tinh nhanh 2011 + 2011 + 2011 + 2011 - 2008 x 4 mik dang can gap
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\(\left(x-2011\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(x-2011=8\)
\(x=2019\)
\(\frac{x-2011}{12}+\frac{x-2011}{20}+\frac{x-2011}{30}+\frac{x-2011}{42}+\frac{x-2011}{56}+\frac{x-2011}{72}=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\frac{2}{9}=\frac{16}{9}\)
\(x-2011=8\Rightarrow x=2019\)
A = |x-2010|+(y+2011)2010 +2011>(=)0+0+2011=2011
dấu = xảy ra khi x-2010=0;y+2011=0
=>x=2010;y=-2011
vậy Min A=2011 khi x=2010 và y=-2011
\(b,S=\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}\)
\(\text{Ta có: }\frac{2007}{2008}< 1\)
\(\frac{2008}{2009}< 1\)
\(\frac{2009}{2010}< 1\)
\(\frac{2010}{2011}< 1\)
\(\Rightarrow\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}< 1+1+1+1\)
\(\Rightarrow\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}< 4\)
a) 100 - 99 + 98 -97 + 96 -95 +...+ 4-3 + 2
= (100 - 99) + (98 -97) + (96 - 95) +...+ (4-3) +2 (gồm 49 cặp và 1 số hạng)
= 1+1+1+....+1 +2
= 49 x 1 + 2 = 51
b) 100 - 5-...-5 - 5 (20 số 5)
= 100 - 20 x 5 = 0
c) 99 - 9 - 9 -... - 9 -9 (11 số 9)
=99 - 11 x 9 = 0
d) 2011 + 2011+2011+2011 - 2008 x 4
= 2011 x 4 - 2008 x 4
= 4 x (2011 - 2008)
= 4 x 3
=12
Bạn chỉ cần lấy : (2008/2009+2009/2010+2010/2011+2011/2008)-4=số dương
vậy (2008+...2008) > 4
Có 4 = 1+1+1+1
Vì 2008/2009<1 ; 2009/2010<1; 2010/2011<1;2011/2012<1
=>2008/2009+2009/2010 + 2010/2011+2011/2012<1+1+1+1=4
Ta có : \(\frac{x+2011}{1}+\frac{x+2008}{2}+\frac{x+2007}{3}+\frac{x+2011}{5}=-15\)
\(\Rightarrow\left(\frac{x+2011}{1}+5\right)+\left(\frac{x+2008}{2}+4\right)+\left(\frac{x+2007}{3}+3\right)+\left(\frac{x+2008}{4}+2\right)+\left(\frac{x+2011}{5}+1\right)\)
\(=0\)
=> \(\frac{x+2016}{1}+\frac{x+2016}{2}+\frac{x+2016}{3}+\frac{x+2016}{4}+\frac{x+2016}{5}=0\)
=> \(\left(x+2016\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=0\)
Vì \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\ne0\)
=> x + 2016 = 0
=> x = -2016
Vậy x = -2016
x+(x+2008)1/2+(x+2007)1/3+(x+2008)1/4+(x+2011)1/5=-15-2011=-2026
<=> x+x/2+1004+x/3+669+x/4+502+x/5+2011/5=-2026
<=>x+x/2+x/3+x/4+x/5+2011/5=-2026-1004-669-502=-4201
<=>x(1+(1)/(2)+(1)/(3)+(1)/(4)+(1)/(5))=-4201-(2011)/(5)=-23016/5
<=>x=-23016/5:(1+1/2+1/3+1/4+1/5)=-2016
(1999/2011-2011/1999) - (-12/1999 - 12/2011)
= 1999/2011 - 2011/1999 + 12/1999 + 12/2011
= (1999/2011 + 12/2011) - 1999/1999 - 12/1999 + 12/1999
= ( 1-1 ) + ( -12/1999 + 12/1999)
= 0 + 0
= 0
(1999/2011-2011/1999)-(12/1999-12/2011)
=1999/2011-2011/1999-12/1999+12/2011
=(1999/2011+12/2011)-(2011/1999-12/1999)
=1-1
=0
tính P sau nó sẽ ra Q, tìm ra P/Q=1
Chi tiết như sau:
Từ đầu bài \(\Rightarrow P=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2012}\right)\)vì cộng vào bên trái 1 lượng bằng những phân số có dấu âm nên trừ đi 2 lần những phân số đó
\(\Rightarrow P=1+\frac{1}{2}+...+\frac{1}{2012}-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)( nhân 2 vào trong)
\(\Rightarrow P=Q\)(trừ đi những cái phân số từ 1 đến 1/1006 sẽ ra Q)
2011 + 2011 + 2011 + 2011 - 2008 x 4
= 2011 x 4 - 2008 x 4
= 4 x ( 2011 - 2008 )
= 4 x 3
= 12
bàng 12 nhé bạn