A=(1/4*9 + 1/9*14 + 1/14*19 + ...+ 1/44*49)*1-3-5-7-..-49/89
Những câu hỏi liên quan
( 1 / 4 . 9 + 1 / 9 . 14 + 1 / 14 . 19 + ... + 1 / 44 .49 ) . (1 - 3 - 5 - 7 - ... - 49) / 89 = ?
TÍNH A=[(1/4*9)+(1/9*14)+(1/14*19)+...+(1/44*49)] x (1-3-5-7-...-49)/89
M=(1/4*9+1/9*14+1/14*19+...1/44*49)*1-3-5-7-9-...-49/89=-9/28
tính 1/4*9+1/9*14+1/14*19+...+1/44*49 * 1-3-5-...-49/89 -19/28
(1/4*9+1/9*14+...+1/44*49)*(1-3-5-7-...-49/89)
Giúp mình vs . Mình cám ơn nhiều
A=(1/4×9+1/9×14+1/14×19+...+1/44×49)×(1-3-5-7-...-49)/89
\(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{99}\right).\dfrac{1-\left(3+5+...+49\right)}{89}\)
\(=\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{49}\right).\dfrac{1-\dfrac{\left(3+49\right).\left[\left(49-3:2+1\right)\right]}{2}}{89}\)
\(=\dfrac{1}{5}.\dfrac{45}{196}.\dfrac{1-\dfrac{52.24}{2}}{89}\)
\(=\dfrac{9}{196}.\dfrac{1-\dfrac{1248}{2}}{89}\)
\(=\dfrac{9}{196}.\dfrac{1-624}{89}\)
\(=\dfrac{9}{196}.\left(-\dfrac{623}{89}\right)\)
\(=-\dfrac{9}{28}\)
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\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+\frac{1}{19\cdot24}+.....+\frac{1}{44\cdot49}\right)1\frac{-3-5-7-...-49}{89}\)
\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+\frac{1}{19.24}+...+\frac{1}{44.49}\right)\frac{1-3-7-...-49}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{1+\left(-3\right)+\left(-7\right)+...+\left(-49\right)}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{1-\left(3+7+...+49\right)}{89}\)
\(=\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\left(49+3\right).24:2\right)}{89}\)
\(=\frac{9}{196}.\frac{-623}{89}\)
\(=\frac{9}{196}.\left(-7\right)\)
\(=\frac{-9}{28}\)
CHÚC BN HỌC TỐT!!!!
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\(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
Ta có: \(\frac{1-3-5-7-...-49}{89}=\frac{1-\left(3+5+7+...+49\right)}{89}=\frac{1-12.52}{89}=-\frac{623}{89}=-7\)
=> \(A=-7\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)=-\frac{7}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{44.49}\right)\)
=>\(A=-\frac{7}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)=-\frac{7}{5}\left(\frac{1}{4}-\frac{1}{49}\right)=-\frac{7}{5}.\frac{45}{196}\)
=> \(A=-\frac{7}{5}.\frac{5.9}{28.7}=-\frac{9}{28}\)
Đáp số: A = -9/28
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Ta có: 1−3−5−7−...−49 /89 =1−(3+5+7+...+49) /89 =1−12.52 /89 =−623 /89 =−7/5
=> A=−7(1/4.9 +1/9.14 +1/14.19 +...+1/44.49 )=−7/5 (5/4.9 +5/9.14 +5/14.19 +...+5/44.49 )
=> A=−75 .5.928.7 =−928
Đáp số: A = -9/28
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Tính \(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(B=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó
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