tim x
4019* x+2010:2010= 2010
Những câu hỏi liên quan
tim x
(2009-x)^2+(2009-x)×(x-2010)+(x-2010)^2/(2009)^2-(2009-x)×(x-2010)+(x-2010)^2=19/49
C/m 3/1*1+2*2+5/2*2+3*3+...+4019/2009*2009+2010*2010<1
CMR: A= 3/1.2^2+5/2^2.3^2+...+4019/2010^2.2009^2<1
Tim x biết |2010 – x | 2011+ |2011 – x | 2010 =1
chờ x,y,z thỏa mãn:x+y+z=2010
1/x+1/y+1/z=1/2010
tính giá trị biểu thức :(x^2010-y^2010)(y^2010-z^2010)(z^2010-x^2010)
tim x y z
\(\left|x-2009\right|^{2009}+\left(y-2010\right)^{2010}+2011\left|z-2011\right|\le0\)
Ta có: /x-2009/2009\(\ge\)0; (y-2010)2010=[(y-2010)1005]2 \(\ge\)0 và 2011/z-2011/\(\ge\)0
Tổng 3 số dương 0 khi và chỉ khi 3 số đó đều=0, khi đó dấu bằng xảy ra.
=> \(\hept{\begin{cases}Ix-2009I^{2009}=0\\\left(y-2010\right)^{2010}=0\\2011Iz-2011I=0\end{cases}}\)
=> x=2009; y=2010; z=2011
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Choa,b,c,d khác 0:
\(\frac{x^{2010}+y^{2010}z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)
Tính\(T=x^{2010}+y^{2010}+z^{2010}+t^{2010}\)
\(\dfrac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\left(1\right)\)
\(Đkxđ:x\ne2009;x\ne2010\)
Đặt \(t=x-2010\left(t\ne0\right)\)
\(\Rightarrow2009-x=-\left(t+1\right)\)
\(\left(1\right)\Leftrightarrow\dfrac{\left(t+1\right)^2-\left(t+1\right)t+t^2}{\left(t+1\right)^2+\left(t+1\right)t+t^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{t^2+2t+1-t^2-t+t^2}{t^2+2t+1+t^2+t+t^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{t^2+t+1}{3t^2+3t+1}=\dfrac{19}{49}\)
\(\Leftrightarrow49t^2+49t+49=57t^2+57t+19\)
\(\Leftrightarrow8t^2+8t-30=0\)
\(\Leftrightarrow4t^2+4t-15=0\)
\(\Leftrightarrow\left(4t^2+4t+1\right)-16=0\)
\(\Leftrightarrow\left(2t+1\right)^2=16=4^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2t+1=4\\2t+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{3}{2}\\t=-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2010=\dfrac{3}{2}\\x-2010=-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4023}{2}\\x=\dfrac{4015}{2}\end{matrix}\right.\)
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A= (1-1/2010) x (1- 2/2010) x (1- 3/2010) x.... x( 1- 2011/2010) =?
Vì ta có 1 - 1/2010 = 0/2010 = 0 nên suy ra biểu thức A = 0
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A=\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right)...\left(1-\frac{2010}{2010}\right)\left(1-\frac{2011}{2010}\right)\)
A=\(\frac{2009}{2010}.\frac{2008}{2010}...0.\frac{-1}{2010}\)
A=0
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