mik biết

a) \(???\)

b) \(123x+877x=2000\)

     \(1000x=2000\)

      \(x=2000:1000\)

      \(x=2\)

c) \(2x.\left(x-10\right)=0\)

    => \(x-10=0\)

         \(x=10\)

d)\(6.\left(x+2\right)-\left(4x+10\right)=100\)

    \(6.x+12-4x+10=100\)

     \(2x+2=100\)

     \(2x=98\)

     \(x=98:2\)

     \(x=49\)

e) \(x.\left(x+1\right)=2+4+6+8+...+2500\)

     \(x.\left(x+1\right)=1563750\)

      mà ta thấy : \(1250.1251=1563750\)

=> \(x=1250\)

g)\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

     \(x.100+5050=5750\)

     \(x.100=5750-5050\)

     \(x.100=700\)

     \(x=7\)

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.

\(\left(2x-1\right)^{10}=\left(2x-1\right)^{100}\)

\(\Rightarrow\left(2x-1\right)^{100}-\left(2x-1\right)^{10}=0\)

\(\Rightarrow\left(2x-1\right)^{10}.\left[\left(2x-1\right)^{90}-1\right]=0\)

\(\Rightarrow\left(2x-1\right)^{10}=0\) hoặc \(\left(2x-1\right)^{90}-1=0\)

\(\Rightarrow2x=1\) hoặc \(\left(2x-1\right)^{90}-1=1\)

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

\(\Rightarrow x=\dfrac{1}{2}\) hoặc \(x=\left\{{}\begin{matrix}1\\0\end{matrix}\right.\)

Mk mới làm

Vào góc học tập là thấy :v

(2x - 1)¹⁰ = (2x - 1)¹⁰⁰
=> (2x -1)¹⁰ - (2x - 1)¹⁰⁰ = 0
=> (2x - 1)¹⁰.[(2x - 1)⁹⁰ - 1] = 0
=> (2x - 1)¹⁰ = 0 hoặc (2x - 1)⁹⁰ - 1 = 0
=> \(\left[{}\begin{matrix}2x=1\\\left(2x-1\right)^{90}-1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x-1=\left[{}\begin{matrix}1\\-1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\left[{}\begin{matrix}1\\0\end{matrix}\right.\end{matrix}\right.\)
@Sarutobi Hùng

a/ x = 4

a) 2^x.(1 + 2³) = 144

2^x . 9 = 144

2^x = 16

=> x = 4

b) (2x - 1)¹⁰ = (2x - 1)¹⁰⁰

(2x - 1)¹⁰⁰ - (2x - 1)¹⁰  = 0

 (2x - 1)¹⁰.[ (2x - 1)⁹⁰ - 1] = 0

=>  (2x - 1)¹⁰ = 0 hoặc  (2x - 1)⁹⁰ - 1 = 0

=> 2x = 1   hoặc    (2x - 1)⁹⁰ = 1

=> x = \(\frac{1}{2}\)  hoặc   \(2x-1=\orbr{\begin{cases}1\\-1\end{cases}}\)

=>                             \(2x=\orbr{\begin{cases}2\\0\end{cases}}\)

=> x = {\(\frac{1}{2};1;0\)}