Tìm x
\(\frac{-11}{6}.x+\frac{3}{2}.x-1\frac{2}{3}=\frac{5}{12}\)
Tìm x biết :
\(\left[\frac{6:\frac{3}{5}-1\frac{1}{16}.\frac{6}{7}}{4\frac{1}{5}.\frac{10}{11}+5\frac{2}{11}}-\frac{\left(\frac{3}{20}+\frac{1}{2}-\frac{1}{15}\right).\frac{12}{49}}{3\frac{1}{3}+\frac{2}{9}}\right].x=2\frac{23}{96}\)
\(x=\frac{903}{391}\)
Bài này sử dụng MTCT đó bạn!
bài 15 tìm x biết
a\(\frac{x}{4}-\frac{3}{7}+\frac{2}{5}=\frac{31}{140}\)
b\(\frac{5}{12}+\frac{5}{x}-\frac{1}{8}=\frac{1}{2}\)
c\(x+\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
d\(\frac{3}{4}-x+\frac{6}{-11}=\frac{5}{6}\)
e\(x-\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
Tìm x biết:
1) \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
2) \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
1) \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
<=> \(\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)
<=> \(x+1=0\) (do 1/2 + 1/3 + 1/4 - 1/5 - 1/6 khác 0)
<=> \(x=-1\)
Vậy...
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
<=> \(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)
<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
<=> \(\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
<=> \(x+2010=0\) (do 1/2009 + 1/2008 + 1/2007 - 1/2000 - 1/1999 - 1/1998 khác 0)
<=> \(x=-2010\)
Vậy....
Tìm x : \(\frac{6:\frac{3}{5}-1\frac{1}{16}\cdot\frac{6}{7}}{4\frac{1}{5}\cdot\frac{10}{11}+5\frac{2}{11}}\)\(-\frac{\left(\frac{3}{20}+\frac{1}{2}-\frac{1}{5}\right)\cdot\frac{12}{49}}{3\frac{1}{3}+\frac{2}{9}}\)
Tìm x biết :
a)\(\frac{2}{3}x-50\%x-\left(-\frac{4}{5}\right):1\frac{3}{5}=-0,12\)\(+1\frac{3}{25}\)
b)\(\left(-1\frac{1}{6}+\frac{2}{3}-\frac{3}{4}\right):x+\left(-1\frac{11}{12}\right).1\frac{21}{23}=-6\frac{1}{3}\)
c)\(50\%x-\frac{1}{3}x-\left(\frac{-2}{3}\right)^2.\left(-1\frac{1}{8}\right)=-119\frac{3}{4}+120\frac{5}{6}\)
Tìm x :
a) \(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
b) \(\left(\frac{9}{11}-x\right):\left(\frac{-10}{11}\right)=1-\frac{4}{5}\)
c) \(\frac{-11}{12}.x+\frac{3}{4}=\frac{-1}{6}\)
d) \(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)
đ) \(\left(\frac{3}{4}-x:\frac{2}{15}\right).\frac{1}{5}=-2,6\)
e) \(3-\left(\frac{1}{6}-x\right).\frac{2}{3}=\frac{2}{3}\)
f) \(\left(1-2x\right).\frac{4}{5}=\left(-2\right)^3\)
g) \(\frac{1}{6}-\left|\frac{1}{2}.x-\frac{1}{3}\right|=\frac{1}{8}\)
Tìm x :
a) \(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
b) \(\left(\frac{9}{11}-x\right):\left(\frac{-10}{11}\right)=1-\frac{4}{5}\)
c) \(\frac{-11}{12}.x+\frac{3}{4}=\frac{-1}{6}\)
d) \(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)
đ) \(\left(\frac{3}{4}-x:\frac{2}{15}\right).\frac{1}{5}=-2,6\)
e) \(3-\left(\frac{1}{6}-x\right).\frac{2}{3}=\frac{2}{3}\)
f) \(\left(1-2x\right).\frac{4}{5}=\left(-2\right)^3\)
g) \(\frac{1}{6}-\left|\frac{1}{2}.x-\frac{1}{3}\right|=\frac{1}{8}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
Vậy x=\(\frac{20}{27}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=1-\frac{4}{5}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=\frac{1}{5}\)
