Đáp án:

$\frac{2489-36}{7467-108}$ `=` $\frac{2453}{7359}$ `=` $\frac{1}{3}$ 

$\frac{2929-303}{8787+1717}$ `=` $\frac{2526}{10504}$ `=` $\frac{1}{4}$ 

\(A=\dfrac{2489-36}{7467-108}=\dfrac{3-1}{3-1}=1\)

\(A=\dfrac{2489-36}{7467-108}=1\)

Bài giải

a) Ta có : \(\frac{4545+101}{6969-303}=\frac{45.101+101}{69.101-101.3}=\frac{101.\left(45+1\right)}{101.\left(69-3\right)}=\frac{101.46}{101.66}=\frac{23}{33}\)

b) Ta có : \(\frac{2929-101}{2.1919+404}=\frac{29.101-101}{2.19.101+4.101}=\frac{101.\left(29-1\right)}{101.\left(19.2+4\right)}=\frac{28}{42}=\frac{2}{3}\)

a)\(\frac{4545+101}{6969-303}\)= \(\frac{\left(4545:45\right)+101}{\left(6969:69\right)-303}\)= \(\frac{101+101}{101-303}\)=\(\frac{202}{-202}\)=-1 

b)\(\frac{2929-101}{2.1919+404}\)= \(\frac{2929-101}{3838+404}\)=\(\frac{\left(2929:29\right)-101}{\left(3838:38\right)+404}\)=\(\frac{101-101}{101+404}\)=\(\frac{0}{505}\)=0

học tốt 

a)\(\frac{4545+101}{6969-303}=\frac{\left(4545:45\right)+101}{\left(6969:69\right)-303}=\frac{101+101}{101-303}=\frac{202}{-202}=-1\)

b)\(\frac{2929-101}{2.1919+404}=\frac{2929-101}{3838+404}=\frac{\left(2929:29\right)-101}{\left(3838:38\right)+404}=\frac{101-101}{101+404}=0\)

Học tốt

@@

Bài toán được giải như sau :

H = 4116-14/10290-35 = 4102/10255 = 2/5

K = 2/3 ( làm như trên nhưng mình làm vậy cho nó nhanh )

L = -4/3

M = -432/17

Oke xong rồi nhé

hi ko nhớ anh à

có đứa làm hộ rồi đó em

\(H=\frac{4116-14}{10290-35}=\frac{14.294-14}{35.294-35}=\frac{14.\left(294-1\right)}{35.\left(294-1\right)}=\frac{14.293}{35.293}=\frac{2}{5}\)
\(K=\frac{29.101-101}{2.19.101+4.101}=\frac{101.\left(29-1\right)}{101.\left(38+4\right)}=\frac{28}{42}=\frac{2}{3}\)

\(I=\frac{1313-1717}{303}=\frac{13.101-17.101}{3.101}=\frac{101.\left(13-17\right)}{3.101}=\frac{-4}{3}\)

\(M=\frac{12-24.3}{1-35}=\frac{12-12.2.3}{-34}=\frac{12.\left(1-6\right)}{-34}=\frac{-60}{-34}=\frac{30}{17}\)

\(H\)\(=\) \(\frac{4116-14}{10290-35}\)

       \(=\) \(\frac{4102}{10255}\) 

       \(=\) \(\frac{4102:2051}{10255:2051}\) 

       \(=\) \(\frac{2}{5}\) 

\(K=\frac{2929-101}{2.1919+404}\)

   \(=\) \(\frac{2828}{4242}\) 

   \(=\) \(\frac{2828:1414}{4242:1414}\) 

   \(=\) \(\frac{2}{3}\)  

\(M=\frac{12-24.3}{1-35}\)

    \(=\) \(\frac{-60}{-34}\)

     \(=\) \(\frac{60}{34}\) 

      \(=\) \(\frac{30}{17}\)

:D

\(H=\frac{14\cdot294-14}{35\cdot294-35}=\frac{14\cdot\left(294-1\right)}{35\cdot\left(294-1\right)}=\frac{14}{35}=\frac{2}{5}\)

\(K=\frac{29\cdot101-101}{2\cdot19\cdot101+4\cdot101}=\frac{101\cdot\left(29-1\right)}{101\cdot\left(2\cdot19+4\right)}=\frac{28}{42}=\frac{2}{3}\)

\(I=\frac{13\cdot101-17\cdot101}{3\cdot101}=\frac{101\cdot\left(13-17\right)}{3\cdot101}=\frac{-4}{3}\)

\(M=\frac{12-72}{-34}=\frac{-60}{-34}=\frac{15}{17}\)

\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{21.23}\)

\(=5-\frac{5}{3}+\frac{5}{3}-\frac{5}{5}+\frac{5}{5}-\frac{5}{7}+...+\frac{5}{21}-\frac{5}{23}\)

\(=5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)

\(=5\left(1-\frac{1}{23}\right)\)

\(=5.\frac{22}{23}\)

\(=\frac{110}{23}\)

\(A=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{21}-\frac{1}{23}\right)\)

\(A=\frac{5}{2}.\left(1-\frac{1}{23}\right)\)

\(A=\frac{5}{2}.\frac{22}{23}\)

\(A=\frac{55}{23}\)

A.\(\frac{2}{5}\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+............+\)\(\frac{2}{21.23}\)

A.\(\frac{2}{5}=1-\frac{1}{23}\)

A.\(\frac{2}{5}=\frac{21}{23}\)

A=\(\frac{21}{23}.\frac{5}{2}\)

Tự tính

\(A=xemlai\) chưa hưa hiểu Quy luật

\(B=\frac{\left(n.\left(n+2\right)+1\right)}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)

\(B=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.5}...\frac{98.98}{97.99}\frac{99.99}{98.100}\frac{100.100}{99.101}\\\)

\(B=\frac{2.100}{1.101}=\frac{200}{101}\)