0,25-30\%x\right).\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)
=> \(\left(0,25-30\%x\right).\frac{1}{3}=5\frac{1}{6}+\frac{1}{4}=\frac{65}{12}\)
=> \(0,25-\frac{30}{100}x=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}.3=\frac{65}{4}\)
=> \(\frac{3}{10}x=0,25-\frac{65}{4}=\frac{1}{4}-\frac{65}{4}=-\frac{64}{4}=-16\)
=> \(x=-16:\frac{3}{10}=-16.\frac{10}{3}=-\frac{160}{3}\)
Phần trong ngăocj của G bạn lấy phần nguyên trừ phân nguyên và phần phân số trừ phần phân số thì ta được:
\(G=\left(153\frac{3}{4}-148\frac{3}{8}\right)\cdot0,3:0,2\)
\(=5\frac{3}{8}\cdot\frac{3}{10}:\frac{2}{10}=\frac{43}{8}\cdot\frac{3}{10}\cdot\frac{2}{10}\)
\(=\frac{43\cdot3\cdot2}{8\cdot10\cdot10}=\frac{43\cdot3\cdot2}{2\cdot4\cdot10\cdot10}=\frac{43\cdot3}{4\cdot10\cdot10}=\frac{129}{400}\)
Ta có:
\(G=\left(153\frac{3}{4}-148\frac{3}{8}\right).0,3:0,2\)
\(G=\left(153\frac{6}{8}-148\frac{3}{8}\right).0,3:0,2\)
\(G=\left[\left(153-148\right)+\left(\frac{6}{8}-\frac{3}{8}\right)\right].\frac{3}{2}\)
\(G=5\frac{3}{8}.\frac{3}{2}\)
\(G=\frac{43}{8}.\frac{3}{2}\)
\(G=\frac{129}{16}\)