\(3^{x+4}=9^{2x-1}\)

\(\Rightarrow3^{x+4}=3^{4x-2}\)

\(\Rightarrow x+4=4x-2\)

\(\Rightarrow3x=6\Rightarrow x=2\)

3x+4=92x−13x+4=92x−1

⇒3x+4=34x−2⇒3x+4=34x−2

a)=1/2 . 8/15 - 3/4.47/9

=4/15 - 47/12

=-73/20

b)=2-1/3 . -21/20

=2+7/20

=47/20

1.

=3/5x(3/7+4/7)+2/5x(13/9-4/9)

=3/5x1+2/5x1

=3/5+2/5

=1

2.Xx(3/4+4/5)=7/10

Xx31/20=7/10

X         =7/10:31/20

X         =14/31

\(\frac{3}{5}\cdot\frac{3}{7}+\frac{3}{5}\cdot\frac{4}{7}+\frac{2}{5}\cdot\frac{13}{9}-\frac{2}{5}\cdot\frac{4}{9}\)

\(=\frac{3}{5}\cdot\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{2}{5}\cdot\left(\frac{13}{9}-\frac{4}{9}\right)\)

\(=\frac{3}{5}\cdot1+\frac{2}{5}\cdot1\)\(=\frac{3}{5}+\frac{2}{5}=1\)

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\(\frac{3}{4}\cdot x+\frac{4}{5}\cdot x=\frac{7}{10}\)

\(\left(\frac{3}{4}+\frac{4}{5}\right)\cdot x=\frac{7}{10}\)

\(\frac{31}{20}\cdot x=\frac{7}{10}\)

\(x=\frac{7}{10}:\frac{31}{20}\)

\(x=\frac{14}{31}\)

\(\left[7\times13+8\times13\right]:\left[9\frac{2}{3}-y\right]=39\)

\(\left[\left(7+8\right)\times13\right]:\left[\frac{29}{3}-y\right]=39\)

\(\left[15\times13\right]:\left[\frac{29}{3}-y\right]=39\)

\(195:\left[\frac{29}{3}-y\right]=39\)

\(\frac{29}{3}-y=195:30\)

\(\frac{29}{3}-y=\frac{13}{2}\)

\(y=\frac{29}{3}-\frac{13}{2}\)

\(y=\frac{19}{6}\)

Học tốt #

[ 7 x 13 + 8 x 13 ] : [ \(9\frac{2}{3}\) - y ] = 39

[ 13 x ( 7 + 8 )]     :  [ \(9\frac{2}{3}\)- y  ]  = 39

[ 13 x 15 ]            :  [\(9\frac{2}{3}\) - y  ]   = 39

195                      : [\(\frac{29}{3}\)  - y  ]   = 39

                              [ \(\frac{29}{3}\) - y  ]   = 195 : 39 (^*1 )

                              [\(\frac{29}{3}-y\)]  = 5

                            y                         = \(\frac{29}{3}-5\)

                           y                          = \(\frac{14}{3}\)

\(\left(7.13+8.13\right)\div\left(9\frac{2}{3}-y\right)=39\)

\(\Rightarrow\left(91+104\right)\div\left(\frac{29}{3}-y\right)=39\)

\(\Rightarrow195\div\left(\frac{29}{3}-y\right)=39\)

\(\Rightarrow\frac{29}{3}-y=195\div39\)

\(\Rightarrow\frac{29}{3}-y=5\)

\(\Rightarrow y=\frac{29}{3}-5\)

\(\Rightarrow y=\frac{14}{3}\)

Vậy \(y=\frac{14}{3}\)

2(x-4)-(x+5)=-13

2x-8-x-5     =-13

2x-x           =-13+5+8

          x     =0

a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)

\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)

\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)

\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)

\(\left(2x+1\right)^2=\frac{25}{16}\)

\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)

\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)

\(x^3-\frac{9}{16}.x=0\)

\(x\left(x^2-\frac{9}{16}\right)=0\)

\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)

Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)

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