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M
1
22 tháng 4 2018
\(\frac{1}{2}\times(x-\frac{1}{3})+\left(\frac{1}{3}-x\right)=0\)
\(\frac{1}{2}\cdot\left(x-\frac{1}{3}\right)-x\left(x-\frac{1}{3}\right)=0\)
\(\left(x-\frac{1}{3}\right)\left(\frac{1}{2}-x\right)=0\)
\(\left(x-\frac{1}{3}\right)=0\)và \(\left(\frac{1}{2}-x\right)=0\)
TH1
S
9 tháng 7 2019
\(1,\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}=6\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=6\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).Tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{11}\right)=6\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{48}{33}=\frac{16}{11}\)
\(2,\left(x+1\right)+\left(x+2\right)+.....+\left(x+211\right)=211x+\left(1+2+....+211\right)=211x+\frac{212.211}{2}=211x+22366=23632\Leftrightarrow211x=23632-22366=1266\Leftrightarrow x=6\)
9 tháng 7 2019
a, \(14:\left(4\frac{2}{3}:1\frac{5}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)
=> \(14:\frac{28}{9}+14:\frac{14}{9}=>14.\frac{9}{28}+14.\frac{9}{14}\)
=> 14. ( \(\frac{9}{28}+\frac{9}{14}\) )
=> \(14.\frac{27}{28}=\frac{419}{28}\)
b, \(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
=> \(\frac{4}{5}+\frac{12}{35}+\frac{4}{21}+\frac{4}{33}\)
=> \(\frac{8}{7}+\frac{24}{77}=\frac{16}{11}\)
bài 2 :
( x + 1 ) + ( x + 2 ) + ... + ( x + 211 ) = 23632
=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 211 ) = 23632
=> 211x + 22366 = 23632
=> 211x = 23632 - 22366
=> 211x = 1266
=> x = 1266 : 211
x = 6
PN
18 tháng 6 2020
\(a,\frac{1}{3}:\left(2x-1\right)=-\frac{4}{21}\)
\(< =>\frac{1}{3}.\frac{1}{2x-1}=-\frac{4}{21}\)
\(< =>\frac{1}{6x-3}=-\frac{4}{21}\)
\(< =>\frac{1}{6x-3}+\frac{4}{21}=0\)
\(< =>21-24x+12=0\)
\(< =>33-24x=0\)
\(< =>x=\frac{33}{24}\)
\(b,\frac{17}{2}-|x-\frac{3}{4}|=\frac{-7}{4}\)
\(< =>|x-\frac{3}{4}|=\frac{17}{2}+\frac{7}{4}=\frac{41}{4}\)
\(< =>\orbr{\begin{cases}x-\frac{3}{4}=\frac{41}{4}\\x-\frac{3}{4}=-\frac{41}{4}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{41}{4}+\frac{3}{4}\\x=-\frac{41}{4}+\frac{3}{4}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=11\\x=-\frac{19}{2}\end{cases}}\)
18 tháng 6 2020
a, \(\frac{1}{3}:\left(2x-1\right)=-\frac{4}{21}\)
\(\Leftrightarrow\frac{1}{3}.\frac{1}{2x-1}=-\frac{4}{21}\)
\(\Leftrightarrow\frac{1}{6x-3}=-\frac{4}{21}\)
\(\Leftrightarrow21=-24x+12\)
\(\Leftrightarrow-24x=9\Leftrightarrow x=-\frac{3}{8}\)
b, \(\frac{17}{2}-\left|x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Leftrightarrow\left|x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{41}{4}\\x-\frac{3}{4}=-\frac{41}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-\frac{19}{2}\end{cases}}}\)
28 tháng 4 2018
Bài 1
a) \(\frac{1}{2}+\frac{3}{8}:3-\frac{3}{16}.2^3 \)
\(=\frac{1}{2}+\frac{3}{8}.\frac{1}{3}-\frac{3}{16}.8 \)
\(=\frac{1}{2}+\frac{1}{8}-\frac{3}{2} \)
\(=\frac{4}{8}+\frac{1}{8}-\frac{12}{8} \)
\(=\frac{-7}{8}\)
30 tháng 4 2019
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};3\right\}\)
e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
30 tháng 4 2019
Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm
0 o là j vậy
\(|x-\frac{1}{2}|=\frac{3}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{3}{4}\\x-\frac{1}{2}=\frac{-3}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{-1}{4}\end{cases}}\)
Vậy...