Question

1. ( 3^x - 5 )⁴ = 2⁸

2. 3^x+2 + 3^x = 10² - 10

3. 3^{x + 1} - 3^x = 4 * 5 - 2

4. 5^{x + 2} - 5^x = 10³ : 2 + 10²

5. 4^x + 4^{x - 1} = 84 : 7

* là dấu nhân , còn x là chữ x chứ ko phải dấu nhân đâu ạ !! MỌI NGƯỜI GIÚP EM NHANH Ạ , MAI EM NỘP Ạ ! Giải theo kiểu đầy đủ , lớp 6 nha !

Answer 1

Quy ước toán học dấu x = dấu .

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1. (3^x - 5)⁴ = 2⁸

<=> (3^x - 5)⁴ = 4⁴

=> Ta có 2 trường hợp :

* TH1 : \(3^x-5=4\Rightarrow3^x=9\Rightarrow x=2\)

* TH2 : \(3^x-5=-4\Rightarrow3^x=1\Rightarrow x=0\)

Vậy x=1 hoặc x=0

Answer 2

a, \(\left(3^x-5\right)^4=2^8\)

\(\Rightarrow\left(3^x-5\right)^4=\left(2^2\right)^4\)

\(\Rightarrow\left(3^x-5\right)^4=4^4\)

\(\Rightarrow3^x-5=4\Rightarrow3^x=1=3^0\)

\(\Rightarrow x=0\)

b, \(3^{x+2}+3^x=10^2-10\)

\(\Rightarrow3^x.\left(3^2+1\right)=100-10\)

\(\Rightarrow3^x.10=90\Rightarrow3^x=9=3^2\)

\(\Rightarrow x=2\)

c, \(3^{x+1}-3^x=4.5-2\)

\(\Rightarrow3^x.\left(3-1\right)=20-2\)

\(\Rightarrow3^x.2=18\Rightarrow3^x=9=3^2\)

\(\Rightarrow x=2\)

Chúc bạn học tốt!!! Mấy câu còn lại làm tương tự!

Answer 3

1) \(\left(3^x-5\right)^4=2^8\)

\(\Leftrightarrow\left[{}\begin{matrix}3^x-5=4\\3^x-5=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3^x=9\\3^x=1\end{matrix}\right.\)

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

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

Vậy \(x_1=0;x_2=3\)

b) \(3^{x+2}+3^x=10^2-10\)

\(\Leftrightarrow\left(3^2+1\right)\cdot3^x=100-10\)

\(\Leftrightarrow\left(9+1\right)\cdot3^x=90\)

\(\Leftrightarrow3^x=9\)

\(\Leftrightarrow3^x=3^2\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

c) \(3^{x+1}-3^x=4\cdot5-2\)

\(\Leftrightarrow\left(3-1\right)\cdot3^x=20-2\)

\(\Leftrightarrow2\cdot3^x=18\)

\(\Leftrightarrow3^x=9\)

\(\Leftrightarrow3^x=3^2\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

d) \(5^{x+2}-5^x=10^3:2+10^2\)

\(\Leftrightarrow\left(5^2-1\right)\cdot5^x=1000:2+100\)

\(\Leftrightarrow\left(25-1\right)\cdot5^x=500+100\)

\(\Leftrightarrow24\cdot5^x=600\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

e) \(4^x+4^{x-1}=84:7\)

\(\Leftrightarrow\left(4+1\right)\cdot4^{x-1}=12\)

\(\Leftrightarrow5\cdot4^{x-1}=12\)

\(\Leftrightarrow4^{x-1}=\dfrac{12}{5}\)

giải tiếp...

Answer 4

2 .

Ta có : 3^x+2 + 3^x = 10² - 10

<=> 3^x.3² +3^x = 10.10-10

<=> 3^x(9+1) = 10(10-1)

<=> 3^x.10=10.9

<=> 3^x = 9

<=> x = 2

Answer 5

3 .

Ta có : 3^x+1 - 3^x = 4.5-2

<=> 3^x . 3 - 3^x = 20 - 2

<=> 3^x ( 3-1) = 18

<=> 3^x = 9

<=> x=2

Answer 6

4 .

