\(1,\\ a,2^x=16=2^4\Rightarrow x=4\\ b,3^{x+1}=9^x=3^{2x}\\ \Rightarrow x+1=2x\Rightarrow x=1\\ c,2^{3x+2}=4^{x+5}=2^{2\left(x+5\right)}\\ \Rightarrow3x+2=2x+10\Rightarrow x=8\\ d,3^{2x-1}=243=3^5\\ \Rightarrow2x-1=5\Rightarrow x=3\\ 2,\\ a,2^{225}=8^{75}< 9^{75}=3^{150}\\ b,2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\\ c,99^{20}=\left(99^2\right)^{10}< \left(99\cdot101\right)^{10}=9999^{10}\\ 3,\\ a,12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}=\left(2\cdot3^2\right)^{16}=18^{16}\\ b,75^{20}=\left(3\cdot5^2\right)^{20}=3^{20}\cdot5^{40}=\left(3^{20}\cdot5^{10}\right)\cdot5^{30}=\left(3^2\cdot5\right)^{10}\cdot5^{30}=45^{10}\cdot5^{30}\)

Bài 1:

a) \(\Rightarrow2^x=2^4\Rightarrow x=4\)

b) \(\Rightarrow3^{x+1}=3^{2x}\Rightarrow x+1=2x\Rightarrow x=1\)

c) \(\Rightarrow2^{3x+2}=2^{2x+10}\Rightarrow3x+2=2x+10\Rightarrow x=8\)

d) \(\Rightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow x=3\)

Bài 2:

a) \(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)

b) \(2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\)

c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)

Bài 3:

a) \(12^8.9^{12}=\left(4.3\right)^8.9^{12}=4^8.3^8.9^{12}=2^{16}.9^4.9^{12}=2^{16}.9^{16}=\left(2.9\right)^{16}=18^{16}\)

b) \(75^{20}=\left(75^2\right)^{10}=5625^{10}=\left(45.125\right)^{10}=45^{10}.125^{10}=45^{10}.5^{30}\)

a) 2x = 16 <=>x=8

b) 3x+1 = 9x <=>9x-3x=1

<=>6x=1 <=>x=1/6

c) 23x+2 = 4x+5 <=>23x-4x=5-2

<=>19x=3 <=>x=3/19

d) 32x-1 = 243 <=>32x=244

<=>x=61/8

a/ 2x=16

x=8

b/ 3x+1=9x

3x-9x=-1

-6x=-1

x=1/6

c/ 23x+2=4x

23x-4x=-2

19x=-2

x=-2/19

d/ 32x-1=243

32x=244

x=61/8

cho c xl nha em :( c ko bt em la ai thật 

.

a)        \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)

\(\Leftrightarrow\)\(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left(-x-2\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)

Vậy...

chẳng có đề bài biết làm ntn

1.

\(\Leftrightarrow\left(1-cos6x\right)cos2x+1-cos2x=0\)

\(\Leftrightarrow cos2x-cos2x.cos6x+1-cos2x=0\)

\(\Leftrightarrow\frac{1}{2}\left(cos8x-cos4x\right)-1=0\)

\(\Leftrightarrow2cos^24x-cos4x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-1\\cos4x=\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow4x=\pi+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

2.

\(\Leftrightarrow1+cos6x+2cos^22x=1-cos2x\)

\(\Leftrightarrow cos6x+cos2x+2cos^22x=0\)

\(\Leftrightarrow cos4x.cos2x+cos^22x=0\)

\(\Leftrightarrow cos2x\left(cos4x+cos2x\right)=0\)

\(\Leftrightarrow cos2x\left(2cos^22x+cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-1\\cos2x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)

3.

Đặt \(\frac{x}{6}=t\Rightarrow\frac{1}{4}+cos^22t=\frac{1}{2}sin^23t\)

\(\Leftrightarrow1+4cos^22t=1-cos6t\)

\(\Leftrightarrow cos6t+4cos^22t=0\)

\(\Leftrightarrow4cos^32t+4cos^22t-3cos2t=0\)

\(\Leftrightarrow cos2t\left(4cos^22t+4cos2t-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2t=0\\cos2t=\frac{1}{2}\\cos2t=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{\pi}{4}+\frac{k\pi}{2}\\t=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}=\frac{\pi}{4}+\frac{k\pi}{2}\\\frac{x}{3}=\frac{\pi}{6}+k\pi\\\frac{x}{3}=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow x=...\)