125. (-8). 38. 5².(-2)

= -1000. 38. 25. (-2)

= -38000. 25. (-2)

= -950000 . (-2)

= 1900 000

\(125.\left(-8\right).38.5^2.\left(-2\right)\)

\(=\left[125.\left(-8\right)\right].38.\left[25.\left(-2\right)\right]\)

\(=-1000.38.\left(-50\right)\)

\(=-\left[-1000.\left(-50\right)\right].38\)

\(=50000.38\)

\(=1900000\)

File: undefined 

a) Ta có \(\lim\limits_{x\rightarrow-\infty}\dfrac{4x+1}{-x+1}=\lim\limits_{x\rightarrow-\infty}\left(\dfrac{-4+\dfrac{1}{x}}{1+\dfrac{1}{x}}\right)=-4\)

b) Ta có \(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{x^2-x-2}{x-2}=\lim\limits_{x\rightarrow2}\left(\dfrac{\left(x+1\right)\left(x-2\right)}{x-2}\right)\)

\(=\lim\limits_{x\rightarrow2}\left(x+1\right)=2+1=3\)

 Để hàm số đã cho liên tục tại \(x=2\) thì \(\lim\limits_{x\rightarrow2}f\left(x\right)=f\left(2\right)=m\) hay \(m=3\).

Lời giải:
Đặt \(\underbrace{111....1}_{100}=a\Rightarrow 9a+1=1\underbrace{000...0}_{100}\)

Khi đó:
\(\underbrace{1111....1}_{100}\underbrace{222....2}=\underbrace{111...1}_{100}\times 1\underbrace{00...0}_{100}+\underbrace{222....2}_{100}\)

\(a(9a+1)+2a=9a^2+3a=3a(3a+1)\) là tích của 2 số
 tự nhiên liên tiếp $3a, 3a+1$

Ta có đpcm.

Lúc đầu thùng 1 chứa số lít dầu là :  (254 - 14) : 2 + 40 = 160(l)

Lúc đầu thùng 2 chứa số lít dầu là : 254 - 160 = 94 (l)

Đ/số:.......

(-2)⁴ - 48: (-3). (120 - 3²) + 12. (-16)

= 16 - 48 : (-3). (120 - 9) + 12. (-16)

= 16 - 48 : (-3). 111 + 12. (-16)

= 16 - (-16). 111 + (-192)

= 16 - (-1776) + (-192)

= 16 + 1776 + (-192)

= 1792 + (-192)

= 1600

   (- 2)⁴ - 48: (-3). (120 - 3²) + 12.(-16)

= 16  + 16. (120 - 9) - 12.16

= 16 + 16. 111 - 12.16

= 16.( 1 + 111 - 12)

= 16. 100

= 1600

1,

Đặt \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\left(2-1\right)A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(1A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(A=2^{32}-1\)

Vậy \(A=2^{32}-1\)

2, \(x^2-6x=-9\)

\(x^2-6x+9=0\)

\(\left(x-3\right)^2=0\)

\(x-3=0\)

\(x=3\)

Vậy \(x=3\)