Bài 3: 

a: Ta có: \(3x^2=75\)

\(\Leftrightarrow x^2=25\)

hay \(x\in\left\{5;-5\right\}\)

b: Ta có: \(2x^3=54\)

\(\Leftrightarrow x^3=27\)

hay x=3

Bài 2: 

b: Ta có: \(30-3\cdot2^n=24\)

\(\Leftrightarrow3\cdot2^n=6\)

\(\Leftrightarrow2^n=2\)

hay n=1

c: Ta có: \(40-5\cdot2^n=20\)

\(\Leftrightarrow5\cdot2^n=20\)

\(\Leftrightarrow2^n=4\)

hay n=2

d: Ta có: \(3\cdot2^n+2^n=16\)

\(\Leftrightarrow2^n\cdot4=16\)

\(\Leftrightarrow2^n=4\)

hay n=2

a) \(2^3.2^2+7^4:7^2\)

\(=2^5+7^2\)

\(=32+49\)

b) \(6^2.47+6^2.53\)

\(=6^2\left(47+53\right)\)

\(=36.100\)

\(=3600\)

1) \(6x+3=0\Leftrightarrow6x=-3\Leftrightarrow x=-\dfrac{1}{2}\)

2) \(-5x-7=0\Leftrightarrow-5x=7\Leftrightarrow x=-\dfrac{7}{5}\)

3) \(\left(3x-2\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\5-x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=5\end{matrix}\right.\)

4) \(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\)

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

6) \(x^4+8=0\)( vô lý do \(x^4+8\ge8>0\))

Vậy \(S=\varnothing\)

Bài 3: 

a: Ta có: 60-3(x-2)=51

\(\Leftrightarrow x-2=3\)

hay x=5

b: Ta có: \(4x-20=25:2^2\)

\(\Leftrightarrow4x=\dfrac{25}{4}+20=\dfrac{105}{4}\)

hay \(x=\dfrac{105}{16}\)

c: Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)

\(\Leftrightarrow288:\left(x-3\right)^2=50-48=2\)

\(\Leftrightarrow\left(x-3\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

a) \(\dfrac{64}{85}\) và \(\dfrac{73}{81}\)

do \(\left\{{}\begin{matrix}64< 73\\85>81\end{matrix}\right.\)\(\Rightarrow\dfrac{64}{85}< \dfrac{73}{81}\)

b) làm tương tự bài a

c) tương tự câu a

d) \(\dfrac{25}{26}\)và\(\dfrac{25251}{26261}\)

ta có \(\dfrac{25}{26}=\dfrac{25.1010}{26.1010}=\dfrac{25250}{26260}\)

so sánh kq vừa tìm đc và 25251/26261 là đc