a) \(\left(2x-3\right)^2=16\)

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

\(2x-3=4\)

\(2x=7\)

\(x=\dfrac{7}{2}=3,5\)

b) \(\left(3x-2\right)^5=-243\)

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

\(3x-2=-3\)

\(3x=-1\)

\(3x=-\dfrac{1}{3}\)

c) \(\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)

\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\left(x-7\right)^{x+1}\times\left[1-\left(x-7\right)^{10}\right]=0\)

\(\left(x-7\right)^{x+1}=0\) ; \(1-\left(x-7\right)^{10}=0\)

\(x-7=0;\left(x-7\right)^{10}=1\)

\(x=7;\left(x-7=1;x-7=-1\right)\)

\(x=7;x=8;x=6\)

a, (2\(x\) - 3)² = 16

     \(\left[{}\begin{matrix}2x-3=-4\\2x-3=4\end{matrix}\right.\)

     \(\left[{}\begin{matrix}2x=-1\\2x=7\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(x\in\){ - \(\dfrac{1}{2}\); \(\dfrac{7}{2}\)}

b, (3\(x\) - 2)⁵   = -243 

    ( 3\(x\) - 2)⁵ =  (-3)⁵

 3\(x\)    -  2     = -3

     3 \(x\)           = -1

        \(x\)          = - \(\dfrac{1}{3}\)

Vậy \(x\) = -\(\dfrac{1}{3}\)

c, \(\left(x-7\right)\)^\(x+1\)  = (\(x-7\))^\(x+11\)

    (\(x-7\))^{\(^{x+1}\)}.( \(\left(x-7\right)^{10}\) -  1 ) = 0

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

         ^{\(\left[{}\begin{matrix}x=7\\x-7=-1\\x-7=1\end{matrix}\right.\)}

         \(\left[{}\begin{matrix}x=7\\x=6\\x=8\end{matrix}\right.\)

Vậy \(x\in\){ 6; 7; 8}

Câu a mình thiếu 2 trường hợp 

\(\left(2x-3\right)^2=16\)

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

\(2x-3=4;2x-3=-4\)

\(2x=7;2x=-1\)

\(x=\dfrac{7}{2};x=-\dfrac{1}{2}\)

(2x - 3)²=16

(2x-3)²=4²

=>2x-3=4 hoặc 2x-3=-4

=>2x=7 hoặc 2x=-1

=>x=7/2 hoặc x=-1/2

( 3x - 2 )⁵ = -24³ đề đúng phải là ( 3x - 2 )⁵=-243

a)  \(\left(2x-3\right)^2=16\)

=>  \(2x-3=4\)

=> \(2x=4+3=7\)

=> \(x=\frac{7}{2}=3,5\)

b) \(\left(3x-2\right)^5=-243\)

=> \(3x-2=-3\)

=> \(3x=-3+2=-1\)

=> \(x=-\frac{1}{3}\)

a) (2x-3)^2=16

có 2 trường hợp:

_ 2x-3=-4 suy ra x=1/2

_ 2x-3=4 suy ra x=7/2

vậy x=1/2 hoặc x=7/2

b) tương tự câu a) nhưng chỉ có một trường hợp là 3x-2=-3 thôi. coi chừng bị lừa

a) 2^x = 16   b) 3^{x + 1} = 9^x

2^x = 2⁴            3^{x + 1} = 3^2x

x = 4                 x + 1 = 2x

                         x      = 1

c) 2^{3x + 2} = 4^{x + 2}

2^{3x + 2} = 2^{2(x + 2)}

3x + 2 = 2(x + 2)

3x + 2 = 2x + 4

x         = 2

d) 3^{2x - 1} = 243

3^{2x - 1} = 3⁵

2x - 1 = 5

2x = 6

x = 3

b) \(\left(3x-2\right)^5=-243\)

\(\Rightarrow\left(3x-2\right)^5=\left(-3\right)^5\)

\(\Rightarrow3x-2=-3\Rightarrow x=\dfrac{-1}{3}\)

c) Vì \(\left(2x-5\right)^{2000}\ge0\forall x;\left(3y+4\right)^{2002}\ge0\forall y\)

\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\ge0\forall x,y\)

Mà theo bài ra \(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)

\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}=0\)

\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right........\)

a)(2x-3)²=16

=>2x-3=4 hoặc 2x-3=-4

<=>2x=7 hoặc 2x=-1

<=>x=7/2 hoặc x=-1/2

b)(3x-2)⁵=243=3⁵

=>3x-2=3

=>3x=5

=>x=5/3

c)(7x+2)^-1=5²

<=>\(\frac{1}{7x+2}=25\)

<=>25(7x+2)=1

<=>175x+50=1

<=>175x=-49

<=>x=-49:175

<=>x=-7/25

d)(x-3/4)⁴=81=3⁴=(-3)⁴

=>x-3/4=3 hoặc x-3/4=-3

<=>x=3+3/4 hoặc x=-3+3/4

<=>x=15/4 hoặc x=-9/4

tìm x,biết:(7x+2)^-1=3^-2

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

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) 3⁵= 243

b)5³=125

tiếp theo bn tự tính nhé

a) \(\left(2x-1\right)^5=243\)

\(\left(2x-1\right)^5=3^5\)

\(\Rightarrow2x-1=3\)

\(2x=3+1\)

\(2x=4\)

\(x=\frac{4}{2}\)

\(x=2\)

Vậy .......

b) \(\left(3x+2\right)^3=125\)

\(\left(3x+2\right)^3=5^3\)

\(\Rightarrow3x+2=5\)

\(3x=5-2\)

\(3x=3\)

\(x=\frac{3}{3}=1\)

Vậy ..........

Mình cần gấp ạ ! * cảm ơn

b) \(3^{x+1}=9^x\)

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

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

\(\Rightarrow x+1=2x\)

\(1=2x-x\)

\(1=x\)

Vậy x=1

c) 2^3x+2 = 4^x+5                                                                  d)3^2x-1 = 243

   2^3x+2 = (2²)^x+5                                                                3^2x-1 = 3⁵

   2^3x+2 = 2^2x+10                                               => 2x-1 = 5

=> 3x+2 = 2x + 10                                                 2x = 6

   3x - 2x = 10-2                                                    x=3

    => x = 8