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NH
9
3 tháng 6 2018
\(\frac{6+x}{33}=\frac{7}{11}\)
\(\Leftrightarrow11\cdot\left(6+x\right)=7\cdot33\)
\(\Leftrightarrow66+11x=231\)
\(\Leftrightarrow11x=231-66\)
\(\Leftrightarrow x=165:11\)
\(\Leftrightarrow x=15\)
Dấu . là dấu nhân
3 tháng 6 2018
\(\frac{6+x}{33}=\frac{7}{11}\)
=> ( 6 + x ) . 11 = 231
=> 6 + x = 231 : 11
6 + x = 21
x = 21 - 6
x = 15
Vậy x = 15
NN
29 tháng 8 2020
Bài 1:
a) \(33^{2x}:11^{2x}=81\)\(\Leftrightarrow\left(33:11\right)^{2x}=81\)
\(\Leftrightarrow3^{2x}=3^4\)\(\Leftrightarrow2x=4\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
b) \(\frac{x}{-5}=\frac{4}{21}\)\(\Leftrightarrow21x=-20\)\(\Leftrightarrow x=\frac{-20}{21}\)
Vậy \(x=\frac{-20}{21}\)
Bài 2:
\(A=\frac{1+3^4+3^8+3^{12}}{1+3^2+3^4+3^6+3^8+3^{10}+3^{12}+3^{14}}\)
\(=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+\left(3^2+3^6+3^{10}+3^{14}\right)}\)
\(=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+3^2.\left(1+3^4+3^8+3^{12}\right)}\)
\(=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right).\left(1+3^2\right)}=\frac{1}{1+3^2}=\frac{1}{1+9}=\frac{1}{10}\)
29 tháng 8 2020
\(33^{2x}:11^{2x}=81\)!
\(\left(33:11\right)^{2x}=81\)
\(3^{2x}=81\)
\(3^{2x}=3^4\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
vậy \(x=2\)
\(\frac{x}{-5}=\frac{4}{21}\)
x.21=-5.4
x.21=-20
x=-20:21
\(x=-\frac{20}{21}\)
vậy \(x=-\frac{20}{21}\)
15 tháng 2 2020
4x-(7+5x)=15+(-29). b,x/11=x+4/33
4x-7-5x=-14. 33x=11(x+4)
4x-5x=14+7. 33x=11x+44
-x=21. 33x-11x=44
x=-21 22x=44
x=44:22=2
NL
22 tháng 4 2017
Bài 1:
\(g.\frac{5}{11}+\frac{6}{11}=\frac{5+6}{11}=\frac{11}{11}=1\)\(\)
\(e.\frac{-17}{25}.\frac{20}{33}+\frac{-17}{25}.\frac{13}{33}+\frac{-3}{25}=\frac{-17}{25}.\left(\frac{20}{33}+\frac{13}{33}\right)+\frac{-3}{25}\)
\(=\frac{-17}{25}.1+\frac{-3}{25}=\frac{-17}{25}+\frac{-3}{25}=\frac{-17-3}{25}=\frac{-20}{25}=\frac{-4}{5}\)
\(d.\frac{5}{7}.\frac{19}{23}+\frac{5}{7}.\frac{5}{23}-\frac{5}{7}.\frac{1}{23}=\frac{5}{7}\left(\frac{19}{23}+\frac{5}{23}-\frac{1}{23}\right)\)
\(=\frac{5}{7}\left(\frac{19+5-1}{23}\right)=\frac{5}{7}.1=\frac{5}{7}\)
\(c.\left(-11\right).\frac{9}{22}=\frac{\left(-11\right).9}{22}=\frac{-99}{22}=\frac{-9}{2}\)
\(b.\frac{5}{6}-\frac{1}{8}=\frac{5.4}{6.4}-\frac{1.3}{8.3}=\frac{20}{24}-\frac{3}{24}=\frac{17}{24}\)
\(a.\frac{2}{3}+\frac{1}{5}-\frac{1}{6}=\frac{2.10}{3.10}+\frac{1.6}{5.6}-\frac{1.5}{6.5}=\frac{20}{30}+\frac{6}{30}-\frac{5}{30}\)
\(=\frac{20+6-5}{30}=\frac{21}{30}=\frac{7}{10}\)
Bài 2:
\(a.\frac{3}{4}+x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{3}{4}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{9}{12}\)
\(\Leftrightarrow x=\frac{2}{12}=\frac{1}{6}\)
\(b.x-\frac{11}{12}=0,5\)
\(\Leftrightarrow x=\frac{1}{2}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{6}{12}+\frac{11}{12}\)
\(\Leftrightarrow x=\frac{17}{12}\)
TD
0
Ta có:
\(2x+33=\left(-11\right)-x\)
\(\Leftrightarrow2x+33=\left(-11\right)+\left(-x\right)\)
\(\Leftrightarrow2x+33=-\left(11+x\right)\)
\(\Leftrightarrow2x=-\left(11+x\right)-33\)
\(\Leftrightarrow2x=-\left(11+x\right)+\left(-33\right)\)
\(\Leftrightarrow2x=-\left(11+33+x\right)\)
\(\Leftrightarrow2x=-\left(44+x\right)\)
\(\Leftrightarrow2x=2\cdot\frac{-\left(44+x\right)}{2}\)
\(\Leftrightarrow2x=2\cdot\left[\left(-44\right):2\right]+\left[\left(-x\right):2\right]\)
\(\Leftrightarrow2x=2\cdot\left(-22\right)+\frac{-x}{2}\)
\(\Leftrightarrow x=\left(-22\right)+\frac{-x}{2}\)
\(\Leftrightarrow\left(-22\right)+\frac{-x}{2}=x\)
\(\Leftrightarrow\frac{-x}{2}=x-22\)
\(\Leftrightarrow\frac{-x}{2}=-22\)
\(\Leftrightarrow-x=\left(-22\right):2\)
\(\Leftrightarrow-x=-11\)
\(\Leftrightarrow x=11\)