c)C=\(\frac{2}{1.2}\)+\(\frac{4}{2.4}\)+\(\frac{6}{4.7}\)+\(\frac{8}{7.11}\)+\(\frac{10}{11.16}\)+\(\frac{12}{16.22}\)
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\(x+\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}=1\)
\(x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}=1\)
\(x+1-\frac{1}{16}=1\)
\(x+\frac{15}{16}=1\)
\(x=1-\frac{15}{16}\)
\(x=\frac{1}{16}\)
a, Ta có \(A=\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)
\(=\frac{3}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}-\frac{3}{102}=\frac{48}{102}=\frac{24}{51}\)
b,Ta có \(\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(=\frac{2-1}{2}+\frac{4-2}{2.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{16-11}{11.16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(=\frac{15}{16}\)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!1111
Đặt \(A=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+\dfrac{6}{16.22}\)
\(1A=1-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{11}+\dfrac{1}{11}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}\right)-\dfrac{1}{22}\)\(1A=1-\dfrac{1}{22}\)
\(1A=\dfrac{22}{22}-\dfrac{1}{22}\)
\(1A=\dfrac{21}{22}\)
\(\dfrac{21}{22}\) không thể rút gọn
\(A=\dfrac{1}{1\cdot2}+\dfrac{2}{2\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{5}{11\cdot16}+\dfrac{6}{16\cdot22}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{22}\\ =1-\dfrac{1}{22}\\ =\dfrac{21}{22}\)
Vậy \(A=\dfrac{21}{22}\)
Ta có: B = \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}\)
=> B = \(\frac{6}{3.5}\)+ \(\frac{6}{5.7}\)+ \(\frac{6}{7.9}\)+ \(\frac{6}{9.11}\)
=>B =\(3.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
=> B = \(3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
=> B = \(3.\left(\frac{1}{3}-\frac{1}{11}\right)\)
=> B = \(3.\frac{8}{33}\)
=> B = \(\frac{8}{11}\)
Vậy: B = \(\frac{8}{11}\)
D= 1/1 - 1 /2 + 1/2 - 1/4 + 1/4 - 1/7 +...+ 1/46 - 1/56
D= 1/1 - 1/56
D= 55/56
vậy D= 55/56
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{29}\)
=1-1/29
=28/29
P=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{46}-\frac{1}{56}\)
P=\(1-\frac{1}{56}\)
P=\(\frac{55}{56}\)
\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{46}-\frac{1}{56}\)
\(P=1-\frac{1}{56}\)
\(P=\frac{55}{56}\)
Ta có :
\(C=\frac{2}{1.2}+\frac{4}{2.4}+\frac{6}{4.7}+\frac{8}{7.11}+\frac{10}{11.16}+\frac{12}{16.22}\)
\(C=2\left(\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}+\frac{6}{16.22}\right)\)
\(C=2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{22}\right)\)
\(C=2\left(1-\frac{1}{22}\right)\)
\(C=2-\frac{1}{11}\)
\(C=\frac{21}{11}\)
Vậy \(C=\frac{21}{11}\)
Chúc bạn học tốt ~