a) 2^x . 4 = 128

<=> 2^x = 32 

<=> 2^x = 2⁵

<=> x = 5

b) x¹⁵ = x¹

<=> x¹⁵ - x = 0

<=> x(x¹⁴ - 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

c) (2x + 1)³ = 125

<=> (2x + 1)³ = 5³

<=> 2x + 1 = 5

<=> 2x = 4

<=> x = 2

d) (x - 5)⁴ = (x - 5)⁶

<=> (x - 5)⁶ - (x - 5)⁴ = 0

<=> (x - 5)⁴[(x - 5)² - 1] = 0

<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)

Khi (x - 5)⁴ = 0 => x - 5 = 0 => x = 5

Khi (x - 5)² - 1 = 0 <=> (x - 5)² = 1² <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

a, 2x . 4 = 128

=> 2x = 128 : 4 = 32

=> x = 32 : 2 = 16

Vậy x = 16

b, x15 = x 1 => Sai đề

c, (2x + 1)³ = 125

=> ( 2x + 1 ) = 5³

=> 2x + 1 = 5 

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2

a, \(\left(x+2\right)\left(x+4\right)-x^2=24\\ \Rightarrow x^2+6x+8-x^2=24\\ \Rightarrow6x+8=24\\ \Rightarrow6x=16\\ \Rightarrow x=\dfrac{8}{3}\)

b, \(\left(x+5\right)\left(x-5\right)=x^2+x\)

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

\(\Rightarrow x^2+x-x^2+25=0\\ \Rightarrow x+25=0\\ \Rightarrow x=-25\)

\(a,< =>x^2+4x+2x+8-x^2=24< =>6x+8=24< =>x=\dfrac{24-8}{6}=\dfrac{8}{3}\)

b,\(< =>x^2-25-x^2-x=0< =>-25-x=0< =>x=-25\)

c,\(< =>4x^2-9-4x^2+4x=0< =>4x-9=0< =>x=\dfrac{9}{4}\)

d,\(< =>x^3+2^3=9< =>x^3=1=>x=1\)

c,(2x+3)(2x-3)=4x(x-1)

⇔ 4x²-9=4x²-4x

⇔ 4x=9

\(\Leftrightarrow x=\dfrac{9}{4}\)

d, (x+2)(x²-2x+4)=9

⇔ x³+8=9

⇔ x³=1

⇔ x=1

Ta có : (x - 5)⁴ = (x - 5)⁶

=>  (x - 5)⁴ - (x - 5)⁶ = 0

=> (x - 5)⁴ (1 - (x - 5)²) = 0

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)=0\\\left(x-5\right)^2=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x-5=1;-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=4;6\end{cases}}\)

Vậy x = {4;5;6}

Tìm x thuộc N:

2x.4=128

2x    =128:4

2x     =32

 x       =32:2

 x       =16

b.x.15=x

   x thỏa mãn điều kiện:0,1

c.(2x+1)³=125

   (2x+1)³=5³

==>2x+1=5

      2x    =5-1

      2x     =4

        x      =4:2=2

phần d mk chưa hiểu lắm

+) Cấm ghi "nhanh lên nhé ai giải được thì tk cho"

a) 2x.4 = 128

    2x    = 128 : 4

    2x    = 32

    2x    = 2⁵

b) x15 = x

    => x = 0

c) (2x + 1)³ = 125

   (2x + 1)³ = 5³

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 2

d) (x - 5).4 = (x - 5).6

=> x = 0

Chia câu hỏi ra cho thành nhiều phần cho dễ trả lời á bạn

Giúp dới

a: Ta có: \(3x-5\ge2\left(x-6\right)-12\)

\(\Leftrightarrow3x-5\ge2x-24\)

hay \(x\ge-19\)

b: Ta có: \(2\left(5-2x\right)\ge3-x\)

\(\Leftrightarrow10-4x-3+x\ge0\)

\(\Leftrightarrow-3x\ge-7\)

hay \(x\le\dfrac{7}{3}\)

a: =>1/3x-2/5x=5

=>-1/15x=5

=>x=-75
b: =>4x=4

=>x=1

c: =>6*3^x-5*3^x=243

=>3^x=243

=>x=5

2: Tìm x

a) Ta có: x+25=40

nên x=40-25=15

Vậy: x=15

b) Ta có: 198-(x+4)=120

\(\Leftrightarrow x+4=198-120=78\)

hay x=78-4=74

Vậy: x=74

c) Ta có: \(\left(2x-7\right)\cdot3=125\)

\(\Leftrightarrow2x-7=\dfrac{125}{3}\)

\(\Leftrightarrow2x=\dfrac{125}{3}+7=\dfrac{125}{3}+\dfrac{21}{3}=\dfrac{146}{3}\)

\(\Leftrightarrow x=\dfrac{146}{3}:2=\dfrac{146}{6}=\dfrac{73}{3}\)

Vậy: \(x=\dfrac{73}{3}\)

d) Ta có: \(x+16⋮x+1\)

\(\Leftrightarrow x+1+15⋮x+1\)

mà \(x+1⋮x+1\)

nên \(15⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(15\right)\)

\(\Leftrightarrow x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

hay \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

Vậy: \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

\(a,x+25=40\\ \Rightarrow x=40-25\\ \Rightarrow x=15\\ b,198-\left(x+4\right)=120\\ \Rightarrow-\left(x+4\right)=120-198\\ \Rightarrow-\left(x+4\right)=-78\\ \Rightarrow x+4=78\\ \Rightarrow x=78-4\\ \Rightarrow x=74\\ c,\left(2x-7\right).3=125\\ \Rightarrow2x-7=\dfrac{125}{3}\\ \Rightarrow2x=\dfrac{125}{3}+7\\ \Rightarrow2x=\dfrac{146}{3}\\ \Rightarrow x=\dfrac{146}{3}:2\Rightarrow x=\dfrac{73}{3}\\ d,\left(x+16\right)⋮\left(x+1\right)\\ \Rightarrow\left[\left(x+1\right)+15\right]⋮\left(x+1\right)\\ mà:\left(x+1\right)⋮\left(x+1\right)\\ \Rightarrow15⋮\left(x+1\right)\\ \Rightarrow\left(x+1\right)\inƯ\left(15\right)\\ \Rightarrow\left(x+1\right)\in\left\{-15;-1;1;15\right\}\\ \Rightarrow x\in\left\{-16;-2;0;14\right\}\)

Tự kết luận nhé bạn