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24 tháng 1 2022
\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,53+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}.\)
\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{53}{100}+\dfrac{1}{2}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+1-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}.\)
\(=\dfrac{\dfrac{263}{440}}{\dfrac{-1487}{1650}}+\dfrac{\dfrac{7}{4}}{\dfrac{35}{12}}=\dfrac{-3945}{5948}+\dfrac{3}{5}=\dfrac{-1881}{29740}.\)
25 tháng 2 2020
Sửa đề: \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
Ta có: \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}+\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)
\(=\frac{-3\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}+\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}\)
\(=\frac{-3}{5}+\frac{3}{5}=0\)
HL
1
DH
27 tháng 12 2019
\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{1}{2}+\frac{5}{11}+\frac{5}{12}}=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}=\frac{\frac{263}{440}}{\frac{263}{264}}=\frac{3}{5}\)
Số chẳng đẹp gì cả.
N
7 tháng 11 2018
\(\frac{0.375-0.3+\frac{3}{11}+\frac{3}{12}}{0.625-0.5+\frac{5}{11}+\frac{5}{12}}\)
\(=\frac{0.075+\frac{3}{11}+\frac{3}{12}}{0.125+\frac{5}{11}+\frac{5}{12}}\)
\(=\frac{\frac{3}{40}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{40}+\frac{5}{11}+\frac{5}{12}}\)
\(=\frac{3.\left(\frac{1}{40}+\frac{1}{11}+\frac{1}{12}\right)}{5.\left(\frac{1}{40}+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
BH
24 tháng 7 2019
\(3x=2y=z\Rightarrow\frac{z}{6}=\frac{x}{2}=\frac{y}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau
\(\frac{z}{6}=\frac{x}{2}=\frac{y}{3}=\frac{x+y+z}{6+2+3}=\frac{99}{11}=9\)
\(\Rightarrow\hept{\begin{cases}z=54\\x=18\\y=27\end{cases}}\)
24 tháng 10 2016
\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
\(A=\left(\frac{\frac{3}{2}+1-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{1}{2}-\frac{5}{11}-\frac{5}{12}}\right):\frac{378}{401}+115\)
\(A=\left(\frac{3.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{-3.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}\right).\frac{401}{378}+115\)
\(A=\left(\frac{3}{5}+\frac{-3}{5}\right).\frac{401}{378}+115\)
\(A=0.\frac{401}{378}+115=115\)
24 tháng 10 2016
A = \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
= \(\left(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3.125}{100}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5.125}{100}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
= \(\left(\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{3\left(\frac{125}{100}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(\frac{125}{100}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\right):\frac{1890}{2005}+115\)
= \(\left(\frac{3}{5}+-\frac{3}{5}\right):\frac{1890}{2005}+115\)
= 115
*TH1:x+0.3=0.5
=>x=0.5-0.3
=>x=0.2
*TH2:x+0.3=-0.5
=>x=-0.5-0.3
=>x=-0.8
Vậy :* x=0.2
* x=-0.8
\(\left|x+0,3\right|=0,5\)
\(\Leftrightarrow x+0,3=\pm0, 5\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+0,3=0,5\\x+0,3=-0,5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,5-0,3\\x=-0,5-0,3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\x=-\left(0,5+0,3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\x=-0,8\end{matrix}\right.\)
Vậy \(x=0,2\) hoặc \(x=-0,8\)