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\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.......+\dfrac{1}{37.38.39}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.....+\dfrac{1}{37.38}-\dfrac{1}{38.39}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{38.39}\)
\(=\dfrac{370}{741}\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+......+\dfrac{1}{37.38.39}\)
Ta có:
\(\dfrac{1}{1.2.3}=\dfrac{1}{1.2}-\dfrac{1}{2.3}\); \(\dfrac{1}{2.3.4}=\dfrac{1}{2.3}-\dfrac{1}{3.4}\);.......
\(\Rightarrow A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...........+\dfrac{1}{37.38}-\dfrac{1}{38.39}\)
\(\Rightarrow A=\dfrac{1}{1.2}-\dfrac{1}{38.39}\)
\(=\dfrac{370}{741}\)
Vậy \(A=\dfrac{370}{741}\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)+1\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+2n+n+2}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+3n+2}\right)\)
\(S_n=\dfrac{1}{4}-\dfrac{1}{2\left(n^2+3n+2\right)}\)
\(S_n=\dfrac{1}{4}-\dfrac{1}{2n^2+6n+4}\)
\(S_n=\dfrac{2n^2+6n+4}{4\left(2n^2+6n+4\right)}-\dfrac{4}{4\left(2n^2+6n+4\right)}\)
\(S_n=\dfrac{2n^2+6n+4}{8n^2+48n+16}-\dfrac{4}{8n^2+48n+16}\)
\(S_n=\dfrac{2n^2+6n}{8n^2+48n+16}\)
\(S_n=\dfrac{2\left(n^2+3n\right)}{2\left(4n^2+24n+8\right)}=\dfrac{n^2+3n}{4n^2+24n+8}\)
\(S_n=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\\ 2S_n=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\\ 2S_n=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ =\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ =\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\\ =>S_n=\dfrac{\left(n+1\right)\left(n+2\right)-2}{4\left(n+1\right)\left(n+2\right)}\)
Giải sai r nhéLinh Nguyễn
Đặt A = 2,7c2 – 3,5c
+Với c = 0,7 ta có :
A = 2,7.(0,7)2 - 3,5.0,7 = 2,7.0,49 - 3,5.0,7 = 1,323 - 2,45 = - 1,127
+Với c = 23 ta có:
A=2,7.(23)2−3,5.(23)=2,7.49−3,5.23
=10,89−73=10,8−219=−10,29
+Với c = 116=76, ta có:
A=2,7.(76)2−3,5.(76)=2,7.4936−3,5.76
=132,336−24,56=132,3−14736=−14,736=−4,912
Chúc bn học tốt
\(A=-5,13:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)
\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{125}{100}+\dfrac{79}{63}\right)\)
\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{5}{4}+\dfrac{79}{63}\right)\)
\(=-5,13:\left(\dfrac{145}{28}-\dfrac{85}{36}+\dfrac{79}{63}\right)\)
\(=-5,13:\dfrac{57}{14}=-5,13:\dfrac{15}{57}\)
\(=\dfrac{-71,82}{57}=1,26\)
Vậy \(A=1,26\)
\(B=\left(3\dfrac{1}{3}.1,9+19,5:4\dfrac{1}{3}\right).\left(\dfrac{62}{75}-\dfrac{4}{25}\right)\)
\(=\left(\dfrac{10}{3}.1,9+19,5:\dfrac{13}{3}\right).\left(\dfrac{62-12}{75}\right)\)
\(=\left(\dfrac{19}{3}+\dfrac{58,5}{13}\right).\dfrac{50}{75}\)
\(=\left(\dfrac{19}{3}+4,5\right).\dfrac{2}{3}\)
\(=\dfrac{32,5}{3}.\dfrac{2}{3}=\dfrac{65}{9}=7\dfrac{2}{9}\)
Vậy \(B=7\dfrac{2}{9}\)
Theo T/C dãy tỉ số bằng nhau
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\frac{a+b}{c}=2\Rightarrow a+b=2c\)
Tương tự ta có
\(b+c=2a\)
\(c+a=2b\)
Xét \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)\)
\(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a\cdot2b\cdot2c}{abc}=8\)
6:
\(4D=2^2+2^4+...+2^{202}\)
=>3D=2^202-1
hay \(D=\dfrac{2^{202}-1}{3}\)
7: \(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)
Sau khi thực hiện phép tính ta được kết quả các giá trị:
\(A=\dfrac{1}{3}\) \(B=-5\dfrac{5}{12}\) \(C=-0,22\)
Sắp xếp: \(-5\dfrac{5}{12}< -0,22< \dfrac{1}{3}\) tức là \(B< C< A\)
Khi tính xong giá trị biểu thức A , B và C ta được kết quả như sau :
\(A=\dfrac{1}{3}\) ; \(B=-5\dfrac{5}{12}\); \(C=-0,22\)
Sắp xếp : \(B< C< A\)\(\left(-5\dfrac{5}{12}< -0,22< \dfrac{1}{3}\right)\)
theo mình thì
Ta chứng minh được bài toán tổng quát sau
2/[(n-1)n(n+1)] = 1/[(n-1)n] - 1/[n(n+1)]
Áp dụng:
ta có 2C = 1/(1.2) - 1/ (2.3) +1/(2.3) - 1/(3.4) + ...+ 1/18.19 - 1/19.20
= 1/(1.2) - 1/(19.20) = [190 - 1] / 19.20 = 189/380
=> C = 189/ 760
mình nhầm