Theo đầu bài ta có:
\(\left(2x-7\right)^{24}+\left(7-2x\right)^{24}=2\)
\(\Rightarrow\left(2x-7\right)^{24}+\left[-\left(2x-7\right)\right]^{24}=2\)
\(\Rightarrow\left(2x-7\right)^{24}+\left(2x-7\right)^{24}=2\)
\(\Rightarrow\left(2x-7\right)^{24}\cdot2=2\)
\(\Rightarrow\left(2x-7\right)^{24}=1\)
\(\Rightarrow\orbr{\begin{cases}2x-7=1\\2x-7=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=8\\2x=6\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)

- Nếu 2x - 7 < -1 hoặc 2x - 7 > 1 thì (2x - 7)² > 2 do đó không thể xảy ra đẳng thức

- Nếu 2x - 7 = 0 thì (2x - 7)²⁴ + (2x - 7)²⁴ = 0 (loại)

- Nếu 2x - 7 = + 1 thì (2x - 7)²⁴ + (2x - 7)²⁴ = 1 (thỏa mãn)

Vậy 2x - 7 = + 1 \(\Leftrightarrow\) x = 4 hoặc x = 3

\(5x-\left(24-2x\right)=\left(-24\right)+20+3x\)

\(5x+\left(-24\right)+2x=\left(-24\right)+20+3x\)

\(7x+\left(-24\right)=\left(-24\right)+20+3x\)

\(7x=20+3x\)

\(\left(7-3\right)x=20\)

\(4x=20\)

\(x=20:4\)

\(x=5\)

5x- (24 - 2x) = -24 + 20 + 3x

5x- 24 + 2x = -4 + 3x

5x + 2x - 3x = 24 - 4

4x = 20

x = 20 \(\div\)4

x = 5

Vậy x = 5

Ta có: \(5x-\left(24-2x\right)=-24+20+3x\)

\(\Leftrightarrow5x-24+2x=-24+20+3x\)

\(\Leftrightarrow5x+2x-3x=24-24+20\)

\(\Leftrightarrow4x=20\)

\(\Leftrightarrow x=\dfrac{20}{4}=5\)

Vậy: \(x=5\)

_Chúc bạn học tốt_

\(2^x+2^{x+1}=24\\ \Rightarrow2^x+2^x\cdot2=24\\ \Rightarrow2^x\left(2+1\right)=24\\ \Rightarrow2^x\cdot3=24\\ \Rightarrow2^x=24:3\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)

1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)

=>\(5^{x-2}=16+9=25\)

=>x-2=2

=>x=4

2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)

=>3^x=0

=>x=0

3: \(\Leftrightarrow2^x+2^x\cdot16=272\)

=>2^x*17=272

=>2^x=16

=>x=4

4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)

=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)

=>2^x-1=8

=>x-1=3

=>x=4

a)(2x+4).8-40=24

(2x+4).8=24+40

(2x+4).8=64

2x+4=64:8

2x+4=8

2x=8+4

2x=12

x=12:2

x=6

b)(2x+4²)-8=64:2

(2x+16)-8=32

2x+16=32+8

2x+16=40

2x=40-16

2x=24

x=24:2

x=12

c)(3x+2).4-20=24

(3x+2).4=24+20

(3x+2).4=44

3x+2=44:4

3x+2=11

3x=11-2

3x=9

x=9:3

x=3

Lần sau bạn nhớ gõ bằng bảng công thức toán ,gõ như vầy khó nhìn quá ,nó sẽ dễ nhìn hơn làm mọi người thích thú hơn nhé!

Ta có : \(2^{2x}.3^{2x}=36^{24}\)

\(\Rightarrow\left(2.3\right)^{2x}=\left(6^2\right)^{24}\)

\(\Leftrightarrow6^{2x}=6^{2.24}\)

\(\Rightarrow x=24\)

Chúc học tốt nhé

Lời giải:

PT $\Leftrightarrow (8x^3-36x^2+54x-27)-(8x^3-2x+12x^2-3)=-24$

$\Leftrightarrow -48x^2+56x-24=-24$

$\Leftrightarrow -48x^2+56x=0$

$\Leftrightarrow 8x(7-6x)=0$

$\Leftrightarrow x=0$ hoặc $7-6x=0$

$\Leftrightarrow x=0$ hoặc $x=\frac{7}{6}$

\(\left(2x-3\right)^3-\left(2x+3\right)\left(4x^2-1\right)=-24\)

\(8x^3-18x^2+27x-27-\left(8x^3-2x+12x^2-3\right)\) =-24

\(8x^3-18x^2+27x-27-8x^3+2x-12x^2-3\) = -24

\(-30x^2+29x-30=-24\)

\(-30x^2+29x=6\)\(x\left(-30x+29\right)=6\)

Ta có: \(\left(2x-3\right)^3-\left(2x+3\right)\left(4x^2-1\right)=-24\)

\(\Leftrightarrow8x^2-36x^2+54x-27-8x^3+2x-12x^2+3=-24\)

\(\Leftrightarrow-48x^2+56x=0\)

\(\Leftrightarrow-8x\left(6x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{6}\end{matrix}\right.\)