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reyms
\(\frac{0,64}{y}=\frac{y}{81}\)
\(\Rightarrow y^2=0,64.81\)
\(\Rightarrow y^2=51,84\)
\(\Rightarrow y=\sqrt{51,84}\)
\(\Rightarrow y=7,2\)
vậy \(y=7,2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{0,09}-\sqrt{0,64}=\frac{-1}{2}=-0,5\)
b) \(0,1\cdot\sqrt{225}-\sqrt{\frac{1}{4}}=0,1\cdot15-\frac{1}{2}=1\)
c) \(\sqrt{0,36}\cdot\sqrt{\frac{25}{16}+\frac{1}{4}}=\frac{3\sqrt{29}}{20}\)
d) đề baì có sai ko ban?
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 5 :
a) \(\dfrac{y}{4}=\dfrac{9}{y}\)
\(\Rightarrow y^2=36\left(y\ne0\right)\)
\(\Rightarrow y=\pm6\)
b) \(\dfrac{y+7}{20}=\dfrac{5}{y+7}\left(y\ne-7\right)\)
\(\Rightarrow\left(y+7\right)^2=100=10^2\)
\(\Rightarrow\left[{}\begin{matrix}y+7=10\\y+7=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=3\\y=-17\end{matrix}\right.\)
c) \(\dfrac{4-5y}{3}=\dfrac{y+2}{5}\)
\(\Rightarrow5\left(4-5y\right)=3\left(y+2\right)\)
\(\Rightarrow20-25y=3y+6\)
\(\Rightarrow28y=14\)
\(\Rightarrow y=\dfrac{14}{28}=\dfrac{1}{2}\)
Bài 4 :
\(\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{c}{10}\)
\(\Rightarrow\dfrac{2a}{10}=\dfrac{3b}{21}=\dfrac{4c}{40}=\dfrac{2a+3b-4c}{10+21-40}=\dfrac{81}{-9}=-9\)
\(\Rightarrow\left\{{}\begin{matrix}a=-9.5=-45\\b=-9.7=-63\\c=-9.10=-90\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$(\frac{-1}{3})^3.x=\frac{1}{81}=(\frac{-1}{3})^4$
$\Rightarrow x=(\frac{-1}{3})^4: (\frac{-1}{3})^3=\frac{-1}{3}$
b.
$2^2.16> 2^x> 4^2$
$\Rightarrow 2^2.2^4> 2^x> (2^2)^2$
$\Rightarrow 2^6> 2^x> 2^4$
$\Rightarrow 6> x> 4$
$\Rightarrow x=5$ (với điều kiện $x$ là số tự nhiên nhé)
c.
$9.27< 3^x< 243$
$3.3^3< 3^x< 3^5$
$\Rightarrow 3^4< 3^x< 3^5$
$\Rightarrow 4< x< 5$
Với $x$ là stn thì không có số nào thỏa mãn.
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\left(x+2\right)^2=81>0\)
\(\orbr{\begin{cases}\left(x+2\right)^2=9^2\\\left(x+2\right)^2=\left(-9\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+2=9\\x+2=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-11\end{cases}}\)
a)\(\left(x+2\right)^2=81\\ =>\left(x+2\right)=9^2hay\left(x+2\right)=\left(-9\right)^2\)
x+2=9 x+2=-9
x=9-2 x=-9-2
x=7(TM) x=-11(TM)
Vậy x=7 hay x=-9
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\1-2x=5\end{matrix}\right.\Leftrightarrow D\\ 2,\Leftrightarrow\left(-3\right)^x=-27\cdot81=-2187=\left(-3\right)^7\\ \Leftrightarrow x=7\left(A\right)\)
\(0,64:c=c:81\)
\(\Rightarrow\frac{0,64}{c}=\frac{c}{81}\Rightarrow2c=0,64\times81\)
\(\Rightarrow2c=51,84\)
\(\Rightarrow c=51,84:2\)
\(\Rightarrow c=25,92\)
0,64 : c = c : 81
=> c . c = 0,63 .81
=> c ^ 2 = 51 , 03
=> c = căn 51 , 03
^-^ ♡ Chúc bạn làm đúng nha !