9: \(A=\dfrac{3^2}{10}+\dfrac{3^2}{40}+...+\dfrac{3^2}{340}\)

\(=3\left(\dfrac{3}{10}+\dfrac{3}{40}+...+\dfrac{3}{340}\right)\)

\(=3\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}\right)\)

\(=3\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)

\(=3\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=3\cdot\dfrac{9}{20}=\dfrac{27}{20}\)

10: \(A=\dfrac{5^2}{1\cdot6}+\dfrac{5^2}{6\cdot11}+...+\dfrac{5^2}{26\cdot31}\)

\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)

\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{25}-\dfrac{1}{31}\right)\)

\(=5\left(1-\dfrac{1}{31}\right)=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)

11: \(A=\dfrac{6}{15}+\dfrac{6}{35}+\dfrac{6}{63}+\dfrac{6}{99}\)

\(=3\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}\right)\)

\(=3\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}\right)\)

\(=3\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)

\(=3\left(\dfrac{1}{3}-\dfrac{1}{11}\right)=3\cdot\dfrac{8}{33}=\dfrac{8}{11}\)

12: \(A=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+...+\dfrac{3}{49\cdot51}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{49\cdot51}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)=\dfrac{3}{2}\cdot\dfrac{16}{51}=\dfrac{8}{17}\)

13: \(A=\dfrac{1}{2}+\dfrac{2}{2\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{5}{11\cdot16}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}\)

\(=1-\dfrac{1}{16}=\dfrac{15}{16}\)

14: \(A=\dfrac{1}{2}+\dfrac{2}{8}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{5}{176}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{2}{2\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{5}{11\cdot16}\)

15: \(A=\dfrac{3}{54}+\dfrac{5}{126}+\dfrac{7}{294}+\dfrac{8}{609}\)

\(=\dfrac{3}{6\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{7}{14\cdot21}+\dfrac{8}{21\cdot29}\)

\(=\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{29}\)

\(=\dfrac{1}{6}-\dfrac{1}{29}=\dfrac{23}{174}\)

16: \(A=\dfrac{5}{24}+\dfrac{5}{104}+\dfrac{5}{234}+\dfrac{5}{414}\)

\(=\dfrac{5}{3\cdot8}+\dfrac{5}{8\cdot13}+\dfrac{5}{13\cdot18}+\dfrac{5}{18\cdot23}\)

\(=\dfrac{1}{3}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}\)

\(=\dfrac{1}{3}-\dfrac{1}{23}=\dfrac{20}{69}\)

17: \(A=\dfrac{\dfrac{3}{54}+\dfrac{5}{126}+\dfrac{7}{294}}{\dfrac{5}{24}+\dfrac{5}{104}+\dfrac{5}{234}}\)

\(=\dfrac{\dfrac{1}{6}-\dfrac{1}{21}}{\dfrac{1}{3}-\dfrac{1}{18}}=\dfrac{15}{126}:\dfrac{15}{54}=\dfrac{54}{126}=\dfrac{3}{7}\)

\(7,75\times x+2,25\times x=50\)

=>\(x\times\left(7,75+2,25\right)=50\)

=>\(10\times x=50\)

=>x=5

atlat hộp bánh là sao em nhỉ

Ta có:

\(\dfrac{24}{24}=24:24=1\)

\(\dfrac{26}{26}=26:26=1\)

\(\dfrac{1122}{1122}=1122:1122=1\)

\(\dfrac{25}{25}=25:25=1\)

\(\dfrac{27}{27}=27:27=1\)

Vậy nên ta có \(\dfrac{24}{24}=1;\dfrac{26}{26}=1;\dfrac{1122}{1122}=1;\dfrac{25}{25}=1;\dfrac{27}{27}=1\)

Vì khi phân số có tử số bằng mẫu số thì phân số đó bằng 1.

1: \(\left(-12,5\right)+17,55+\left(-3,5\right)-\left(-2,45\right)\)

\(=\left(-12,5-3,5\right)+17,55+2,45\)

=-16+20

=4

2: \(\dfrac{-3}{5}\cdot\dfrac{2}{7}+2\dfrac{3}{5}-\dfrac{3}{5}\cdot\dfrac{5}{7}\)

\(=-\dfrac{3}{5}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{13}{5}\)

\(=-\dfrac{3}{5}+\dfrac{13}{5}=\dfrac{10}{5}=2\)

3: \(\dfrac{2}{3}:x=2,4-\dfrac{4}{5}\)

=>\(\dfrac{2}{3}:x=2,4-0,8=1,6\)

=>\(x=\dfrac{2}{3}:1,6=\dfrac{2}{4,8}=\dfrac{1}{2,4}=\dfrac{5}{12}\)

Bài 10:

\(C=\dfrac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}\)

\(=\dfrac{\left|x-2017\right|+2019-1}{\left|x-2017\right|+2019}=1-\dfrac{1}{\left|x-2017\right|+2019}\)

\(\left|x-2017\right|+2019>=2019\forall x\)

=>\(\dfrac{1}{\left|x-2017\right|+2019}< =\dfrac{1}{2019}\forall x\)

=>\(-\dfrac{1}{\left|x-2017\right|+2019}>=-\dfrac{1}{2019}\forall x\)

=>\(C=-\dfrac{1}{\left|x-2017\right|+2019}+1>=-\dfrac{1}{2019}+1=\dfrac{2018}{2019}\forall x\)

Dấu '=' xảy ra khi x-2017=0

=>x=2017

số dư khi số gồm 1000 chữ số 7 được chia cho 15 là 10.

dư 4 bạn oi

\(\dfrac{-5}{6}\cdot\dfrac{14}{19}+\dfrac{-9}{12}\cdot\dfrac{14}{19}-\dfrac{5}{18}\)

\(=\dfrac{14}{19}\left(-\dfrac{5}{6}-\dfrac{9}{12}\right)-\dfrac{5}{18}\)

\(=\dfrac{14}{19}\cdot\dfrac{-10-9}{12}-\dfrac{5}{18}\)

\(=\dfrac{14}{19}\cdot\dfrac{-19}{12}-\dfrac{5}{18}=\dfrac{-7}{6}-\dfrac{5}{18}\)

\(=\dfrac{-26}{18}=-\dfrac{13}{9}\)

Câu 4:

a: Xét ΔKBC vuông tại K và ΔHCB vuông tại H có

BC chung

\(\widehat{KBC}=\widehat{HCB}\)(ΔABC cân tại A)

Do đó: ΔKBC=ΔHCB

b: Ta có: ΔKBC=ΔHCB

=>\(\widehat{KCB}=\widehat{HBC}\)

=>\(\widehat{EBC}=\widehat{ECB}\)

=>ΔEBC cân tại E