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a: \(B=\left(-\dfrac{1}{5}-\dfrac{5}{7}+\dfrac{-3}{35}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{2}\right)+\dfrac{1}{41}\)
\(=\dfrac{-7-25-3}{35}+\dfrac{3+2+1}{6}+\dfrac{1}{41}=\dfrac{42}{41}-1=\dfrac{1}{41}\)
Chào bạn, bạn hãy theo dõi bài giải của mình nhé!
\(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}=\frac{\frac{88}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}=\frac{\frac{115}{132}}{\frac{103}{132}}=\frac{115}{132}:\frac{103}{132}=\frac{115}{132}\cdot\frac{132}{103}=\frac{115\cdot132}{132\cdot103}=\frac{115}{103}\)
Chúc bạn học tốt!
M= \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}\)
=\(\frac{3.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}\)
=\(\frac{3}{4}\)
\(-\frac{3}{5}.\frac{2}{7}+-\frac{3}{5}.\frac{5}{7}+2\frac{3}{5}\)
\(=-\frac{3}{5}\left(\frac{2}{7}+\frac{5}{7}\right)+2\frac{3}{5}\)
\(=-\frac{3}{5}+\frac{13}{5}\)
\(=\frac{10}{5}=2\)
ta có : ( -5/28 +7/4 + 8/35 ) : (- 69/20)
= ( -25/140 + 245/140 + 32/140 ) x (-20/69)
= (252/140) x (-20/69)
= (9/5) x (-20/69)
= (- 12/23)
tính nhanh:
2 x 3/7 + (2/9 - 10/7) - 5/3 x 9
= 6/7 + 2/9 - 10/7 - 5/3 x 9 = 6/7 + 2/9 - 10/7 - 15
= (6/7 - 10/7 ) + (2/9 - 135/9) = ( - 4/7 ) + (-133/9 )
= (- 36/63) + (-931/63)
= (- 967/63)
đặt A= \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)
B=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{10000}{10001}\)
Lấy A.B= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10000}{10001}=\frac{1}{10001}\)
mặt khác
Ta có
\(\frac{1}{2}< \frac{2}{3}\\\)
\(\frac{3}{4}< \frac{4}{5}\)
....
\(\frac{9999}{10000}< \frac{10000}{10001}\)
=> A<B
=> A.A<A.B
=>A2<\(\frac{1}{10001}< \frac{1}{10000}\)
=>A<\(\sqrt{\frac{1}{10000}}=\frac{1}{100}\)
Vậy \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)<\(\frac{1}{100}\)
ĐPCM
Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)
Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)
Vậy B < 1
A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}=\frac{3}{5}-\frac{1}{7}\)
\(=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)