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NH
1
6 tháng 10 2018
\(2018^2-2017.2019\)
\(=2018^2-\left(2018-1\right)\left(2018+1\right)\)
\(=2018^2-\left(2018^2-1\right)=1\)
\(56^2+56.88+44^2\)
\(=56^2+2.56.44+44^2\)
\(=\left(56+44\right)^2\)
\(=100^2=10000\)
\(\frac{2018^3+1}{2018^2-2017}\)
\(=\frac{\left(2018+1\right)\left(2018^2-2018+1\right)}{2018^2-2017}\)
\(=\frac{2019\left(2018^2-2017\right)}{2018^2-2017}=2019\)
Chúc bạn học tốt.
28 tháng 6 2017
Sửa đề: \(\dfrac{100+\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\left(\dfrac{99}{2}+1\right)+\left(\dfrac{98}{3}+1\right)+...+\left(\dfrac{1}{100}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}+\dfrac{101}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}}-2\)
\(=\dfrac{101\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}}-2\)
\(=101-2=99\)
Vậy...
18 tháng 10 2017
Ta có: \(\dfrac{n^3-1}{n^3+1}=\dfrac{\left(n-1\right)\left(n^2+n+1\right)}{\left(n+1\right)\left(n^2-n+1\right)}=\dfrac{\left(n-1\right)[\left(n+0,5\right)^2+0,75]}{\left(n+1\right)[\left(n-0,5\right)^2+0,75]}\)
Thay vào M ta có:
\(M=\dfrac{2,5^2+0.75}{3.\left(1,5^2+0,75\right)}.\dfrac{2.\left(3,5^2+0,75\right)}{4.\left(2,5^2+0,75\right)}...\dfrac{99[\left(100,5\right)^2+0,75]}{101.[\left(99,5\right)^2+0,75}\)
\(=\dfrac{1.2.3...99}{3.4.5...101}.\dfrac{\left(2,5^2+0,75\right).\left(3,5^2+0,75\right)...[\left(100,5\right)^2+0,75]}{\left(1,5^2+0,75\right).\left(2,5^2+0,75\right)...[\left(99,5\right)^2+0,75]}\)\(=\dfrac{1.2}{100.\left(101\right)}.\dfrac{\left(100,5\right)^2+0,75}{1,5^2+0,75}=\dfrac{2}{3}.\dfrac{\left(100^2+100+1\right)}{3.100.101}>\dfrac{2}{3}\left(đpcm\right)\)
28 tháng 5 2017
Nguyễn Trần Thành ĐạtXuân Tuấn TrịnhHung nguyenHoang HungQuan Ace Legona giúp với
NP
2
8 tháng 6 2016
a) \(56^2+44^2+2.56.44=56^2+2.56.44+44^2=\left(56+44\right)^2=100^2=10000\)
b) \(36^2+64^2+72.64=36^2+2.36.64+64^2=\left(36+64\right)^2=100^2=10000\)
c) \(136^2+36^2-72.136=136^2-2.36.136+36^2=\left(136-36\right)^2=100^2=10000\)
9 tháng 6 2016
a) $56^2+44^2+2.56.44=56^2+2.56.44+44^2=\left(56+44\right)^2=100^2=10000$562+442+2.56.44=562+2.56.44+442=(56+44)2=1002=10000
b) $36^2+64^2+72.64=36^2+2.36.64+64^2=\left(36+64\right)^2=100^2=10000$362+642+72.64=362+2.36.64+642=(36+64)2=1002=10000
c) $136^2+36^2-72.136=136^2-2.36.136+36^2=\left(136-36\right)^2=100^2=10000$
3
PT
5 tháng 7 2018
\(\dfrac{x+1}{58}+\dfrac{x+2}{57}=\dfrac{x+3}{56}+\dfrac{x+4}{55}\)
\(\Leftrightarrow\left(\dfrac{x+1}{58}+1\right)+\left(\dfrac{x+2}{57}+1\right)=\left(\dfrac{x+3}{56}+1\right)+\left(\dfrac{x+4}{55}+1\right)\)
\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}-\dfrac{x+59}{56}-\dfrac{x+59}{55}=0\)
\(\Leftrightarrow\left(x+59\right)\left(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\right)=0\)
\(\Leftrightarrow x+59=0\)
\(\Leftrightarrow x=-59\)
5 tháng 7 2018
\(\dfrac{x+1}{58}+\dfrac{x+2}{59}=\dfrac{x+3}{56}+\dfrac{x+4}{55}\)
\(\Leftrightarrow\dfrac{x+1}{58}+1+\dfrac{x+2}{57}+1=\dfrac{x+3}{56}+1+\dfrac{x+4}{55}+1\)
\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}=\dfrac{x+59}{56}+\dfrac{x+59}{55}\)
\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}-\dfrac{x+59}{56}-\dfrac{x+59}{55}=0\)
\(\Leftrightarrow\left(x+59\right)\left(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\right)=0\)
Mà \(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\ne0\)
\(\Rightarrow x+59=0\)
\(\Leftrightarrow x=-59\)
Vậy: \(S=\left\{-59\right\}\)
ND
3
TR
11 tháng 3 2017
\(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)
\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}=\dfrac{x-4}{56}-1+\dfrac{x-5}{55}-1+\dfrac{x-6}{54}-1\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{56}+\dfrac{x-60}{55}+\dfrac{x-60}{54}\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}-\dfrac{1}{54}\right)=0\)
\(\Leftrightarrow x-60=0\)
\(\Rightarrow x=60\)
vậy \(S=\left\{60\right\}\)
22 tháng 6 2022
b: \(\Leftrightarrow\left(\dfrac{29-x}{21}+1\right)+\left(\dfrac{27-x}{23}+1\right)+\left(\dfrac{25-x}{25}+1\right)+\left(\dfrac{23-x}{27}+1\right)+\left(\dfrac{21-x}{29}+1\right)=0\)
=>50-x=0
hay x=50
c: \(\Leftrightarrow\dfrac{x-2}{2001}+1=\dfrac{x-1}{2002}+\dfrac{x}{2003}\)
\(\Leftrightarrow\left(\dfrac{x-2}{2001}-1\right)=\left(\dfrac{x-1}{2002}-1\right)+\left(\dfrac{x}{2003}-1\right)\)
=>x-2003=0
hay x=2003
\(\dfrac{56^3+44^3}{100}-14.56\)
\(=\dfrac{\left(56+44\right)\left(56^2-56.44+44^2\right)}{100}-14.56\)
\(=\dfrac{100.\left(56^2-56.44+44^2\right)}{100}-14.56\)
\(=\)\(56^2-44.56+44^2-14.56\)
\(=\left(56^2-14.56\right)-\left(44.56-44^2\right)\)
\(=56\left(56-14\right)-44\left(56-44\right)\)