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\(A=1+\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}\)
\(< =>A=1+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)
\(< =>2A=2+\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)
\(< =>2A=2+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}\)
\(< =>2A=\frac{5}{2}-\frac{1}{12}=\frac{29}{12}\)
\(< =>A=\frac{29}{12}.\frac{1}{2}=\frac{29}{24}\)
Bài 1:
gọi số đó là x
ta có : \(\frac{-1}{12}< x< \frac{-1}{2}\)
hay :
\(\frac{-1}{12}< x< \frac{-6}{12}\)
vậy \(x\in\left\{\frac{-2}{12};\frac{-3}{12};\frac{-4}{12};\frac{-5}{12}\right\}\)
Tính tổng tất cả các phân số có mẫu số là 12 là :
\(\frac{-2}{12}+\frac{-3}{12}+\frac{-4}{12}+\frac{-5}{12}=\frac{-14}{12}=\frac{-7}{6}\)
bài 2:
\(A=1+\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}\)
\(A=1+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)
\(2A=2+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}\)
\(2A=3-\frac{1}{12}\)
\(A=\left(\frac{35}{12}\right):2=\frac{35}{24}\)
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)....\left(1+\frac{1}{224}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}......\frac{225}{224}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.....\frac{15.15}{14.16}\)
\(=\frac{\left(2.3.4....15\right)\left(2.3.4....15\right)}{\left(2.3.4....14\right)\left(3.4.5....16\right)}\)
\(=\frac{15.2}{16}=\frac{15}{8}\)
#It's the moment when you're in good mood, you accidentally click back =.=
1) Calculate
\(P=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{63}.1\frac{1}{80}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{64}{63}.\frac{81}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.9}{10}=\frac{9}{5}\)
ta có: 10010 + 1 > 10010 - 1
⇒ A = \(\frac{100^{10}+1}{100^{10}-1}< \frac{100^{10}+1-2}{100^{10}-1-2}=\frac{100^{10}-1}{100^{10}-3}=B\)
vậy A < B
1)
a) \(-\frac{8}{15}< \frac{x}{45}< -\frac{2}{5}\)
Lại có: \(-\frac{8}{15}=\frac{-24}{45};-\frac{2}{5}=\frac{-18}{45}\)
=> \(-\frac{24}{45}< \frac{x}{45}< -\frac{18}{45}\)
=> -24 < x < - 18
=> x \(\in\){ - 23; -22; -21; -20 ; -19 } ( thử lại thỏa mãn )
b) \(x=\frac{-4}{3}+\frac{-7}{5}=-\frac{4.5}{3.5}+\frac{-7.3}{5.3}=-\frac{41}{15}\)
c) \(\frac{83}{x}=\frac{13}{4}+\frac{9}{10}=\frac{83}{20}\)
=> x = 20 ( thử lại thỏa mãn)
d) \(x=\frac{10}{8}+\frac{-24}{48}+\frac{105}{-120}=-\frac{1}{8}\)
e) \(\left|x-\frac{1}{2}\right|=\left|-\frac{2}{7}\right|+\frac{5}{4}\)
\(\left|x-\frac{1}{2}\right|=\frac{2}{7}+\frac{5}{4}\)
\(\left|x-\frac{1}{2}\right|=\frac{43}{28}\)
TH1: \(x-\frac{1}{2}=\frac{43}{28}\)
\(x=\frac{57}{28}\)
TH2: \(x-\frac{1}{2}=-\frac{43}{28}\)
\(x=-\frac{29}{28}\)
A=1/8+1/24+1/48+1/80+1/120+1/168+1/224=>2A=2/8+2/24+2/48+2/80+2/120+2/168+2/224
2A=2/2*4+2/4*6+2/6*8+2/8*10+2/10*12+2/12*14+2/14*16
2A=1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12+1/12-1/14+1/14-1/16
2A=1/2-1/16
2A=7/16
A=7/16:2
A=7/32