sắp xếp :
\(\dfrac{1001}{2002};\dfrac{-1003}{2003};\dfrac{-1002}{2003};\dfrac{-1003}{-2002};\dfrac{1004}{-2003}\)
giải hương trình:\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}\)=4
\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)
\(\Leftrightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)
\(\Leftrightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}=0\right)\)
\(\Leftrightarrow x-2007=0\)
\(\Leftrightarrow x=2007\)
chứng minh \(\dfrac{2002}{\sqrt{2003}}+\dfrac{2003}{\sqrt{2002}}>\sqrt{2002}+\sqrt{2003}\)
Đặt 2002=a; 2003=b
Theo đề, ta có:
\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}>\sqrt{a}+\sqrt{b}\)
\(\Leftrightarrow\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}>\sqrt{a}+\sqrt{b}\)
\(\Leftrightarrow a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{a}>0\)
\(\Leftrightarrow a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)>0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(\sqrt{a}+\sqrt{b}\right)>0\)(luôn đúng)
Tìm x biết: \(\dfrac{x-1016}{1001}+\dfrac{x-13}{1002}+\dfrac{x+992}{1003}=\dfrac{x+995}{1004}+\dfrac{x-7}{1005}+1\)
\(\dfrac{x-1016}{1001}+\dfrac{x-13}{1002}+\dfrac{x+992}{1003}=\dfrac{x+995}{1004}+\dfrac{x-7}{1005}+1\)
<=>\(\dfrac{x-1016}{1001}-1+\dfrac{x-13}{1002}-2+\dfrac{x+992}{1003}-3=\dfrac{x+995}{1004}-3+\dfrac{x-7}{1005}-2\)
<=>\(\dfrac{x-2017}{1001}+\dfrac{x-2017}{1002}+\dfrac{x-2017}{1003}=\dfrac{x-2017}{1004}+\dfrac{x-2017}{1005}\)
<=>\(\left(x-2017\right)\left(\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}-\dfrac{1}{1004}-\dfrac{1}{1005}\right)=0\)
vì 1/1001+1/1002+1/1003-1/1004-1/1005 khác 0 nên x-2017=0<=>x=2017
vậy..........
Chứng minh rằng:
\(\dfrac{2002}{\sqrt{2003}}+\dfrac{2003}{\sqrt{2002}}>\sqrt{2002}+\sqrt{2003}\)
Đặt \(\sqrt{2002}=a,\sqrt{2003=b}\)
Ta có:
VT = \(\dfrac{a^2}{b}+\dfrac{b^2}{a}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng engel ta có:
\(\dfrac{a^2}{b}+\dfrac{b^2}{a}\ge\dfrac{\left(a+b\right)^2}{a+b}=a+b\)
hay \(\dfrac{2002}{\sqrt{2003}}+\dfrac{2003}{\sqrt{2002}}\ge\sqrt{2002}+\sqrt{2003}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b\)
Mà \(a\ne b\)
\(\Rightarrow\)\(\dfrac{2002}{\sqrt{2003}}+\dfrac{2003}{\sqrt{2002}}>\sqrt{2002}+\sqrt{2003}\)(đpcm)
chứng minh bất đẳng thức
\(\dfrac{2002}{\sqrt{2003}}+\dfrac{2003}{\sqrt{2002}}>\sqrt{2002}+\sqrt{2003}\)
\(\dfrac{2002}{\sqrt{2003}}+\dfrac{2003}{\sqrt{2002}}\)
\(=\dfrac{2002+1}{\sqrt{2003}}+\dfrac{2013-1}{\sqrt{2002}}+\dfrac{1}{\sqrt{2002}}-\dfrac{1}{\sqrt{2003}}\)
\(=\sqrt{2003}+\sqrt{2002}+\dfrac{1}{\sqrt{2002}}-\dfrac{1}{\sqrt{2003}}\)
\(>\sqrt{2003}+\sqrt{2002}+\dfrac{1}{\sqrt{2003}}-\dfrac{1}{\sqrt{2003}}=\sqrt{2003}+\sqrt{2002}\left(đpcm\right)\)
so sánh A= \(\dfrac{2003^{2003}+1}{2003^{2004}+1}\)
B=
\(\dfrac{2003^{2002}+1}{2003^{2003}+1}\)
Ta có: \(2003^{2003}+1=2003^{2002+1}+1và2003^{2004}+1=2003^{2003+1}+1\)
\(\Rightarrow A>B\)
Giải phương trình sau :
\(\dfrac{x-4}{2001}+\dfrac{x-3}{2002}+\dfrac{x-2}{2003}=\dfrac{x-2003}{2}+\dfrac{x-2002}{3}+\dfrac{x-2001}{4}\)
