a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)

Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1

\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=8-\frac{2}{5}=\frac{38}{5}\)

b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)

\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)

Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1

\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=10-\left(1-\frac{1}{11}\right)\)

\(=10-\frac{10}{11}=\frac{100}{11}\)

100/11 nha

câu b bằng 891/110

câu a bằng 76/10

1.x=120

110% của 180 là:

\(180:100.110=198\)

Đáp số :\(198\)

120 - [7 . 20 - (134 - 110) . 5]

= 120 - [140 - 24 . 5]

= 120 - (140 - 120)

= 120 - 20

= 100

\(120-\left[7.20-\left(134-110\right).5\right]\)

\(=120-\left[140-24.5\right]\)

\(=120-\left[140-120\right]\)

\(=120-20\)

\(=100\)

_Chúc bạn học tốt_

120 - [7 . 20 - (134 - 110) . 5]

= 120 - [140 - 24 . 5]

= 120 - (140 - 120)

= 120 - 20

= 100

Ta có : 

\(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-...-\frac{1}{110}\)

\(=\)\(1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)

\(=\)\(1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

\(=\)\(1-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=\)\(1-\left(1-\frac{1}{11}\right)\)

\(=\)\(1-1+\frac{1}{11}\)

\(=\)\(\frac{1}{11}\)

Chúc bạn học tốt ~ 

\(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{110}\)

\(=1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)

\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=1-\left(1-\frac{1}{11}\right)\)

\(=1-\frac{10}{11}\)

\(=\frac{1}{11}\)

Chúc bạn học tốt !!! 

Đặt \(A=1-\frac{1}{2}-\frac{1}{6}-...-\frac{1}{110}\)

\(\Rightarrow-A=-1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)

\(\Rightarrow-A=-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow-A=-\frac{1}{11}\Rightarrow A=\frac{1}{11}\)

bài số mấy hả bn, hai bài đấy mk lm hết rồi

nếu là bài 4:

                                                               giải

Diện tích hình thang AIGE là :

                (15+5)x15:2=150(m2)

Diện tích hình vuông  IBEH là :

                   15x15=225(m2)

Diện tích hình thang BCHD là

                   (15+10)x15:2=187,5(m2)

Diện tích mảnh đất là:

                      150+225+187,5=562,5(m2)

                          Đ/S:........

bài cuối cùng của tuần 20 trong quyển ôn luyện toán lớp 5

A = \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+......+\frac{109}{110}\)

A = \(1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}....+1-\frac{1}{110}\)

A = \(10-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)

A = \(10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)

A = \(10-\left(1-\frac{1}{11}\right)\)

A = \(10-\frac{10}{11}\)

A = \(\frac{100}{11}\)

ai giup mk voi!!!! mk can gap

1/2 + 5/6 + 11/12 + 19/20 + 29/30 + 41/42 + 55/56 + 51/72 + 89/90 + 109/110

= 1 - 1/2 + 1 - 1/6 + 1 - 1/12+ ...... +  1- 1/110 

= 10 - ( 1/2 + 1/6 + ........ + 1/110 )

= 10 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/10 - 1/11 ) 

= 10 - ( 1 - 1/11 )

= 10 - 10/11

= 100/11

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=110\)

\(\left(x+x+...+x\right)+\left(1+2+3+...+10\right)=110\)

\(10x+55=110\)

\(10x=110-55\)

\(10x=55\)

\(x=55:10\)

\(x=5,5\)

\(10xX+\left(1+2+3+...+10\right)=110\)

\(1+2+3+...+10=\dfrac{10\left(1+10\right)}{2}=55\)

\(\Rightarrow10xX+55=110\Rightarrow10xX=55\)

\(\Rightarrow X=55:10=5,5\)

Ta có : \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|\ge0\forall x\)

=> 11x \(\ge\)0

=> x  \(\ge\)0 

Khi đó \(\orbr{\begin{cases}x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=11x\left(10\text{ số hạng x }\right)\\x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=-11x\left(10\text{ số hạng x}\right)\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=11x\\10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=-11x\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\\10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=-11x\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\\10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=-11x\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(1-\frac{1}{11}\right)=11x\\10x+\left(1-\frac{1}{11}\right)=-11x\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{10}{11}\\21x=-\frac{10}{11}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{10}{11}\left(\text{tm}\right)\\x=-\frac{10}{231}\left(\text{loại}\right)\end{cases}}}\)

Vậy \(x=\frac{10}{11}\)