Câu hỏi của Nguyễn Hồ Yến Ngân - Toán lớp 6 - Học toán với OnlineMath

Em tham khảo nhé!

G= \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}\)

G= \(\frac{\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{98.99}{2}}{1.2+2.3+3.4+...+98.99}\)

G = \(\frac{\frac{1.2+2.3+...+98.99}{2}}{1.2+2.3+3.4+...+98.99}\)

G= \(\frac{1}{2}\)

mh chịu thôi

mình chiụ thôi, bó tay luôn

\(2A=\frac{1.2+2.3+3.4+...+98.99}{1.2+2.3+3.4+...+98.99}\)

\(2A=1\)

\(A=\frac{1}{2}\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)

Câu hỏi của Nguyễn Hồ Yến Ngân - Toán lớp 6 - Học toán với OnlineMath

Em tham khảo bài bạn làm :)

b) Em tham khảo: Câu hỏi của lê chí dũng - Toán lớp 6 - Học toán với OnlineMath

vâng ạ nhưng e cx đg cần câu tl phần a

\(D=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3..+98\right)}{1.98+2.97+...+98.1}\)

\(=\frac{\left(1+1+1...+1\right)+\left(2+2+...2\right)+\left(3+...+3\right)+...+\left(97+97\right)+98}{1.98+2.97+...+98.1}\)

( có 99 số 1; 98 số 2; 87 số 3;...; 2 số 97; 1 số 98)

\(=\frac{1.98+2.97+3.96+...+97.2+98.1}{1.98+2.97+...+98.1}=1\)

a, đặt đề bài là A

Ta có : A=( 1-1/2+1/2-1/3+...+1/9-1/10).(x-1)+1/10.x=x-9/10

= (1-1/10).(x-1)+1/10.x

= 9/10 .( x-1 )+1/10.x

=1.x-9/10

nên x= 0 hoặc 1

với -1 nữa nha

Gọi tổng trên là A
A=1/1.2.3+1/2.3.4+1/3.4.5+...1/98.99.100
Ta xét :
1/1.2 ‐ 1/2.3 = 2/1.2.3; 1/2.3 ‐ 1/3.4 = 2/2.3.4;...; 1/98.99 ‐ 1/99.100 = 2/98.99.100
tổng quát: 1/n﴾n+1﴿ ‐ 1/﴾n+1﴿﴾n+2﴿ = 2/n﴾n+1﴿﴾n+2﴿.
Do đó: 2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/98.99.100
= ﴾1/1.2 ‐ 1/2.3﴿ + ﴾1/2.3 ‐ 1/3.4﴿ +...+ ﴾1/98.99 ‐ 1/99.100﴿
= 1/1.2 ‐ 1/2.3 + 1/2.3 ‐ 1/3.4 + ... + 1/98.99 ‐ 1/99.100
= 1/1.2 ‐ 1/99.100
= 1/2 ‐ 1/9900
= 4950/9900 ‐ 1/9900
= 4949/9900.
Vậy A = 4949 / 9900

Bn làm sai r . kết quả là \(\frac{101}{297}\) nhưng mik ko bt cách giải thôi

xin lỗi nha,gửi lời giải nhầm người

\(A=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{2n+1}{n^2\left(n^2+2n+1\right)}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)

\(=1-\dfrac{1}{n^2+2n+1}\)

\(=\dfrac{n^2+2n}{n^2+2n+1}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

Xét thừa số tổng quát:

\(\dfrac{k}{\left(\dfrac{k-1}{2}.\dfrac{k+1}{2}\right)^2}\)\(=\dfrac{k}{\left(\dfrac{\left(k-1\right)\left(k+1\right)}{4}\right)^2}=\dfrac{k}{\left(\dfrac{\left(k-1\right)\left(k+1\right)}{4}\right)^2}\)

\(=\dfrac{k}{\dfrac{\left[\left(k-1\right)\left(k+1\right)\right]^2}{16}}=\dfrac{k}{\dfrac{\left(k^2-1\right)^2}{16}}=\dfrac{16k}{\left(k^2-1\right)^2}\)

Thay \(k=3;5;....2n+1\) ta được:

\(\dfrac{16.3}{\left(3^2-1\right)^2}+\dfrac{16.5}{\left(5^2-1\right)^2}+....+\dfrac{16.n}{\left(n^2-1\right)^2}\)

\(=16.\left(\dfrac{3}{\left(3^2-1\right)^2}+\dfrac{5}{\left(5^2-1\right)^2}+...+\dfrac{n}{\left(n^2-1\right)^2}\right)\)

\(=16.\left(\dfrac{3}{\left[\left(3-1\right)\left(3+1\right)\right]^2}+\dfrac{5}{\left[\left(5-1\right)\left(5+1\right)\right]^2}+...+\dfrac{n}{\left[\left(n-1\right)\left(n+1\right)\right]^2}\right)\)

\(=16.\left(\dfrac{3}{4.16}+\dfrac{5}{16.36}+...+\dfrac{n}{\left(n-1\right)^2.\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{12}{4.16}+\dfrac{20}{16.36}+...+\dfrac{4n}{\left(n-1\right)^2.\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{1}{4}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{36}+...+\dfrac{1}{\left(n-1\right)^2}-\dfrac{1}{\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{\left(n+1\right)^2}{4\left(n+1\right)^2}-\dfrac{4}{4\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{\left(n+1\right)^2-4}{4\left(n+1\right)^2}\right)=\dfrac{4\left(n+1\right)^2-16}{4\left(n+1\right)^2}\)

\(=\dfrac{4\left[\left(n+1\right)^2-4\right]}{4\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-4}{\left(n+1\right)^2}\)

Chúc bạn học tốt!!!