Giúp mình với mọi người ai làm đc câu nào thì làm :)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
NV
1
1 tháng 4
\(A=\dfrac{5\cdot4^6\cdot9^4-3^9\cdot\left(-8\right)^4}{4\cdot2^{13}\cdot3^8+2\cdot8^4\cdot\left(-27\right)^3}\)
\(=\dfrac{5\cdot2^{12}\cdot3^8-3^9\cdot2^{12}}{2^{15}\cdot3^8-2^{13}\cdot3^9}\)
\(=\dfrac{2^{12}\cdot3^8\left(5-3\right)}{2^{13}\cdot3^8\left(2^2-3\right)}\)
\(=\dfrac{2^{13}}{2^{13}}=1\)
1 tháng 4
Gọi số học sinh của trường là x(bạn)
(Điều kiện: \(x\in Z^+\))
Số học khi xếp hàng 13 thì dư 4 em nên \(x-4\in B\left(13\right)\)
=>\(x-4\in\left\{...;247;260;273;...;598;...\right\}\)
=>\(x\in\left\{...;251;264;277;...;602;...\right\}\)
mà 250<=x<=600
nên \(x\in\left\{251;264;277;...;589\right\}\left(1\right)\)
Số học sinh khi xếp hàng 17 thì dư 9 em nên \(x-9\in B\left(17\right)\)
=>\(x-9\in\left\{...;255;272;...;595;...\right\}\)
=>\(x\in\left\{...;264;281;...;604;...\right\}\)
mà 250<=x<=600
nên \(x\in\left\{264;281;...;587\right\}\left(2\right)\)
Số học sinh khi xếp hàng 5 thì vừa hết nên \(x\in B\left(5\right)\)
mà 250<=x<=600
nên \(x\in\left\{250;255;260;...;600\right\}\left(3\right)\)
Từ (1),(2),(3) suy ra
\(\left\{{}\begin{matrix}x\in\left\{251;264;...;589\right\}\\x\in\left\{264;281;...;587\right\}\\x\in\left\{250;255;260;...;600\right\}\end{matrix}\right.\)
=>x=485
Vậy: Số học sinh là 485 bạn
T
1 tháng 4
d) \(\dfrac{9}{14}\cdot\dfrac{92}{121}+\dfrac{-2}{121}\cdot\dfrac{9}{14}+\dfrac{31}{121}\cdot\dfrac{9}{14}\) (sửa đề)
\(=\dfrac{9}{14}\cdot\left(\dfrac{92}{121}+\dfrac{-2}{121}+\dfrac{31}{121}\right)\)
\(=\dfrac{9}{14}\cdot\left(\dfrac{90}{121}+\dfrac{31}{121}\right)\)
\(=\dfrac{9}{14}\cdot1=\dfrac{9}{14}\)
e) \(\dfrac{2023}{2024}\cdot\dfrac{2^4}{3^7}\cdot\dfrac{-3^9}{2^3}\cdot\dfrac{-2024}{2023}\)
\(=\dfrac{2023\cdot2^4\cdot3^9\cdot2024}{2024\cdot3^7\cdot2^3\cdot2023}=2\cdot3^2=18\)
KV
1 tháng 4
a) Các điểm có trong hình: A. B, C, D, E, K
b) 5 cặp đường thẳng cắt nhau:
AB, IB
AC, KC
AC, AI
AB, AE
DE, BI
d) Các tia trùng nhau chung gốc A:
- AB, AC.
- AD, AE.
- AI, AK.
31 tháng 3
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: \(\dfrac{2}{3}:x=2,4-\dfrac{4}{5}\)
=>\(\dfrac{2}{3}:x=2,4-0,8=1,6\)
=>\(x=\dfrac{2}{3}:1,6=\dfrac{2}{4,8}=\dfrac{1}{2,4}=\dfrac{5}{12}\)
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: \(A=\dfrac{\dfrac{3}{54}+\dfrac{5}{126}+\dfrac{7}{294}}{\dfrac{5}{24}+\dfrac{5}{104}+\dfrac{5}{234}}\)
\(=\dfrac{\dfrac{1}{6}-\dfrac{1}{21}}{\dfrac{1}{3}-\dfrac{1}{18}}=\dfrac{15}{126}:\dfrac{15}{54}=\dfrac{54}{126}=\dfrac{3}{7}\)