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a)\(8\sqrt{x}-3\sqrt{\frac{4}{81}}=5,2\)
\(\Rightarrow8\sqrt{x}-3.\frac{2}{9}=5,2\)
\(\Rightarrow8\sqrt{x}-\frac{2}{3}=5,2\)
\(\Rightarrow8\sqrt{x}=5,2+\frac{2}{3}\)
\(\Rightarrow8\sqrt{x}=\frac{40}{3}\)
\(\Rightarrow\sqrt{x}=\frac{40}{3}:8\)
\(\Rightarrow\sqrt{x}=\frac{5}{3}\)
\(\Rightarrow x=\frac{25}{9}\)
b)\(12-3x^2=10+\sqrt{\frac{25}{16}}\)
\(\Rightarrow12-3x^2=10+\frac{5}{4}\)
\(\Rightarrow12-3x^2=11,25\)
\(\Rightarrow3x^2=12-11,25\)
\(\Rightarrow3x^2=0,75\)
\(\Rightarrow x^2=0,25\)
\(\Rightarrow x=\sqrt{0,25}\)
\(\Rightarrow x=0,5\)

e, Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=k\left(k\in Z\right)\)
\(\Leftrightarrow x=4k,y=5k\) (1)
Theo bài ra ta có: xy = 80
Từ (1) \(\Rightarrow4k.5k=80\Rightarrow20.k^2=80\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
+ Với k = 2 \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
+ Với k = -2 \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-10\end{matrix}\right.\)
Vậy \(\left(x,y\right)\in\left\{\left(8,10\right);\left(-8,-10\right)\right\}\)
a) \(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=\dfrac{-16}{4}=-4\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-4\\\dfrac{y}{5}=-4\\\dfrac{z}{-2}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-12\\y=-20\\z=8\end{matrix}\right.\)

\(8.\sqrt{x} -3\sqrt{\dfrac{4}{81}}= 5,2 \)
\(8.\sqrt{x} - 3. \dfrac{2}{9}=5,2\)
\(8.\sqrt{x} - \dfrac{2}{3}= 5,2\)
\(8.\sqrt{x}= 5,2+ \dfrac{2}{3}\)
\(8.\sqrt{x}= \dfrac{88}{15}\)
\(\sqrt{x}= \dfrac{88}{15}: 8\)
\(\sqrt{x}= \dfrac{11}{15}\)
\(\sqrt{x}= \sqrt{ {\dfrac{121}{225}}}\)
\(\Rightarrow x= \dfrac{121}{225}\)
Chúc bạn học tốt

a, 24-x=32=25
=> 4-x=5
<=> x=-1
b, (x+1,5)2+(y-2,5)10=0
Vì (x+1,5)2\(\ge\)0, (y-2,5)10\(\ge\)0
\(\Rightarrow\hept{\begin{cases}x+1,5=0\\y-2,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1,5\\y=2,5\end{cases}}}\)
a)\(2^{4-x}\)=32
=>\(2^{4-x}\)=32=\(2^5\)
=>4-x=5
=>x=4-5=-1
=>x=-1

Theo bài ra , ta có :
\(\frac{x}{4}+\frac{x}{8}+\frac{x}{16}=\frac{x}{9}+\frac{x}{27}+\frac{x}{81}\)
\(\Rightarrow\frac{x}{4}+\frac{x}{8}+\frac{x}{16}-\frac{x}{9}-\frac{x}{27}-\frac{x}{81}=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)

Ta có : \(\frac{\left(4^x\right)^2}{2^x}=8\)
\(\Rightarrow4^{2x}=8.2^x\)
\(\Rightarrow4^{2x}=2^3.2^x\)
\(\Rightarrow\left(2^2\right)^{2x}=2^{x+3}\)
\(\Rightarrow2^{4x}=2^{x+3}\)
=> 4x = x + 3
=> 3x = 3
=> x = 1
Vậy x = 1.

1/ 2x = 45.46
=> 2x = 45 + 6
=> 2x = 411
=> 2x = (22)11
=> 2x = 222
=> x = 22
vậy_
2/ 2x = 46.163
=> 2x = (22)6.(24)3
=> 2x = 212.212
=> 2x = 212 + 12
=> 2x = 224
=> x = 24
3/ 2x = 45.162
=> 2x = (22)5.(24)2
=> 2x = 210.28
=> 2x = 210 + 8
=> 2x = 218
=> x = 18
vậy_
\(\frac{1}{2^x}=4^5.4^3=4^{5+3}=4^8\)
\(\Rightarrow1=4^8.2^x=2^{2.8+x}=2^{16+x}\)
ta có 1 < 21 => 216+x < 21
=> 216+x = 20
=> 16+x=0
=> x= -16
Sửa đề đi bạn ơi
vãi cả tìm x mà dell thấy x đâu :v