Ta có : 5^x+2 - 5^x = \(\dfrac{10^3}{2}\) + 10²

<=> 5^x.5² -5^x = 10^{2 .\(\dfrac{10}{2}+10^2\)}

<=> 5^x ( 5² -1 ) = 10² ( 5+1)

<=> 5^x . 24 = 10².6

<=> 5^x = \(\dfrac{600}{24}\)

<=> 5^x = 25

<=> x=2

5 .

Ta có : 4^x - 4^x-1 = \(\dfrac{84}{7}\)

<=> 4^x - 4^x : 4 = 12

<=> 4^x - 4^x . \(\dfrac{1}{4}=12\)

<=> 4^x ( 1- \(\dfrac{1}{4}\)) = 12

<=> 4^x = 12: ( 1- \(\dfrac{1}{4}\) )

<=> 4^x = 16

<=> x=2

Answer 7

1)\(\left(3^x-5\right)^4=2^8\Leftrightarrow\left(3^x-5\right)^4=\left(2^2\right)^4=4^4\)

\(\Leftrightarrow3^x-5=4\Leftrightarrow3^x=9\Leftrightarrow x=2\)

2) \(3^{x+2}+3^x=10^2-10\Leftrightarrow3^x.3^2+3^x=90\)

\(\Leftrightarrow3^x\left(3^2+1\right)=90\Leftrightarrow3^x.10=90\Leftrightarrow3^x=9\Leftrightarrow x=2\)
3) \(3^{x+1}-3^x=4.5-2\Leftrightarrow3^x.3-3^x=18\)

\(\Leftrightarrow3^x\left(3-1\right)=18\Leftrightarrow3^x.2=18\Leftrightarrow3^x=9\Leftrightarrow x=2\)

4)\(5^{x+2}-5^x=10^3:2+10^2\Leftrightarrow5^x.5^2-5^x=600\)

\(\Leftrightarrow5^x\left(5^2-1\right)=600\Leftrightarrow5^x.24=600\Leftrightarrow5^x=25\Leftrightarrow x=2\)

5) \(4^x+4^{x-1}=84:7\Leftrightarrow4^x+4^x:4=12\)

\(\Leftrightarrow4^x+4^x=12.4=48\Leftrightarrow2\left(4^x\right)=48\Leftrightarrow4^x=24\Leftrightarrow x\in\left\{\varnothing\right\}\)

Answer 8

a) \(\left(3^x-5\right)^4=2^8\)

\(\Rightarrow\left(3^x-5\right)^4=\left(2^2\right)^4\)

\(\Rightarrow\left(3^x-5\right)^4=4^4\)

\(\Rightarrow3^x-5=4\)

\(\Rightarrow3^x=4+5\)

\(\Rightarrow3^x=9\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

Vậy \(x=3\).

b) \(3^{x+2}+3^x=10^2-10\)

\(\Rightarrow3^x\left(3^2+1\right)=10\left(10-1\right)\)

\(\Rightarrow3^x\cdot10=90\)

\(\Rightarrow3^x=\dfrac{90}{10}=9\)

\(\Rightarrow3^x=3^2\)

\(\Rightarrow x=2\)

Vậy \(x=2\).

c) \(3^{x+1}-3^x=4\cdot5-2\)

\(\Rightarrow3^x\left(3-1\right)=20-2\)

\(\Rightarrow3^x\cdot2=18\)

\(\Rightarrow3^x=\dfrac{18}{2}=9\)

\(\Rightarrow3^x=3^2\)

\(\Rightarrow x=2\)

Vậy \(x=2\).

d) \(5^{x+2}-5^x=10^3:2+10^2\)

\(\Rightarrow5^x\left(5^2-1\right)=1000:2+100\)

\(\Rightarrow5^x\cdot24=600\)

\(\Rightarrow5^x=\dfrac{600}{24}=25\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

Vậy \(x=2\).

e) \(4^x+4^{x-1}=84:7\)

\(\Rightarrow4^x:1+4^x:4=12\)

\(\Rightarrow4^x\cdot1+4^x\cdot\dfrac{1}{4}=12\)

\(\Rightarrow4^x\left(1+\dfrac{1}{4}\right)=12\)

\(\Rightarrow4^x\cdot\dfrac{5}{4}=12\)

\(\Rightarrow4^x=12:\dfrac{5}{4}=12\cdot\dfrac{4}{5}=9.6\)

\(\Rightarrow x\in\left\{\varnothing\right\}\)

Vậy \(x\in\left\{\varnothing\right\}\).