5/3 + -2/ 7 - (- 1.2)
5/3 + -2/ 7 - (- 1.2)
\(\dfrac{5}{3}+\dfrac{2}{7}-\left(-1.2\right)=\dfrac{5}{3}+\dfrac{2}{7}+2=\dfrac{83}{21}\)
5/3 + -2/ 7 - (- 1.2)
Tính:
a) \(\dfrac{-7}{8}\) . \(\dfrac{3}{5}\) - \(\dfrac{2}{5}\) . \(\dfrac{7}{8}\) + \(3\dfrac{7}{8}\)
b) -1,6 : ( 1 + \(\dfrac{2}{3}\) )
c) \(\dfrac{6}{7}\) + \(\dfrac{5}{8}\) : 5 - \(\dfrac{3}{16}\) . ( -2 )\(^2\)
d) \(\dfrac{1^2}{1.2}\) . \(\dfrac{2^2}{2.3}\) . \(\dfrac{3^2}{3.4}\) . \(\dfrac{4^2}{4.5}\)
a: \(=\dfrac{-7}{8}\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+3+\dfrac{7}{8}=\dfrac{-7}{8}+\dfrac{7}{8}+3=3\)
b: \(=-\dfrac{8}{5}:\dfrac{5}{3}=-\dfrac{24}{25}\)
c: \(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{6}{8}=\dfrac{6}{7}-\dfrac{5}{8}=\dfrac{48}{56}-\dfrac{35}{56}=\dfrac{13}{56}\)
tính
a) P = 1 / 1.2 + 2 / 2.4 + 3 / 4.7 + ...+ 10 / 46.56
b) A= 3 / 1.2 + 3 / 2.3 + 3 / 3.4 + ....+ 3 / 99.100 chú ý : / là phần nha
c) B = 3 / 1.4 + 3 / 4.7 + 3 / 7.10 + ... + 3 / 100.103
d) C= 5 / 1.4 + 5 / 4.7 + 5 / 7.10 + ...+ 5 / 100.103
e) D= 7 / 1.5 + 7 / 5.9 + 7 / 9.13 +...+ 7 / 101.105
a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)
\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)
\(P=1-\dfrac{1}{56}\)
\(P=\dfrac{55}{56}\)
b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)
\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=3\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}\)
\(A=\dfrac{297}{100}\)
c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(B=1-\dfrac{1}{103}\)
\(B=\dfrac{102}{103}\)
d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)
\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}.\dfrac{102}{103}\)
\(C=\dfrac{170}{103}\)
e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)
\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}.\dfrac{104}{105}\)
\(D=\dfrac{26}{15}\)
tính
a) P = 1 / 1.2 + 2 / 2.4 + 3 / 4.7 + ...+ 10 / 46.56
b) A= 3 / 1.2 + 3 / 2.3 + 3 / 3.4 + ....+ 3 / 99.100 chú ý : / là phần nha
c) B = 3 / 1.4 + 3 / 4.7 + 3 / 7.10 + ... + 3 / 100.103
d) C= 5 / 1.4 + 5 / 4.7 + 5 / 7.10 + ...+ 5 / 100.103
e) D= 7 / 1.5 + 7 / 5.9 + 7 / 9.13 +...+ 7 / 101.105
a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)
=\(1-\frac{1}{56}=\frac{55}{56}\)
b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)
= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)
=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)
=> \(A=\frac{99}{100}.3=\frac{297}{100}\)
c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)
=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)
e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)
=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)
=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
=\(1-\frac{1}{105}=\frac{104}{105}\)
=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)
Bài 1: Tính giá trị của biểu thức.
a) 3- 1 4/5 : ( -3/4)
b) 5. | -1/10 + 7/15| - 2/13. 4 1/3
c) 7^9: 7^7 -3^2+ 2^1.2^2
CMR: A=3/(1.2)^2+5/(2.3)^2+7/(3.4)^2+...+4033/(2016.2017)^2<1
\(A=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{4033}{\left(2016.2017\right)^2}\)
\(=\dfrac{3}{1.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{4033}{2016^2.2017^2}\)
\(=\dfrac{1}{1}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{2016^2}-\dfrac{1}{2017^2}\)
\(=1-\dfrac{1}{2017^2}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
Vậy...
Giá trị của tổng A= 3/(1.2)^2 +5/(2.3)^2+ 7/(3.4)^2+...+ 89/(44.45)^2 là ...
\(\frac{1}{n^2\left(n+1\right)^2}=\frac{1}{2n+1}.\left[\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\right]\)
\(A_n=\frac{2n+1}{n^2\left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\\ \)
\(A=1-\frac{1}{\left(45\right)^2}\)
Giá trị của tổng A= 3/(1.2)^2 + 5/(2.3)^2 + 7/(3.4)^2 + 9/(4.5)^2 + ... + 89/(44.45)^2?
Lời giải đây bn nhé :
\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{89}{\left(44.45\right)^2}\)
=\(\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{89}{1936.2025}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{1936}-\frac{1}{2025}\)
=\(1-\frac{1}{2025}\)
=\(\frac{2024}{2025}\)
xong r nhé