Tìm nghiệm dương của phương trình:
\(\left(1+x-\sqrt{x^2-1}\right)^{2008}+\left(1+x+\sqrt{x^2-1}\right)^{2008}=2^{2009}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-2008=X;y-2009=Y;z-2010=Z\)
\(\sqrt{X}+\sqrt{Y}+\sqrt{Z}+3012=\frac{1}{2}\left(X+Y+Z+2008+2009+2010\right)\)
\(2.\sqrt{X}+2\sqrt{Y}+2\sqrt{Z}+2.3012=X+Y+Z+2009\cdot3\)
\(\left(X-2\sqrt{X}+1\right)+\left(Y-2\sqrt{Y}+1\right)+\left(Z-2\sqrt{Z}+1\right)+3.2008=2.3012\)
\(\left(\sqrt{X}-1\right)^2+\left(\sqrt{Y}-1\right)^2+\left(\sqrt{Z}-1\right)^2=2.3012-3.2008=0\)
\(X=1;Y=1;Z=1\Rightarrow x=2009;y=2010;z=2011\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\sqrt{x^2+x+1}+\sqrt{x^2-x-1}=2\left|x\right|\)
bien doi ve trai ta co:
\(=\sqrt{x^2+2.\frac{1}{2}x+\frac{1}{2}-\frac{1}{2}+1}+\sqrt{x^2-2.\frac{1}{2}x-\frac{1}{2}+\frac{1}{2}-1}\)
\(=\sqrt{\left(x+\sqrt{\frac{1}{2}}\right)^2-\left(\frac{1}{2}-1\right)}+\sqrt{\left(x-\sqrt{\frac{1}{2}}\right)^2-\left(\frac{1}{2}+1\right)}\)
\(=\sqrt{\left(x+\sqrt{\frac{1}{2}}\right)^2+\frac{1}{2}}+\sqrt{\left(x-\sqrt{\frac{1}{2}}\right)^2-\frac{3}{2}}\)
den day thi mk chiu
a)Đặt \(x+\frac{4017}{2}=t\) thì pt <=> \(\left(t-\frac{1}{2}\right)^4+\left(t+\frac{1}{2}\right)^4=\frac{1}{8}\)
<=>\(\left[\left(t+\frac{1}{2}\right)^2-\left(t-\frac{1}{2}\right)^2\right]^2+2\left(t-\frac{1}{2}\right)^2\left(1+\frac{1}{2}\right)^2-\frac{1}{8}=0\)
<=>\(\left[\left(t+\frac{1}{2}-t+\frac{1}{2}\right)\left(t+\frac{1}{2}+t-\frac{1}{2}\right)\right]^2+2\left(t^2-\frac{1}{4}\right)^2-\frac{1}{8}=0\)
<=>\(\left(2t\right)^2+2\left(t^4-\frac{1}{2}t^2+\frac{1}{16}\right)-\frac{1}{8}=0\Leftrightarrow4t^2+2t^4-t^2+\frac{1}{8}-\frac{1}{8}=0\)
<=>\(2t^4+3t^2=0\Leftrightarrow t^2\left(2t^2+3\right)=0\Leftrightarrow t^2=0\)(do \(2t^2+3\ge3>0\))<=>t=0
<=>\(x+\frac{4017}{2}=0\Leftrightarrow x=-\frac{4017}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Để (1) có 2 nghiệm dương \(\Rightarrow\left\{{}\begin{matrix}\Delta'=\left(m+3\right)^2-m-1\ge0\\x_1+x_2=2\left(m+3\right)>0\\x_1x_2=m+1>0\end{matrix}\right.\) \(\Rightarrow m>-1\)
\(P=\left|\dfrac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{x_1x_2}}\right|>0\Rightarrow P^2=\dfrac{\left(\sqrt{x_1}-\sqrt{x_2}\right)^2}{x_1x_2}\)
\(P^2=\dfrac{x_1+x_2-2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{2\left(m+3\right)-2\sqrt{m+1}}{m+1}=\dfrac{4}{m+1}-\dfrac{2}{\sqrt{m+1}}+2\)