\(\dfrac{x-4}{2001}\)- 1 +\(\dfrac{x-3}{2002}\)-1 + \(\dfrac{x-2}{2003}\)-1 =\(\dfrac{x-2003}{2}\)-1 + \(\dfrac{x-2002}{3}\)-1 +\(\dfrac{x-2001}{4}\)-1 <=> \(\dfrac{x-2005}{2001}\)+\(\dfrac{x-2005}{2002}\)+\(\dfrac{x-2005}{2003}\)-\(\dfrac{x-2005}{2}\)-\(\dfrac{x-2005}{3}\)-\(\dfrac{x-2005}{4}\)= 0 <=> (x-2005). (\(\dfrac{1}{2001}\)+\(\dfrac{1}{2002}\)+\(\dfrac{1}{2003}\)-\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)) =0 <=> x-2005=0 ( vì \(\dfrac{1}{2001}\) +\(\dfrac{1}{2002}\) +\(\dfrac{1}{2003}\)- \(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\) khác 0) =>x = 2005
x-4/2001+ x-3/2002 + x-2/2003= x-2003/2 + x-2002/3 + x-2001/4
<=>(x-4/2001 -1)+(x-3/2002 -1)+(x-2/2003 -1)-(x-2003/2 -1)+
(x-2002/3 -1)+(x-2001/4 -1) =0
<=>x-2005/2001+ x-2005/2002+ x-2005/2003- x-2005/2-
x-2005/3- x-2005/4 =0
<=>(x-2005).(1/2001+1/2002+1/2003- 1/2- 1/3- 1/4)=0
<=>x-2005=0 (vì 1/2001+1/2002+1/2003-1/2-1/3-1/4)
<=>x=2005
Vậy pt có nghiệm là x=2005
\(\dfrac{x-4}{2001}+\dfrac{x-3}{2002}+\dfrac{x-2}{2003}=\dfrac{x-2003}{2}+\dfrac{x-2002}{3}+\dfrac{x-2001}{4}\)
\(\Leftrightarrow\dfrac{x-4}{2001}-1+\dfrac{x-3}{2002}-1+\dfrac{x-2}{2003}-1=\dfrac{x-2003}{2}-1+\dfrac{x-2002}{3}-1+\dfrac{x-2001}{4}-1\)
\(\Leftrightarrow\dfrac{x-2005}{2001}+\dfrac{x-2005}{2002}+\dfrac{x-2005}{2003}-\dfrac{x-2005}{2}-\dfrac{x-2005}{3}-\dfrac{x-2005}{4}=0\)
\(\Leftrightarrow\left(x-2005\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}\ne0\right)=0\)
\(\Leftrightarrow x-2005=0\)
\(\Leftrightarrow x=2005\)
Vậy nghiệm của PT là \(x=2005\)
2003 x 4 + 1998 + 2001 x 2002
------------------------------------------------
2002 + 2002 x 1002 + 2002 x 1003
\(\frac{2003\times4+1998+2001\times2002}{2002+2002\times1002+2002\times1003}\)
\(=\frac{2003\times4+2\times999+2001\times2\times1001}{2002.\left(1+1002+1003\right)}\)
\(=\frac{2\times\left(2003\times2+999+2001\times1001\right)}{1001\times2\times\left(1+1002+1003\right)}\)
\(=\frac{2003\times2+999+2001\times1001}{1001\times\left(1+1002+1003\right)}\)
\(=1\)
mk ko bít
Chứng minh 1-1/2+1/3-1/4+...+1/2002-1/2003 = 1/1002+1/1003+...+1/2003
Đáp án của tớ là:
\(\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}=\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)-\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)=\)\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...-\frac{1}{2002}\)\(-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...-\frac{1}{2002}\)
Vậy:\(1+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}=\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}\)
xin chòa hôm nay mình sẽ giúp bạn lam bài toán này
ta có
1/1002+1/1003+....+1/2003=(1+1/2+1/3+.....+1/2003)-(1+1/2+1/3+....+1/1001)
1/1002+1/1003+....+1/2003=(1+1/2+1/3+.....+1/2003)-(1/2+1/4+1/6+....+1/2002)-(1/2+1/4+1/6+......+1/2002)
1/1002+1/1003+.....+1/2003=1+1/2+1/3+....+1/2003-1/2+1/4+1/6+....+1/2002-1/2-1/4-1/6-....-1/2002
Vậy1/1002+1/1002+.....+1/2003=1-1/2+1/3-1/4+....-2/2002-1/2003
Sửa: Vậy: \(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2003}=\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}\)