Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{5}{39}\)
e) \(\dfrac{1}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{1}{5}-\dfrac{1}{5}=\dfrac{1}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{1}{5}=\dfrac{1}{5}.1-\dfrac{1}{5}=0\)
=32.233+67.233+68.233
=233.(68+32+67)
=233.167
=38911
hok tốt k nhé bạn
\(32\times\left(233+67\right)+68\times233\)
=\(32\times233+67\times233+68\times233\)
=\(233\left(32+67+68\right)\)
=\(233\times167\)
=\(38911\)
#hoctot
A = 1.2 + 2.3 + 3.4 + ... + 1999.2000
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 1999.2000.3
3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 1999.2000.(2001 - 1998)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 1999.2000.2001 - 1998.1999.2000
3A = 1999.2000.2001
A = 1999.2000.2001 : 3
A = 2 666 666 000
T=\(\frac{2,85\times\left(156,23-56,23\right)}{\left(1,8+4,2\right)+\left(2,1+3,9\right)+\left(2,4+3,6\right)+\left(2,7+3,3\right)+\left(3+1,5\right)}\)=\(\frac{2,85\times100}{6+6+6+6+4,5}\)=\(\frac{285}{28,5}\)=10
chúc bn học tốt
)
A = 1.2 + 2.3 + 3.4 + ... + 1999.2000
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 1999.2000.3
3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 1999.2000.(2001 - 1998)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 1999.2000.2001 - 1998.1999.2000
3A = 1999.2000.2001
A = 1999.2000.2001 : 3
A = 2 666 666 000
= 11/20 + 163/20
=49/15
= 11/20 + 163/20
=49/15