Tính nhanh 6/5×5/6 - 1/5×7/6×3+3/5 ÷ 6/3
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Đặt P = ... ( biểu thức đề bài )
Nhận xét: Với \(k\inℕ^∗\) ta có:
\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)
\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)
1 + 2 + 3 + 5 + 6 + 7 - 7 - 6 - 5 - 3 - 2 - 1
= ( 1 - 1 ) + ( 2 - 2 ) + ( 3 - 3 ) + ( 5 - 5 ) + ( 6 - 6 ) + ( 7 - 7 )
= 0
nha
= (1/7+7/1)+(2/6+6/2)+(3/5+5/3)+4/4
= 7/1/7+3/1/3+2/4/15+1
= 13/26/35
b: \(B=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{3}{8}+\dfrac{6}{8}+\dfrac{-6}{11}-\dfrac{5}{11}=-2-1+\dfrac{9}{8}=\dfrac{9}{8}-3=-\dfrac{15}{8}\)
c: \(C=\left(\dfrac{4}{3}+\dfrac{7}{3}+\dfrac{1}{3}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)=4+1=5\)
d: \(D=\dfrac{4}{19}\left(\dfrac{-5}{6}-\dfrac{7}{12}\right)-\dfrac{40}{57}\)
\(=\dfrac{4}{19}\cdot\dfrac{-17}{12}-\dfrac{40}{57}=-1\)
e: \(E=\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{9}{5}\right)+\dfrac{2}{3}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
a, \(\dfrac{3}{5}\) \(\times\) \(\dfrac{6}{7}\) + \(\dfrac{3}{5}\) : 7 + \(\dfrac{6}{5}\)
= \(\dfrac{3}{5}\) \(\times\) \(\dfrac{6}{7}\) + \(\dfrac{3}{5}\) \(\times\) \(\dfrac{1}{7}\) + \(\dfrac{6}{5}\)
= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{6}{7}\) + \(\dfrac{1}{7}\)) + \(\dfrac{6}{5}\)
= \(\dfrac{3}{5}\) + \(\dfrac{6}{5}\)
= \(\dfrac{9}{5}\)
b, 2018 \(\times\) ( \(\dfrac{1}{2}\) + \(\dfrac{1212}{2424}\))
= 2018 \(\times\) ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\))
= 2018 \(\times\)1
= 2018
a) 3/5 x 6/7 + 3/5 x 1/7 + 6/5
3/5 x ( 6/7 + 1/7 ) + 6/5
3/5 x 7/7 + 6/5
3/5 x 1 + 6/5
3/5 + 6/5
9/5
b) 2018 x ( 1/2 + 1212/ 2424 )
2018 x ( 1/2 + 12/24 )
2018 x ( 1/2 + 1/2 )
2018 x 2/2
2018 x 1
2018
Học tốt
= 51/20
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= 3/14
------------------------------------
= 4/7 - 2/7
= 2/7
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= 17/45
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= 2/3
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= 2 x 1/4
= 1/2