DỄ

sai

sai

\(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{\frac{3}{4}+\frac{3}{24}+\frac{3}{124}}\) + \(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{127}}{\frac{3}{7}+\frac{3}{17}+\frac{3}{127}}\)

= \(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{3.\left(\frac{1}{4}+\frac{1}{24}+\frac{1}{124}\right)}\) + \(\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{127}\right)}{3.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{127}\right)}\)

= \(\frac{1}{3}\) + \(\frac{2}{3}\) = 1

Fan Karry, mình giúp liền nè.

Đáp án là (1+2+3+...+n)^2

sao lại mũ 2 jay ?

\(\frac{x+1}{97}+\frac{x+1}{98}=\frac{x+1}{99}+\frac{x+1}{100}\)

\(=>\frac{x+1}{97}+\frac{x+1}{98}-\frac{x+1}{99}-\frac{x+1}{100}=0\)

\(=>\left(x+1\right).\left(\frac{1}{97}+\frac{1}{98}-\frac{1}{99}-\frac{1}{100}\right)=0\)

Vì \(\frac{1}{97}>\frac{1}{98}>\frac{1}{99}>\frac{1}{100}\)

Nên \(\frac{1}{97}+\frac{1}{98}-\frac{1}{99}-\frac{1}{100}\) khác 0

=>x+1=0

=>x=-1

Vậy x=-1

Hu hu,giúp Mk đi mừ.3 tick lun

Mai chủ nhật mà

thi xong nói đề nhé

\(\left(2\frac{4}{5}.x+50\right):\frac{2}{3}=-51\)

\(\left(2\frac{4}{5}.x+50\right)\)      = (- 51) x \(\frac{2}{3}\)

\(\left(2\frac{4}{5}.x+50\right)\)      = - 34

\(2\frac{4}{5}.x\)                   = -34 - 50

\(2\frac{4}{5}.x\)                   = - 84

x                           = -84 : \(2\frac{4}{5}\)

x                           = -30

(14/5 . x + 50) : 2/3 = -51

(14/5 . x + 50) = -51 . 2/3

(14/5 . x + 50) = -34

14/5 . x = -34 - 50

14/5 . x = -84

x = -84 : 14/5

x = -30