\(\frac{3}{3}\)+ \(\frac{2}{5}\) = ?
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HQ
Hà Quang Minh
Giáo viên
8 tháng 10 2023
(A) \(\frac{2}{3} + \frac{{ - 4}}{6} = \frac{4}{6} + \frac{{ - 4}}{6} = 0\) => A sai
(B) \(\frac{2}{3}.\frac{{ - 1}}{5} = \frac{{ - 2}}{{15}}\) mà \(\frac{{3 - 2}}{5} = \frac{1}{5}\) => B sai
(C) \(\frac{2}{3} - \frac{3}{5} = \frac{{10}}{{15}} - \frac{9}{{15}} = \frac{1}{{15}}\) => C đúng
(D) \(\frac{3}{5}:\frac{3}{{ - 5}} = \frac{3}{5}.\frac{{ - 5}}{3} = \frac{{ - 15}}{{15}} = - 1\) => D sai
=> Chọn C.
24 tháng 1 2022
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
3 tháng 8 2017
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}+\frac{\frac{3}{2}+1+\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}+\frac{5}{4}}\)
\(=\frac{3.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{3}{5}\)
\(=\frac{6}{5}\)
6 tháng 7 2019
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
27 tháng 8 2020
\(A=\frac{\frac{-11}{2}+\frac{\frac{-5}{3}}{1-\frac{4}{3}}}{\frac{3}{5}-\frac{\frac{-2}{5}}{\frac{4}{5}-\frac{2}{3}}}\)
\(A=\frac{\frac{-11}{2}+\frac{\frac{-5}{3}}{\frac{-1}{3}}}{\frac{3}{5}-\frac{\frac{-2}{5}}{\frac{2}{15}}}\)
\(A=\frac{\frac{-11}{2}+5}{\frac{3}{5}-\left(-3\right)}=\frac{\frac{-1}{2}}{\frac{3}{5}+3}=\frac{\frac{-1}{2}}{\frac{18}{5}}=\frac{-5}{36}\)
cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt:>
\(\frac{3}{3}+\frac{2}{5}=\frac{3}{3}=1\)vậy \(\frac{3}{3}+\frac{2}{5}=1+\frac{2}{5}=\frac{1}{1}+\frac{2}{5}=\frac{5}{5}+\frac{2}{5}=\frac{7}{5}\)7 : 5 = 1,4 , kết quả là \(1,4\)