\(\dfrac{16^x}{8}=2^x\)

\(\Leftrightarrow16^x=2^x.8\)

\(\Leftrightarrow\left(2.8\right)^x=2^x.8\)

\(\Leftrightarrow2^x.8^x=2^x.8\)

\(\Leftrightarrow x=1\)

\(\dfrac{2}{3}-5x^2=\dfrac{-58}{3}\)

\(\Leftrightarrow5x^2=\dfrac{2}{3}+\dfrac{58}{3}\)

\(\Leftrightarrow5x^2=20\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

\(\dfrac{2}{3}-5x^2=-\dfrac{58}{3}\)

\(5x^2=\dfrac{2}{3}+\dfrac{58}{3}\)

\(5x^2=\dfrac{60}{3}=20\)

\(x^2=\dfrac{20}{5}=4\)

\(x^2=2^2\)

\(x=2\)

Gọi chiều dài nền nhà là x ( m ) ( x ϵ N*  ; x < 48 )

Chiều rộng nền nhà là : \(\dfrac{x}{3}\left(m\right)\)

Theo bài ra , ta có :

 \(x.\dfrac{x}{3}=48\)

\(\Leftrightarrow x^2=24\)

\(\Leftrightarrow x=2\sqrt{6}\)

MÌnh nhầm sửa đoạn cuối nhé : \(x.\dfrac{x}{3}=48\)

\(\Leftrightarrow x^2=144\)

\(\Leftrightarrow x=12\)

x = 2

$3^x+4^x=5^x\iff {\left(\frac{3}{5}\right)}^x+{\left(\frac{4}{5}\right)}^x=1$

TH 1 : x<2 $\Rightarrow {\left(\frac{3}{5}\right)}^x+{\left(\frac{4}{5}\right)}^x>{\left(\frac{3}{5}\right)}^2+{\left(\frac{4}{5}\right)}^2=1$

TH2 : x=2 thỏa

TH3 : x>2

$\Rightarrow {\left(\frac{3}{5}\right)}^x+{\left(\frac{4}{5}\right)}^x<{\left(\frac{3}{5}\right)}^2+{\left(\frac{4}{5}\right)}^2=1$

\(\left(2x-1\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

(2x - 1)² = 81

(2x - 1)² = 9²

2x - 1 = 9

2x = 9 + 1

2x = 10

x = 10 : 2

x = 5

\(\left(2x-3\right)^2=9\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\)

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

(2x-3)² = 9

(2x-3)² = 3²

2x-3=3

2x=3+3

2x=6

x=6:2

x=3

Lời giải:

$(2x-3)^2=49=7^2=(-7)^2$

$\Rightarrow 2x-3=7$ hoặc $2x-3=-7$

$\Rightarrow x=5$ hoặc $x=-2$

(2x-3)²=49
(2x-3)²=7²
2x-3=7
​2x=7+3
2x=10
x=10:2
x=5
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