\(\frac{9}{11}-x=\frac{1}{5}\cdot\frac{-10}{11}\)
\(\frac{9}{11}-x=\frac{-2}{11}\)
\(x=\frac{9}{11}-\frac{-2}{11}\)
\(x=1\)
Vậy x=1
\(\frac{-11}{12}\cdot x+\frac{3}{4}=\frac{-1}{6}\)
\(\frac{-11}{12}\cdot x=\frac{-1}{6}-\frac{3}{4}\)
\(\frac{-11}{12}\cdot x=\frac{21}{12}\)
\(x=\frac{-21}{11}\)
Vậy x=\(\frac{-21}{11}\)
\(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)
\(\frac{3}{2}+x=\frac{-5}{4}-\frac{9}{2}\)
\(\frac{3}{2}+x=\frac{23}{4}\)
\(x=\frac{17}{4}\)
Vậy x=\(\frac{17}{4}\)
\(\left(\frac{3}{4}-x:\frac{2}{15}\right)\cdot\frac{1}{5}=-2,6\)
\(\frac{3}{4}-x:\frac{2}{15}=\frac{-13}{5}:\frac{1}{5}\)
\(\frac{3}{4}-x:\frac{2}{15}=-13\)
\(x:\frac{2}{15}=\frac{3}{4}-\left(-13\right)\)
\(x:\frac{2}{15}=\frac{45}{4}\)
\(x=\frac{3}{2}\)
Vậy x=\(\frac{3}{2}\)
\(3-\left(\frac{1}{6}-x\right)\cdot\frac{2}{3}=\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=\frac{2}{3}:\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=1\)
\(\frac{1}{6}-x=2\)
\(x=\frac{1}{6}-2\)
\(x=\frac{-11}{6}\)
Vậy x=\(\frac{-11}{6}\)
\(\left(1-2x\right)\cdot\frac{4}{5}=\left(-2\right)^3\)
\(1-2x=\frac{-1}{10}\)
\(2x=1-\frac{-1}{10}\)
\(2x=\frac{11}{10}\)
\(x=\frac{11}{20}\)
Vậy x=\(\frac{11}{20}\)
\(\frac{1}{6}-\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{8}\)
\(\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{7}{12}\) \(\frac{1}{2}x-\frac{1}{3}=\frac{-7}{12}\)
\(\frac{1}{2}x=\frac{11}{12}\) \(\frac{1}{2}x=\frac{-1}{4}\)
\(x=\frac{11}{6}\) \(x=\frac{-1}{2}\)
Vậy \(x\in\left\{\frac{11}{6};\frac{-1}{2}\right\}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{6}{9}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
tk mình đi mình làm nốt cho hjhj ^^
Tìm x biết :
a) \(-\frac{2}{3}.x+4=-12\)
b) \(-\frac{3}{4}+\frac{1}{4}:x=-3\)
c) \(\frac{x+1}{2}+\frac{x+2}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
d)\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
d) \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
<=> \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}-\frac{x+10}{2000}-\frac{x+11}{1999}-\frac{x+12}{1998}=0\)
<=> \(\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+10}{2000}+1\right)-\left(\frac{x+11}{1999}+1\right)-\left(\frac{x+12}{1998}+1=0\right)\)
<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
<=>\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
<=> x+2010 = 0 vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\)
<=> x = -2010
\(\frac{4}{9}:\frac{-1}{7}+6\frac{5}{9}.\frac{2}{3}\)
\(\left(\frac{-1}{3}\right)^4.\frac{4}{11}+\frac{7}{11}.\left(\frac{-1}{3}\right)^2\)
\(\left(\frac{-1}{7}\right)^0-2\frac{4}{9}:\left(\frac{2}{3}\right)^2\)
\(\left(\frac{1}{3}-\frac{5}{6}\right)^2+\frac{5}{6}:2\)
Tìm x
\(\left(\frac{3}{5}-\frac{1}{2}x\right)^2+8=12\)
\(\frac{7}{12}-\left|\frac{1}{3}-\frac{1}{2}x\right|=\frac{1}{3}\)
\(\frac{13}{12}-\frac{1}{3}.\left|\frac{1}{5}-\frac{1}{3}x\right|=\frac{1}{2}\)
Nhanh hộ mình nha
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