a) \(8x+56:14=60\)

\(\Rightarrow8x+4=60\)

\(\Rightarrow8x=56\)

\(\Rightarrow x=\dfrac{56}{8}\)

\(\Rightarrow x=7\)

b) Mình làm rồi nhé !

c) \(41-2^{x+1}=9\)

\(\Rightarrow2^{x+1}=41-9\)

\(\Rightarrow2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

d) \(3^{2x-4}-x^0=8\)

\(\Rightarrow3^{2x-4}-1=8\)

\(\Rightarrow3^{2x-4}=9\)

\(\Rightarrow3^{2x-4}=3^2\)

\(\Rightarrow2x-4=2\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

g) \(65-4^{x+2}=2014^0\)

\(\Rightarrow65-4^{x+2}=1\)

\(\Rightarrow4^{x+2}=64\)

\(\Rightarrow4^{x+2}=4^3\)

\(\Rightarrow x+2=3\)

\(\Rightarrow x=1\)

i) \(120+2\left(4x-17\right)=214\)

\(\Rightarrow2\left(4x-17\right)=214-120\)

\(\Rightarrow2\left(4x-17\right)=94\)

\(\Rightarrow4x-17=47\)

\(\Rightarrow4x=47+17\)

\(\Rightarrow4x=64\)

\(\Rightarrow x=16\)

a: \(8x+56:14=60\)

=>8x+4=60

=>8x=60-4=56

=>x=56/8=7

b: \(5^{2x-3}-2\cdot5^2=5^2\cdot3\)

=>\(5^{2x-3}=5^2\cdot3+2\cdot5^2=5^3\)

=>2x-3=3

=>2x=6

=>x=3

c: \(41-2^{x+1}=9\)

=>\(2^{x+1}=41-9=32\)

=>x+1=5

=>x=4

d: \(3^{2x-4}-x^0=8\)

=>\(3^{2x-4}-1=8\)

=>\(3^{2x-4}=8+1=9\)

=>2x-4=2

=>2x=6

=>x=3

g: \(65-4^{x+2}=2014^0\)

=>\(65-4^{x+2}=1\)

=>\(4^{x+2}=65-1=64\)

=>x+2=3

=>x=1

i: 120+2(4x-17)=214

=>2(4x-17)=214-120=94

=>4x-17=94/2=47

=>4x=64

=>\(x=\dfrac{64}{4}=16\)

\(\frac{sin^22x+4sin^2x-4}{1-8sin^2x-cos4x}=\frac{4sin^2x.cos^2x-4\left(1-sin^2x\right)}{1-8sin^2x-\left(1-2sin^22x\right)}=\frac{4sin^2x.cos^2x-4cos^2x}{2sin^22x-8sin^2x}\)

\(=\frac{-4cos^2x\left(1-sin^2x\right)}{8sin^2x.cos^2x-8sin^2x}=\frac{-4cos^2x.cos^2x}{-8sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)

\(\frac{cos2x}{cot^2x-tan^2x}=\frac{cos2x.sin^2x.cos^2x}{cos^4x-sin^4x}=\frac{\left(cos^2x-sin^2x\right).\left(2sinx.cosx\right)^2}{4\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)}=\frac{1}{4}sin^22x\)

a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)

=3x+4

b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)

\(=\dfrac{x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=x+3\)

c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)

d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)

=7x+1

e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)

\(=\dfrac{x^2\left(2x+1\right)+3x\left(2x+1\right)+6\left(2x+1\right)}{2x+1}=x^2+3x+6\)

f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)

g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)

Chọn đáp án A

8: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

3) \(...\Rightarrow2^x\left(2^3+1\right)=36\)

\(\Rightarrow2^x.9=36\)

\(\Rightarrow2^x=4\)

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

4) \(...\Rightarrow4^{x+1}-4^x=12\)

\(\Rightarrow4^x\left(4-1\right)=12\)

\(\Rightarrow4^x.3=12\)

\(\Rightarrow4^x=4=4^1\Rightarrow x=1\)

5) \(...\Rightarrow5^{x+1}\left(5^2-1\right)=3000\)

\(\Rightarrow5^{x+1}.24=3000\)

\(\Rightarrow5^{x+1}=125\)

\(\Rightarrow5^{x+1}=5^3\)

\(\Rightarrow x+1=3\)

\(\Rightarrow x=2\)

6) Bạn xem lại đề

a. \(2^x.2^3+2^x=36\)

\(2^x\left(2^3+1\right)=36\)

\(2^x.9=36\)

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

b. \(4^x.4^1-\left(2^2\right)^x=12\)

\(4^x.4-4^x=12\)

\(4^x\left(4-1\right)=12\)

\(4^x.3=12\)

\(4^x=4\)

x = 1

c. \(5^x.5^3-5^x.5^1=3000\)

\(5^x\left(5^3-5^1\right)=3000\)

\(5^x.120=3000\)

\(5^x=25\)

x = 2

d. \(4^{x+1}=2^{2x}\)

\(4^x.4=\left(2^2\right)^x\)

\(4^x.4=4^x\)

Có vẻ như câu 4 này để bài thiếu 

1a.

Đặt \(5x+6=u\)

\(cos2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

\(\Leftrightarrow1-2sin^2u+3sinu=2\)

\(\Leftrightarrow2sin^2u-3sinu+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)

2a.

\(cos^2x-sin^2x+sin^2x+2cosx+1=0\)

\(\Leftrightarrow cos^2x+2cosx+1=0\)

\(\Leftrightarrow\left(cosx+1\right)^2=0\)

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\)

a/

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x-2\left(1-sin^22x\right)=0\)

\(\Leftrightarrow1-\frac{1}{2}\left(cos6x+cos2x\right)-2cos^22x=0\)

\(\Leftrightarrow1-cos4x.cos2x-2cos^22x=0\)

\(\Leftrightarrow2cos^22x-1+cos4x.cos2x=0\)

\(\Leftrightarrow cos4x+cos4x.cos2x=0\)

\(\Leftrightarrow cos4x\left(cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

b/

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}\right)+\sqrt{3}sin2x=1\)

\(\Leftrightarrow cos2x.cos\left(\frac{\pi}{3}\right)-sin2x.sin\left(\frac{\pi}{3}\right)+\frac{1}{2}+\sqrt{3}sin2x=1\)

\(\Leftrightarrow\frac{1}{2}cos2x+\frac{\sqrt{3}}{2}sin2x=\frac{1}{2}\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

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

c/

\(\Leftrightarrow5+5cosx=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)

\(\Leftrightarrow3+5cosx=sin^2x-cos^2x\)

\(\Leftrightarrow3+5cosx=1-cos^2x-cos^2x\)

\(\Leftrightarrow2cos^2x+5cosx+2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)