C= 4/4+ 4/28+ 4/70+ 4/130+...+ 4/418+ 4/550
Ghi rõ cách làm giúp mk vs
Cảmon
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Đặt A = 9/4 + 9/28 +.. + 9/550
A = 9/1.4 + 9/4.7 +... + 9/22.25
A = 3( 3/1.4 + 3/4.7 + .. + 3/22.25)
A = 3 . (1/1 - 1/4 + 1/4 - 1/7 + ... +1/22 - 1/25)
A = 3 (1 - 1/25)
A = 3. 24 / 25
A = 72/25
A=5/4+5/28+5/70+5/130+...+5/418+5/550
Xét:5/1.4+5/4.7+5/7.10+5/10.13+...+5/19.22+5/22.25
=5-5/25
=24/5
\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{418}+\frac{3}{550}\)
\(\Leftrightarrow\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{19.22}+\frac{3}{22.25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
Nhớ k cho m nhé!
= \(\frac{9}{1x4}+\frac{9}{4x7}+\frac{9}{7x10}+.........+\frac{9}{19x22}+\frac{9}{22x25}\)
= \(\frac{1}{3}x\left(\frac{9}{1}-\frac{9}{4}\right)+\left(\frac{9}{4}-\frac{9}{7}\right)x\frac{1}{3}+........+\left(\frac{9}{22}-\frac{9}{25}\right)x\frac{1}{3}\)
= \(\frac{1}{3}\left(\frac{9}{1}-\frac{9}{4}+\frac{9}{4}-\frac{9}{7}+....+\frac{9}{22}-\frac{9}{25}\right)\)
= \(\frac{1}{3}x\left(\frac{9}{1}-\frac{9}{25}\right)\)
= \(\frac{1}{3}x\frac{216}{25}\)
= \(\frac{72}{25}\)
nhớ ********** nha bn thân
\(\frac{9}{4}+\frac{9}{28}+\frac{9}{70}+\frac{9}{130}+...+\frac{9}{418}+\frac{9}{550}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{19.22}+\frac{3}{22.25}\right)\)
\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{22}-\frac{1}{25}\right)\)
\(=3\left(1-\frac{1}{25}\right)\)
\(=3.\frac{24}{25}=\frac{72}{25}\)
Lướt qua rồi! không phải bạn k mà ấn tượng "đừng lướt qua"
\(A=\frac{3a}{4.1}+\frac{3a}{7.4}+\frac{3a}{10.7}+\frac{3a}{13.10}+..+\frac{3a}{22.19}+\frac{3a}{25.22}=\frac{48}{25}\)
\(a.\left(\frac{3}{4.1}+\frac{3}{7.4}+\frac{3}{10.7}+\frac{3}{13.10}+..+\frac{3}{22.19}+\frac{3}{25.22}\right)=\frac{48}{25}\)
\(B=\left(\frac{3}{4.1}+\frac{3}{7.4}+\frac{3}{10.7}+\frac{3}{13.10}+..+\frac{3}{22.19}+\frac{3}{25.22}\right)\)
\(B=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{22}-\frac{1}{25}\)
\(B=\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
\(A=a.B=\frac{24a}{25}=\frac{48}{25}\Rightarrow a=2\)
\(\frac{3a}{4}+\frac{3a}{28}+\frac{3a}{70}+...+\frac{3a}{418}+\frac{3a}{550}=\frac{48}{25}\)
\(\Rightarrow a\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{19.22}+\frac{3}{22.25}\right)=\frac{48}{25}\)
\(\Rightarrow a\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\right)=\frac{48}{25}\)
\(\Rightarrow a\left(1-\frac{1}{25}\right)=\frac{48}{25}\)
\(\Rightarrow a.\frac{24}{25}=\frac{48}{25}\)
\(\Rightarrow a=2\)
\(\dfrac{4}{3}\times3+\dfrac{2}{5}=4+\dfrac{2}{5}=\dfrac{22}{5}\)
Bạn tải photomath về là giải được ngay !
Nhớ k cho mình nhé !
Gọi biểu thức này là A
Ta có :
\(A=\frac{4}{45}+\frac{4}{105}+\frac{4}{189}+\frac{4}{297}+\frac{4}{929}\)
\(\frac{3}{2}A=\frac{3}{2}\times\left(\frac{4}{45}+\frac{4}{105}+\frac{4}{189}+\frac{4}{297}+\frac{4}{929}\right)\)
\(\frac{3}{2}A=\frac{6}{45}+\frac{6}{105}+\frac{6}{189}+\frac{6}{297}+\frac{6}{929}\)
\(\frac{3}{2}A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{???}\)
Bạn nên xem lại bài
\(C=\dfrac{4}{4}+\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{418}+\dfrac{4}{550}\\ =\dfrac{4}{1\times4}+\dfrac{4}{4\times7}+\dfrac{4}{7\times10}+\dfrac{4}{10\times13}+...+\dfrac{4}{19\times22}+\dfrac{4}{22\times25}\\ =\dfrac{4}{3}\times\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{19}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{25}\right)\\ =\dfrac{4}{3}\times\left(1-\dfrac{1}{25}\right)=\dfrac{4}{3}\times\dfrac{24}{25}=\dfrac{32}{25}\)
C = \(\dfrac{4}{4}\) + \(\dfrac{4}{28}\)+ \(\dfrac{4}{70}\) + \(\dfrac{4}{130}\)+.....+\(\dfrac{4}{418}\)+ \(\dfrac{4}{550}\)
C = \(\dfrac{4}{3}\)( \(\dfrac{3}{4}\)+\(\dfrac{3}{28}\)+ \(\dfrac{3}{70}\)+ \(\dfrac{3}{130}\)+.....+\(\dfrac{3}{418}\)+ \(\dfrac{3}{550}\))
C = \(\dfrac{4}{3}\)( \(\dfrac{3}{1.4}\)+ \(\dfrac{3}{4.7}\)+ \(\dfrac{3}{7.10}\)+......+\(\dfrac{3}{19.22}\)+\(\dfrac{3}{22.25}\))
C = \(\dfrac{4}{3}\)( \(\dfrac{1}{1}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)- \(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\)- \(\dfrac{1}{10}\)+ ......+\(\dfrac{1}{19}\)- \(\dfrac{1}{22}\)+ \(\dfrac{1}{22}\)- \(\dfrac{1}{25}\))
C = \(\dfrac{4}{3}\)(1- \(\dfrac{1}{25}\))
C = \(\dfrac{4}{3}\) . \(\dfrac{24}{25}\)
C =\(\dfrac{32}{25}\)