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28 tháng 10 2023

a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)

=>\(\left(x-2\right)\left(x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)

mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)

nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)

d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

=>\(2^x\left(1+2+2^2+2^3\right)=120\)

=>\(2^x\cdot15=120\)

=>\(2^x=8\)

=>x=3

e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)

=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)

=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

22 tháng 6 2016

\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

   =  \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

    = \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

     = \(\frac{1}{4}+\frac{1}{2}\)

      =  \(\frac{3}{4}\)

b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)

    =\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)

   = \(-\frac{35}{27}+\frac{47}{21}\)

   =        \(\frac{178}{189}\)

c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)

  = \(\frac{117}{13}-\frac{311}{65}\)

 =       \(\frac{274}{65}\)

d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)

\(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)

\(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)

=     \(\frac{1}{3}+\frac{5}{2}\)

=         \(\frac{17}{6}\)

7 tháng 9 2021

\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)

\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)

Bài 2: 

a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)

\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)

\(=-\dfrac{3}{5}\)

b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)

\(\Leftrightarrow8x-1=5\)

\(\Leftrightarrow8x=6\)

hay \(x=\dfrac{3}{4}\)

2 tháng 10 2017

câu E

\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )

4 tháng 10 2023

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`@` `\text {Ans}`

`\downarrow`

\(\left(\dfrac{3}{5}\right)^{2x-1}=\left(\dfrac{9}{25}\right)^3\)

`=>`\(\left(\dfrac{3}{5}\right)^{2x-1}=\left(\dfrac{3^2}{5^2}\right)^3\)

`=>`\(\left(\dfrac{3}{5}\right)^{2x-1}=\left(\dfrac{3}{5}\right)^6\) 

`=> 2x - 1 = 6`

`=> 2x=6+1`

`=> 2x=7`

`=> x = 7/2`

`=> x = 3,5`

Vậy, `x = 3,5`

`@` `\text {Kaizuu lv uuu}`

2 tháng 10 2016

a) x^2 = 9   =>  x=3 hoặc x = -3

b) x^2 = 5   =>  \(x=\sqrt{5}\)

c) x^2 - 4 = 0

 => x^2 = 4             =>   x = 2     hoặc      x = -2

d) x^2 + 1 = 82

=>  x^2 = 81     =>     x = 9 hoặc  x = -9

e)  (2x)^2 = 6 

=>  4 . x^2 = 6     

=> x^2 = 3/2           

=> \(x=\sqrt{\frac{3}{2}}\)

f) (x-1)^2 = 9

=> x-1 = 3     hoặc x - 1 = -3

=> x = 4             hoặc  -2

g) (2x+3)^2  =  25

=> 2x + 3 = 5               hoặc        2x + 3 = -5

=> x = 1                      hoặc          x = -4

2 tháng 10 2016

Ta có: 

a, \(x^2=9\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

b, \(x^2=5\Rightarrow\orbr{\begin{cases}x=2,5\\x=-2.5\end{cases}}\)

Các câu còn lại tương tự nhé bn

16 tháng 10 2023

\(A=\dfrac{9}{8}-\dfrac{8}{9}+\dfrac{3}{24}+\dfrac{1}{4}-\dfrac{5}{16}+\dfrac{19}{25}-\dfrac{1}{9}+\dfrac{2}{25}-\dfrac{1}{81}\)

\(=\dfrac{9}{8}+\dfrac{1}{4}-\dfrac{5}{16}+\dfrac{1}{8}-\dfrac{8}{9}-\dfrac{1}{9}-\dfrac{1}{81}+\dfrac{19}{25}+\dfrac{2}{25}\)

\(=\dfrac{10}{8}+\dfrac{1}{4}-\dfrac{5}{16}-1-\dfrac{1}{81}+\dfrac{21}{25}\)

\(=\dfrac{20+4-5}{16}-\dfrac{82}{81}+\dfrac{21}{25}\)

\(=\dfrac{19}{16}-\dfrac{82}{81}+\dfrac{21}{25}\)

\(=\dfrac{32891}{16\cdot81\cdot25}\)

b: \(B=-\dfrac{1}{3}-\dfrac{8}{35}-\dfrac{2}{9}-\dfrac{1}{35}+\dfrac{4}{5}-\dfrac{4}{9}+\dfrac{3}{7}\)

\(=\dfrac{-1}{3}-\dfrac{2}{9}-\dfrac{4}{9}-\dfrac{8}{35}-\dfrac{1}{35}+\dfrac{4}{5}+\dfrac{3}{7}\)

\(=\dfrac{-3-2-4}{9}+\dfrac{-9}{35}+\dfrac{28+15}{35}\)

\(=-1+\dfrac{-9+43}{35}=-1+\dfrac{34}{35}=-\dfrac{1}{35}\)