c/ 2x - 1 = \(5^{98}:5^{96}\)

2x - 1 = \(5^2\) = 25

2x = 25 + 1 = 26

x = 26 : 2

x = 13

d/ 7x + 3 = \(3^5.2^3.9\)

7x + 3 = \(3^5.3^2.8=3^7.8=2187.8\)

7x + 3 = \(17496\)

7x = 17496 - 3 = 17493

x = 17493 : 7

x = 2499

e/\(2^{2x+6}=1\)

\(2^{2x+6}=2^0\)

2x + 6 = 0

2x = 0 - 6 = - 6

x = - 6 : 2

x = - 3

j/ \(2^x=8\)

\(2^x=2^3\)

x = 3

g/ \(2^x:2^3=16\)

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

x - 3 = 4

x = 4 + 3

x = 7

h/ \(2^x+2^{x+1}+2^{x+2}=56\)

\(2^x\left(1+2+2^2\right)\) = 56

\(2^x.7=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

x = 3

Bài a, b thiên phong giải r, mk chỉ làm những bài còn lại thôi. Chúc bạn học tốt!!!

a, \(x+2^3=5^2\Rightarrow x+8=25\Rightarrow x=17\)

b, \(2x+3=7^2\Rightarrow2x+3=49\Rightarrow x=\dfrac{49-3}{2}=23\)

Còn lại tự làm.

A.(x+2)^x-1=15⁰

=>A.(x+2)^x-1=1

=> x + 2 = 1 hoặc x + 2 = -1 hoặc x - 1 = 0

=> x = -1 hoặc x = -3 hoặc x = 1.

B. (5-x)^x=1(x<5)

=> 5 - x = 1 hoặc 5 - x = -1 hoặc x = 0

=> x = 4 hoặc x = 6 hoặc x = 0.

C.15^x-2=225

=> 15^x-2=15²

=> x - 2 = 2 => x = 4.

D.(x+2)².(x+1)=64

=>(x+2).(x+2).(x+1)=64 = 1.2.32 = 2.2.16 = ...

Mà x + 2 và x + 2 và x + 1 chỉ hơn kém nhau 1 đơn vị nên không có x nào thỏa mãn.

E.(x-5)³.(x-5)=16

=>(x-5)⁴=16=2⁴

=>x-5=2=>x=7.

a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)

\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)

=>x=10

b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)

hay \(x\in\left\{0;1;2\right\}\)

c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)

\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)

\(\Leftrightarrow6-x=0\)

hay x=6

Bài 1 tự làm!

Bài 2:

a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)

b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)

\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)

c, Đề chưa rõ

d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)

e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)

\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)

f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x⁰ = 1)

\(\Rightarrow x+1=1\Rightarrow x=0\)

Làm phần c nè!

c, \(\left(x-5\right)^4=\left(x-5\right)^6\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)

\(\Rightarrow\left(x-5\right)^4.\left(x-5\right)^2=0\)

\(\Rightarrow x-5=0\Rightarrow x=5\)

làm con trai cũng có nhìu cái ko được như con gái đâu kiss_rain_and_you

6)\(3.\left(x+1\right)-2^3.2=11\)

\(3\left(x+1\right)-2^4=11\)

\(3.\left(x+1\right)=11+16\)

3.(x+1)=27

x+1=27:3

x+1=9

x=9-1

x=8

Vậy x=8

7) \(180-2.\left(x+5\right)^2=130\)

\(2.\left(x+5\right)^2=180-130\)

2.(x+5)²=50

(x+5)²=50:2

(x+5)²=25

x+5=5  hoặc x+5=-5

x=5-5                x=-5-5

x=0                     x=-10

Vậy x=0 hoặc x=-10

\(a,\left(7x-11\right)^3=2^5.5^2+200.\)

\(\left(7x+11\right)^3=32.25+200.\)

\(\left(7x+11\right)^3=800+200.\)

\(\left(7x-11\right)^3=1000.\)

\(\left(7x-11\right)^3=10^3.\)

\(\Rightarrow7x-11=10.\)

\(\Rightarrow x=\left(10+11\right):3=7\in Z.\)

Vậy.....

\(b,3^x+25=26.2^2+2.3^0.\)

\(3^x+25=26.4+2.\)

\(3^x+25=104+2.\)

\(3^x+25=106.\)

\(3^x=106-25.\)

\(3^x=81.\)

\(3^x=3^4\Rightarrow x=4\in Z.\)

Vậy.....

\(c,2^x+3.2=64.\)(có vấn đề).

\(d,5^{x+1}+5^x=750.\)

\(5^x.5^1+5^x+1=750.\)

\(5^x\left(5^1+1\right)=750.\)

\(5^x\left(5+1\right)=750.\)

\(5^x.6=750.\)

\(5^x=750:6.\)

\(5^x=125.\)

\(5^x=5^3\Rightarrow x=3\in Z.\)

Vậy.....

\(e,x^{15}=x.\)

\(\Rightarrow x\left(x^{14}-1\right)=0\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right..\)

\(f,\left(x-5\right)^4=\left(x-5\right)^6.\)

\(\Leftrightarrow\left(x-5\right)^4-\left(x-5^6\right)=0.\)

\(\Leftrightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0.\)

\(\Leftrightarrow\left(x-5\right)^4\left(1-x+5\right)\left(1+x-5\right)=0.\)

\(\Leftrightarrow\left(x-5\right)^4\left(6-x\right)\left(x-4\right)=0.\)

\(\Leftrightarrow\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\in Z.\)

\(6-x=0\Rightarrow x=6\in Z.\)

\(x-4=0\Rightarrow x=4\in Z.\)

Vậy.....

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