3. \(\frac{\left(1+2^2+3^2+...+2005^2\right).12}{2.3+4.6+6.9+...+4010.6015}\)
GIẢI CHI TIẾT NHA. LẸ NHA. THANKS
Những câu hỏi liên quan
(2.1.2+1/1.2) + (2.2.3+1/2.3) + (2.3.4+1/3.4) +...+ (2.9.10/9.10) = ?
Tính :
a) (1+2+3+..+2017+2018).(\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{2018.2019}\)) . \(\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
b) A= \(1.2.3+2.3.4+3.4.5+...+x.\left(x+1\right).\left(x+2\right)\)
a)Xét 1/2-1/3-1/6=3/6-2/6-1/6=0
=> (1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).(1/2-1/3-1/6)=(1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).0=0
b) 4A=1.2.3.4+2.3.4.4+..+x(x+1)(x+2)4
=1.2.3.4+2.3.4.5-1.2.3.4+...+x(x+1)(x+2)(x+3)-x(x+1)(x+2)(x-1)
= (x-1)x(x+1)(x+2)
=> A=x(x+1)(x+2)(x-1)/4
Đúng 0
Bình luận (0)
(2.1.2 + 1/1.2) + (2.2.3 + 1/2.3) + (2.3.4 + 1/3.4) + (2.4.5 + 1/4.5)+...+(2.9.10 + 9/10) = ?
help me! (ngu toàn tập)a)frac{3}{left(1.2right)^2}+frac{5}{left(2.3right)^2}+...+frac{2n+1}{left[nleft(n+1right)right]^2}b)left(1-frac{1}{2^2}right)left(1-frac{1}{3^2}right)left(1-frac{1}{4^2}right)....left(1-frac{1}{n^2}right)c)frac{150}{5.8}+frac{150}{8.11}+frac{150}{11.14}+...+frac{150}{47.50}d)frac{1}{1.2.3}+frac{1}{2.3.4}+frac{1}{3.4.5}+...+frac{1}{left(n-1right)nleft(n+1right)}
Đọc tiếp
help me! (ngu toàn tập)
a)\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
b)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{n^2}\right)\)
c)\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+...+\frac{150}{47.50}\)
d)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
Đúng 0
Bình luận (0)
Thực hiện phép tính
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\left(ĐA:A=\frac{1}{6}-\frac{-10}{3}=\frac{7}{2}\right)\)
Gợi ý : Phân tích hết ra thành tích các thừa số nguyên tố rồi đặt cái chung ra ngoài
-> rút gọn
-> kết quả
P/S : bài này cx ko dài lắm nhưg lười ^^
Đúng 1
Bình luận (0)
Xem thêm câu trả lời
\(\frac{2.1+1}{\left(1+1\right)^2}+\frac{2.2+1}{\left(2^2+2\right)^2}+\frac{2.3+1}{\left(3^3+3\right)^2}+....+\frac{2.2015+1}{\left(2015^2+2015\right)^2}+\frac{2.1016+1}{\left(2016^2+2016\right)^2}\)
tính tổng . ai giúp vs
\(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.6\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(C=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
\(D=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
1. Cho \(\frac{a}{c}=\frac{c}{b}\) Chứng minh \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\)
2. Tính B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6-8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
Bài 1: \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\) (1)
Từ \(\frac{a}{c}=\frac{c}{b}\Rightarrow ab=c^2\)
Thay vào (1) ta có:
\(\frac{a^2+ab}{b^2+ab}=\frac{a}{b}\Rightarrow\frac{a\left(a+b\right)}{b\left(a+b\right)}=\frac{a}{b}\) (luôn đúng)
Vậy ta có điều phải chứng minh
Đúng 0
Bình luận (0)
\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^68^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6.2^{12}.3^5}\)- \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)= \(\frac{2^{12}.3^4.\left(3-1\right)}{2^{24}.3^{11}}\)-\(\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+8\right)}\)=\(\frac{1}{2^{11}.3^7}\)-\(\frac{-10}{3}\)
Đúng 0
Bình luận (0)
hình như bạn viết sai đề bài rồi
Đúng 0
Bình luận (0)