\(2.3^{x+1}=10.3^{12}+8.3^{12}\\ \Rightarrow2.3^{x+1}=3^{12}\left(10+8\right)\\ \Rightarrow2.3^{x+1}=3^{12}.18\\ \Rightarrow3^{x+1}=9.3^{12}\\ \Rightarrow3^{x+1}=3^{14}\\ \Rightarrow x+1=14\\ \Rightarrow x=13\)

\(2.3^x=10.3^{12}+8.27^4\)

\(2.3^x=10.3^{12}+8.\left(3^3\right)^4\)

\(2.3^x=18.3^{12}\)

\(3^x=9.3^{12}=3^{14}\)

\(x=14\)

2.3ˣ = 10.3¹² + 8.27⁴

2.3ˣ = 10.3¹² + 8.(3³)⁴

2.3ˣ = 10.3¹² + 8.3¹²

2.3ˣ = 3¹².(10 + 8)

2.3ˣ = 3¹².18

3ˣ = 3¹².18 : 2

3ˣ = 3¹² . 9

3ˣ = 3¹² . 3²

3ˣ = 3¹⁴

x = 14

Bài này đã giải rồi nha, e coi lại câu hỏi trước nha

1^2+3^4+5^6+7^8+9^10=3492564909

1.200.400.0=0

=-45*312-312+312*(-54)

=312(-45-1-54)

=-31200

312 * 425 + 312 * 574 + 312

= 312 x ( 425 + 574 + 1 )

= 312 x 1000

= 312000

k mik nha

=312 x 425 + 312 x 574 + 312 x 1

= 312 x ( 425 + 574 + 1 )

= 312 x 1000 = 312000

312 x 425 + 312 x 574 + 312

= 312 x 425 + 312 x 574 + 312 x 1

= 312 x ( 425 + 574 + 1 )

= 312 x 1000

= 312000

Ta có: \(\dfrac{312}{x}-\dfrac{312}{x+4}=\dfrac{1}{2}\)

\(\Leftrightarrow x\left(x+4\right)=624x+2496-624x\)

\(\Leftrightarrow x^2+4x-2496=0\)

\(\Leftrightarrow\left(x+2\right)^2=2500\)

\(\Leftrightarrow\left[{}\begin{matrix}x=48\\x=-52\end{matrix}\right.\)

Lời giải:
1. Bạn coi lại đề

2. $=234\times (1257-257)=234\times 1000=234000$

3. $=175\times (1274-273-1)=175\times 1000=175000$

4. $=(4\times 25)\times (125\times 8)=100\times 1000=100000$

1. ...............................

2. $=234\times (1257-257)=234\times 1000=234000$

3. $=175\times (1274-273-1)=175\times 1000=175000$

4. $=(4\times 25)\times (125\times 8)=100\times 1000=100000$

=312x425+312x574+312x1

=312x(425+574+1)

=312x1000

=312000