Tìm x ,biết:
\(2-\left(\frac{37}{7}+x-3\right)\times\frac{49}{11}=0\)
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DP
25 tháng 6 2017
\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)
\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(x=1\div\left(-\frac{3}{5}\right)\)
\(x=-\frac{5}{3}\)
\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)
\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)
\(-\frac{4}{21}\cdot x=\frac{3}{5}\)
\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)
\(x=-\frac{63}{20}\)
DP
25 tháng 6 2017
\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)
\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)
\(\frac{5}{7}\cdot x=1\)
\(x=1\div\frac{5}{7}\)
\(x=\frac{7}{5}\)
\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)
\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)
\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)
\(\frac{17}{33}\cdot x=\frac{9}{14}\)
\(x=\frac{9}{14}\div\frac{17}{33}\)
\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)
10 tháng 8 2015
\(\frac{7}{5}+\left(\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right)\cdot x=\frac{16}{5}\)
\(\frac{2}{5}x=\frac{16}{5}-\frac{7}{5}\)
\(\frac{2}{5}x=\frac{9}{5}\)
x = \(\frac{9}{5}:\frac{2}{5}\)
x = 9/2
10 tháng 8 2015
\(1\frac{2}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)
\(\frac{7}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)
\(\frac{7}{5}+\frac{2}{5}.x=\frac{16}{5}\)
\(\frac{2}{5}.x=\frac{16}{5}-\frac{7}{5}\)
\(\frac{2}{5}.x=\frac{9}{5}\)
\(x=\frac{9}{5}:\frac{2}{5}\)
\(x=\frac{9}{5}.\frac{5}{2}\)
\(x=\frac{9}{2}\)
17 tháng 7 2015
a,(x-7)x+1-(x-7)x+11=0
=>(x-7)x+1.[1-(x-7)10]=0
=>(x-7)x+1=0
=>x-7=0
=>x=7
hoặc 1-(x-7)10=1
=>(x-7)10=1
=>x-7=-1;1
=>x=8;6
vậy x=6;7;8
b,(x-1)2=36/49
=>x-1=6/7;-6/7
=>x=13/7;1/7
vậy x=1/7;13/7
9 tháng 2 2019
a. \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{16}{5}+\frac{2}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{14}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\frac{14}{5}-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{5}{3}\end{cases}.}\)
Vậy \(x\in\left\{-\frac{5}{3};\frac{7}{3}\right\}.\)
b. \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}\times\left(x-7\right)^{10}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}.}\)Xét 2 trường hợp:
- \(\left(x-7\right)^{x+1}=0\)\(\Leftrightarrow x-7=0\Leftrightarrow x=7.\)
- \(1-\left(x-7\right)^{10}=0\Leftrightarrow\left(x-7\right)^{10}=1\Leftrightarrow\left(x-7\right)^{10}=\left(\pm1\right)^{10}\)\(\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=6\end{cases}.}}\)
Vậy \(x\in\left\{6;7;8\right\}.\)
\(2-\left(\frac{37}{7}+x-3\right)x\frac{49}{11}=0\)
\(\left(\frac{37}{7}+x-3\right)x\frac{49}{11}=2\)
\(\frac{37}{7}+x-3=\frac{22}{49}\)
\(x=-\frac{90}{49}\)
\(2-\left(\frac{37}{7}+x-3\right)\cdot\frac{49}{11}=0\)
\(\left(\frac{37}{7}+x-3\right)\cdot\frac{49}{11}=0+2\)
\(\left(\frac{37}{7}+x-3\right)\cdot\frac{49}{11}=2\)
\(\frac{37}{7}+x-3=2:\frac{49}{11}\)
\(\frac{37}{7}+x-3=\frac{22}{49}\)
\(\frac{37}{7}+x=\frac{22}{49}+3\)
\(\frac{37}{7}+x=\frac{169}{49}\)
\(x=\frac{169}{49}-\frac{37}{7}\)
\(x=-\frac{90}{49}\)