\(P^2=\left(\dfrac{2}{\sqrt{m+1}}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\Rightarrow P\ge\dfrac{\sqrt{7}}{2}\)
Dấu "=" xảy ra khi \(\sqrt{m+1}=4\Rightarrow m=15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta nhân cả 2 vế với \(x+\sqrt{x^2+2008}\)
hay \(y+\sqrt{y^2+2008}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-\sqrt{x^2-1}=\frac{x^2-\left(x^2-1\right)}{x+\sqrt{x^2-1}}=\frac{1}{x+\sqrt{x^2-1}}=t\)\(\Rightarrow x+\sqrt{x^2-1}=\frac{1}{t}\)
Ta có: \(\left(1+t\right)^{2015}+\left(1+\frac{1}{t}\right)^{2015}=2^{2016}\)(1)
Áp dụng Côsi ta có:
\(1+t\ge2\sqrt{t}\Rightarrow\left(1+t\right)^{2015}\ge2^{2015}.\sqrt{t^{2015}}\)
\(1+\frac{1}{t}\ge\frac{2}{\sqrt{t}}\Rightarrow\left(1+\frac{1}{t}\right)^{2015}\ge\frac{2^{2015}}{\sqrt{t^{2015}}}\)
\(\Rightarrow\left(1+t\right)^{2015}+\left(1+\frac{1}{t}\right)^{2015}\ge2^{2015}\left(\sqrt{t^{2015}}+\frac{1}{\sqrt{t^{2015}}}\right)\)
\(\ge2^{2015}.2\sqrt{\sqrt{t^{2015}}.\frac{1}{\sqrt{t^{2015}}}}=2^{2016}\)
Dấu "=" xảy ra khi và chỉ khi t = 1.
Do đó, từ (1) => \(t=\frac{1}{x+\sqrt{x^2-1}}=1\Rightarrow x+\sqrt{x^2-1}=1\)
\(\Rightarrow1-x=\sqrt{x^2-1}\Rightarrow\left(1-x\right)^2=x^2-1\Leftrightarrow2-2x=0\Leftrightarrow x=1\)
Vậy: \(x=1\text{ là nghiệm (nguyên) duy nhất của phương trình.}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
M = x.√[(2008+y²).(2008+z²)\(2008+x²)] + y.√[(2008+x²).(2008+z²)\(2008+y²)] + z.√[(2008+y²).(2008+x²)\(2008+z²)]
ta có:
2008 + x² = xy + xz + yz + x²
2008 + x² = (x+y).(x+z)
tương tự: 2008 + y² = (x+y).(y+z) và 2008 + z² = (z+y).(x+z)
chỉ việc thay vào rùi rút gọn thui
=> M = x.√[(x+y).(y+z).(x+z).(z+y)\ (x+y).(x+z)] + y.√[(x+y).(x+z).(x+z).(z+y)\(y+x).(y+z)] + z.√[(x+y).(x+z).(y+z).(y+x)\(x+z).(z+y)]
=> M = x.|y+z| + y.|z+x| + z.|x+y|
=> M = 2.2008
Thay \(xy+yz+xz=2018\) ta được:
\(\left\{{}\begin{matrix}2018+x^2=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\\2018+y^2=y^2+xy+yz+xz=\left(y+z\right)\left(x+y\right)\\2018+z^2=z^2+xy+yz+xz=\left(x+z\right)\left(y+z\right)\end{matrix}\right.\)
Sau đó thay vào lần lượt đề bài là được
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>|x-2009|=2009-x
=>x-2009<=0
=>x<=2009
b: =>2x-1=0 và y-2/5=0 và x+y-z=0
=>x=1/2 và y=2/5 và z=x+y=1/2+2/5=5/10+4/10=9/10