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14 tháng 7 2019
1. Ta có: \(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
=> \(x\left(6-x\right)^{2003}-\left(6-x\right)^{2003}=0\)
=> \(\left(6-x\right)^{2003}\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}\left(6-x\right)^{2003}=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}6-x=0\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)
14 tháng 7 2019
Bài 2. Ta có: (3x - 5)100 \(\ge\)0 \(\forall\)x
(2y + 1)100 \(\ge\)0 \(\forall\)y
=> (3x - 5)100 + (2y + 1)100 \(\ge\)0 \(\forall\)x;y
Dấu "=" xảy ra khi: \(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}}\) => \(\hept{\begin{cases}3x=5\\2y=-1\end{cases}}\) => \(\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{1}{2}\end{cases}}\)
Vậy ...
8 tháng 6 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(-\frac{1}{x+1}=\frac{2001}{4006}-\frac{1}{2}\)
\(-\frac{1}{x+1}=-\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(\Rightarrow x=2012\)
NM
8 tháng 6 2016
Ta có: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{2003}:2\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{2003}{4006}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2003}{4006}-\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2}{4006}=\frac{1}{2003}\)
=> x + 1 = 2003
=> x = 2002
Vậy x = 2002
Duyệt nha !!!
chúc hk tốt!!!
29 tháng 8 2016
\(2003-\left|x-2003\right|=x\)
\(\Leftrightarrow\left|x-2003\right|=2003-x\left(1\right)\)
+ ) Nếu : \(x\ge2003\) thì ( 1 ) \(\Leftrightarrow x-2003=2003-x\)
\(\Leftrightarrow2x=2.2003\)
\(\Leftrightarrow x=2003\left(nhận\right)\)
+ ) Nếu \(x< 2003\) thì ( 1 ) \(\Leftrightarrow2003-x=2003-x\)
\(\Leftrightarrow0.x=0\)
Vậy pt có vô số nghiệm với \(x< 2003\)
8 tháng 6 2016
13 +16 +110 +....+1x(x+1):2 =20012003
26 +212 +220 +....+2x(x+1) =20012003
2(12.3 +13.4 +14.5 +....+1x(x+1) )=20012003
12 −13 +13 −14 +14 −15 +....+1x −1x+1 =20012003 :2=20014006
12 −1x+1 =20014006
1x+1 =12 −20014006 =12003
=> x+1 = 2003
=> x = 2003 - 1
=> x = 2002
Xin 1 tích đúng
8 tháng 6 2016
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{x.\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(\Rightarrow\frac{x-1}{x+1}=\frac{2001}{2003}\)
\(\Rightarrow2x=4004\)
\(\Rightarrow x=2002\)
26 tháng 5 2016
\(PT\Leftrightarrow\frac{x+4+2000}{2000}+\frac{x+3+2001}{2001}=\frac{x+2+2002}{2002}+\frac{x+1+2003}{2003}\)
<=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
<=> \(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
<=> x + 2004 = 0
<=> x = -2004.
26 tháng 5 2016
\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\right)\)
\(\Rightarrow x=-2004\)
26 tháng 5 2016
(x+4/2000 + 1)+(x+3/2001 + 1) = (x+2/2002 + 1)+(x+1/2003)+1
(x+2004/2000) + (x+2004/2001) = (x+2004/2002) + (x+2004/2003)
(x+2004).(1/2000+1/2001) = (x+2004).(1/2002+1/2003)
+ Với x+2004=0 suy ra x=-2004. Ta có 0.(1/2000+1/2001)=0.(1/2002+1/2003), đúng
+ Với x+2004 khác 0 thì (x+2004).(1/2000+1/2001) = (x+2004).(1/2002+1/2003)
1/2000+1/2001 = 1/2002+1/2003, vô lí vì 1/2000+1/2001 > 1/2002+1/2003
Vậy x=-2004
TH1: \(6-x=0\)
\(\Rightarrow x=6-0=6\)
TH2: \(6-x\ne0\)
\(\Rightarrow x=\frac{\left(6-x\right)^{2003}}{\left(6-x\right)^{2003}}=1\)
Vậy \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)
x = 6 và x = 1
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