Đề là: \(2sin^22x-3cos2x+6sin^2x-9=0\) đúng không nhỉ?

\(\Leftrightarrow2\left(1-cos^22x\right)-3cos2x+3\left(1-cos2x\right)-9=0\)

\(\Leftrightarrow-2cos^22x-6cos2x-4=0\)

\(\Leftrightarrow cos^22x+3cos2x+2=0\)

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

\(\Rightarrow...\)

⇔2^x=2⁸:2

⇔2^x=2⁷

⇒x=7

⇔2^x=2⁸:2

⇔2^x=2⁷

⇒x=7

2^x=2⁸:2

2^x=2⁷

 x=7

Vậy x = 7

2^7

2^x+1 = 2⁸

=> x + 1 = 8

=> x = 8 - 1 = 7

C1

\(2.2^x=2^8\)

\(2^x=2^7\)

\(\Rightarrow x=7\)

C2:

\(2.2^x=2^8\)

\(\Rightarrow1+x=8\)

  \(\Rightarrow x=7\)

a: =>3x-2x=-11-9

=>x=-20

c: \(\Leftrightarrow\left(2x+3\right)\left(x^2+3\right)=2\left(2x+3\right)\)

=>2x+3=0

hay x=-3/2

\(5^{x+1}-5^x=2.2^x+8.2^x\\ \Leftrightarrow5^x.5-5^x=2.2^x+8.2^x\\ \Leftrightarrow5^x\left(5-1\right)=2^x\left(2+8\right)\\ \Leftrightarrow5^x.4=2^x.10\\ \Leftrightarrow5^x:2^x=\dfrac{5}{2}\\ \Leftrightarrow\left(\dfrac{5}{2}\right)^x=\dfrac{5}{2}\\ \Rightarrow x=1\)

a: 3x+9=2x-11

=>3x-2x=-11-9

=>x=-20

b: \(\dfrac{2x-3}{5}-2=\dfrac{2-x}{4}\)

=>4(2x-3)-20=5(2-x)

=>8x-12-20=10-5x

=>8x-32=10-5x

=>13x=42

hay x=42/13

d: Ta có: \(3^{x+2}\cdot2^{x-1}=36^3\)

\(\Leftrightarrow3^x\cdot9\cdot2^x\cdot\dfrac{1}{2}=6^6\)

\(\Leftrightarrow6^x=6^6:\dfrac{9}{2}=10368\)

=> Đề sai rồi bạn

1/2.2x+4.2x=9.2x

2x.(1/2+4)=9.2x

2x.9/2=9.2x

2x:2x=9:9/2

x=2

vậy x=2

\(a,\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{5}{4}\\\dfrac{2}{3}-x=\dfrac{5}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{23}{12}\\x=-\dfrac{7}{12}\end{matrix}\right.\\ b,\Rightarrow0,1x\cdot1,35=0,2\cdot1,25=0,25\\ \Rightarrow0,135x=0,25\Rightarrow x=\dfrac{50}{27}\\ c,ĐK:x\ge0\\ PT\Leftrightarrow-2\sqrt{x}=-6\Leftrightarrow x=9\left(tm\right)\\ d,\Leftrightarrow3^{x+2}\cdot2^{x-1}=\left(3^2\cdot2^2\right)^3=3^6\cdot2^6\\ \Leftrightarrow\left\{{}\begin{matrix}x+2=6\\x-1=6\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

a) I\(x - \dfrac{2}{3} \)I \(- \dfrac{5}{4} = 0\)

    I\(x - \dfrac{2}{3} \)I = 0 +\(\dfrac{5}{4} \)

    I\(x - \dfrac{2}{3} \)I = \(\dfrac{5}{4} \)

    I\(x - \dfrac{2}{3} \)I = \(\begin{cases} \dfrac{5}{4} + \dfrac{2}{3}\\ \dfrac{-5}{4} + \dfrac{2}{3}\\ \end{cases} \)

    \(x \) = \(\begin{cases} \dfrac{23}{12} \\ \dfrac{-7}{12} \end{cases} \)

b) \(\dfrac{1,35}{0,2} = \dfrac{1,25}{0,1x}\)

     1,35 . 0,1x = 1,25 . 0,2

     1,35 . 0,1x = 0,25

      0,1x = 0,25 : 1,35 

      \(\dfrac{1}{10}\)x = \(\dfrac{5}{27}\)

       x = \(\dfrac{5}{27} : \dfrac{1}{10}\)

       x = \(\dfrac{5}{27} . \dfrac{10}{1}\)

       x = \(\dfrac{50}{27}\)