c)(2x-3)^2 = 25

   (2x-3)^2 = 5^2 

=> 2x-3 = 5

    2x = 8

  x= 4

a) ( x-2)³ = 27

(x-2)^3 = 3^3

x-2 = 3

x= 5

b) 2^x + 2^x+3= 144

2^x + 2^x * 8 = 144

2^x (1+8) =144

2^x = 144:9

2^x = 16

2^x = 2^4

x=4

c) ( 2-3)²= 25

Câu c sao ko có x.

d) ( 2x-1)¹⁰= ( 2x-1) ¹⁰⁰

(ko bt)

a. (x-2)³=27

<=> (x-2)³=3³

=> x-2=3 =>x=5

c.(2x-3)²=25

<=> (2x-3)²=5²

=>2x-3=5 => 2x=8 =>x=4

a) ( x-2)³ = 27

(x-2)3 = 3³

x-2 = 3

x = 5

b) 2^x + 2^x+3= 144

2^x ( 1 + 2³ ) = 144

2^x . 9 = 144

2^x = 16

2^x = 2⁴

x = 4

c) ( 2x-3)²= 25

(2x-3)² = 5²

2x-3 = 5

2x = 8

x = 4

d) ( 2x-1)¹⁰= ( 2x-1) ¹⁰⁰

Dễ thấy 2x - 1 = 0 vì 0¹⁰ = 0¹⁰⁰

Vậy x = 0,5

Nếu 2x - 1 \(\ne\)0 

( 2x - 1 )¹⁰ : ( 2x - 1 )¹⁰ = ( 2x - 1 )¹⁰⁰ : ( 2x - 1 )¹⁰

( 2x - 1 )⁹⁰ = 1

Vậy 2x - 1 = 1 hoặc - 1.

Vậy x = 1 ; 0 ; 0,5.

\(a,\left(x-5\right)\left(2x+3\right)=x^2-25\\ \Leftrightarrow a,\left(x-5\right)\left(2x+3\right)-\left(x-5\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+3-x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-8\end{matrix}\right.\\ b,\dfrac{2x}{3}+\dfrac{2x-1}{6}=\dfrac{x-1}{2}\\ \Leftrightarrow\dfrac{4x}{6}+\dfrac{2x-1}{6}-\dfrac{3\left(x-1\right)}{6}=0\\ \Leftrightarrow4x+2x-1-3x+3=0\\ \Leftrightarrow3x+2=0\\ \Leftrightarrow x=-\dfrac{2}{3}\)

a, x=2=35/2

x=log(35/2)

x=log(35)-log(20)

x=log(35)-1

b) \(\left(\sqrt{9}+\sqrt{4}\right).\sqrt{x}=10\)

\(\left(3+2\right).\sqrt{x}=10\)

\(5.\sqrt{x}=10\)

\(\sqrt{x}=2\)

\(x=\sqrt{2}\)

Áp dụng tc dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x=45\\y=60\end{matrix}\right.\\ \Leftrightarrow x+y=105\left(C\right)\)

giúp me được ko