bài toán khó thế kỉ 22

mk ko bt

k mk nhé

Tao không biết và tao cũng chẳng quan tâm

1. Tính:

a) 23 + 24 - ( 65 x 2 - 4 ) x 4 + 66 + 44 - ( 54 : 9 x 9 : 9 + 45 ) : 1 + 1 - 1 x ( 23 + 23 - 23 x 2 +1 ) = 51

b) 4444444444444444444444444444444444444444444444444444444444444444444 : 1

= 4444444444444444444444444444444444444444444444444444444444444444444

c) 3⁰ = 0

d) 1/1 = 1

e)

a) - 530

b) 44444444444444444444444444444444444444444444444444444444444444444444

c) 0

d) 1

e) Em là em

\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}\)

\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}=0\)

\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\right)=0\)

\(\Leftrightarrow x-23=0\left(vì\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\ne0\right)\)

\(\Leftrightarrow x=23\)

vậy................

\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)

\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)

\(\Leftrightarrow300-x=0\left(vì\frac{1}{99}+\frac{1}{97}+\frac{1}{95}>0\right)\)

\(\Leftrightarrow x=300\)

vậy..........

Có 6 chữ số 0 ở tận cùng.

• Câu a:
  $$\frac{x-23}{24} + \frac{x-23}{25} = \frac{x-23}{26} + \frac{x-23}{27}$$$$\Leftrightarrow \frac{x-23}{24} + \frac{x-23}{25} - \frac{x-23}{26} - \frac{x-23}{27} = 0$$$$\Leftrightarrow (x-23)\left(\frac{1}{24} + \frac{1}{25} - \frac{1}{26} - \frac{1}{27}\right) = 0$$Vì $\frac{1}{24} + \frac{1}{25} - \frac{1}{26} - \frac{1}{27} \neq 0$, nên:
  $$x - 23 = 0 \Leftrightarrow x = 23$$Vậy tập nghiệm của phương trình là $S = \{23\}$.
• Câu b:
  $$\left(\frac{x+2}{98} + 1\right) + \left(\frac{x+3}{97} + 1\right) = \left(\frac{x+4}{96} + 1\right) + \left(\frac{x+5}{95} + 1\right)$$$$\Leftrightarrow \frac{x+100}{98} + \frac{x+100}{97} = \frac{x+100}{96} + \frac{x+100}{95}$$$$\Leftrightarrow \frac{x+100}{98} + \frac{x+100}{97} - \frac{x+100}{96} - \frac{x+100}{95} = 0$$$$\Leftrightarrow (x+100)\left(\frac{1}{98} + \frac{1}{97} - \frac{1}{96} - \frac{1}{95}\right) = 0$$Vì $\frac{1}{98} + \frac{1}{97} - \frac{1}{96} - \frac{1}{95} \neq 0$, nên:
  $$x + 100 = 0 \Leftrightarrow x = -100$$Vậy tập nghiệm của phương trình là $S = \{-100\}$.