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ĐK : x>0
Đặt \(\sqrt{2010+\sqrt{x}}=t\left(t>0\right)\Rightarrow t^2=2010+\sqrt{x}\)
\(Pt\Rightarrow x+\sqrt{x}=t^2+t\)
Xét hàm số \(f\left(a\right)=a^2+a\) là hàm đồng biến \(\forall a>0\)
\(f\left(\sqrt{x}\right)=f\left(t\right)\Rightarrow x=t^2\Leftrightarrow x-\sqrt{x}-2010=0\\ \Leftrightarrow x=\left(\dfrac{1+\sqrt{8041}}{2}\right)^2\)
e: \(f\left(-x\right)=\dfrac{\left(-x\right)^4+3\cdot\left(-x\right)^2-1}{\left(-x\right)^2-4}=\dfrac{x^4+3x^2-1}{x^2-4}=f\left(x\right)\)
Vậy: f(x) là hàm số chẵn
\(c,f\left(-x\right)=\sqrt{-2x+9}=-f\left(x\right)\)
Vậy hàm số lẻ
\(d,f\left(-x\right)=\left(-x-1\right)^{2010}+\left(1-x\right)^{2010}\\ =\left[-\left(x+1\right)\right]^{2010}+\left(x-1\right)^{2010}\\ =\left(x+1\right)^{2010}+\left(x-1\right)^{2010}=f\left(x\right)\)
Vậy hàm số chẵn
\(g,f\left(-x\right)=\sqrt[3]{-5x-3}+\sqrt[3]{-5x+3}\\ =-\sqrt[3]{5x+3}-\sqrt[3]{5x-3}=-f\left(x\right)\)
Vậy hàm số lẻ
\(h,f\left(-x\right)=\sqrt{3-x}-\sqrt{3+x}=-f\left(x\right)\)
Vậy hàm số lẻ
\(\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right):\dfrac{456}{123}\)
\(=\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\cdot\dfrac{123}{456}\)
\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\right]\)
\(=\dfrac{123}{456}\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}-\dfrac{2009}{2010}+\dfrac{1}{2011}\right)\)
\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}+\dfrac{1}{2011}\right)-\left(\dfrac{2011}{2010}+\dfrac{2009}{2010}\right)\right]\)
\(=\dfrac{123}{456}\left(1-2\right)\)
\(=-\dfrac{123}{456}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(\sqrt{x-2}=\sqrt{(x-2).1}\leq \frac{x-2+1}{2}\)
\(\sqrt{y+2009}=\sqrt{(y+2009).1}\leq \frac{y+2009+1}{2}\)
\(\sqrt{z-2010}=\sqrt{(z-2010).1}\leq \frac{z-2010+1}{2}\)
Cộng theo vế suy ra :
\(\sqrt{x-2}+\sqrt{y+2009}+\sqrt{z-2010}\leq \frac{x+y+z}{2}\)
Dấu bằng xảy ra khi \(x-2=y+2009=z-2010=1\Leftrightarrow \left\{\begin{matrix} x=3\\ y=-2008\\ z=2011\end{matrix}\right.\)
a)\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\left(1\right)\)
ĐK:\(x\ne0\)
\(\left(1\right)\Leftrightarrow\dfrac{x^3+1-\left(x^3-1\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2}{\left(x^2+1\right)^2-x^2}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2x-3}{x\left(x^4+x^2+1\right)}=0\Rightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)
\(\dfrac{9-x}{2009}+\dfrac{11-x}{2011}=2\Leftrightarrow\left(\dfrac{9-x}{2009}-1\right)+\left(\dfrac{11-x}{2011}-1\right)=0\Leftrightarrow\dfrac{-2000-x}{2009}+\dfrac{-2000-x}{2011}=0\\ \Leftrightarrow\left(-2000-x\right)\left(\dfrac{1}{2009}+\dfrac{1}{2011}\right)=0\Rightarrow x=-2000\)
a) \(\dfrac{3x^2+1}{\sqrt{x-1}}=\dfrac{4}{\sqrt{x-1}}\)
ĐKXĐ: \(x>1\)
\(3x^2+1=4\)
\(3x^2=3\)
\(x^2=1\)
\(x=\pm1\)
=> Pt vô nghiệm
b) ĐKXĐ: x>-4
\(x^2+3x+4=x+4\)
\(x^2+2x=0\)
\(x\left(x+2\right)=0\)
\(\left[{}\begin{matrix}x=0\\x+2=0\Leftrightarrow x=-2\end{matrix}\right.\)
Câu a:
ĐKXĐ: \(x\neq \pm 3\)
\(\left|\frac{x+5}{-x^2+9}\right|=2\Rightarrow \left[\begin{matrix} \frac{x+5}{-x^2+9}=2\\ \frac{x+5}{-x^2+9}=-2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x+5=2(-x^2+9)\\ x+5=-2(-x^2+9)\end{matrix}\right.\Rightarrow \left[\begin{matrix} 2x^2+x-13=0\\ 2x^2-x-23=0\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=\frac{-1\pm \sqrt{105}}{4}\\ x=\frac{1\pm \sqrt{185}}{4}\end{matrix}\right.\) (đều thỏa mãn )
Vậy.......
Câu b:
ĐKXĐ: \(x< 2\)
Ta có: \(\frac{4}{\sqrt{2-x}}-\sqrt{2-x}=2\)
\(\Rightarrow 4-(2-x)=2\sqrt{2-x}\)
\(\Leftrightarrow 4=(2-x)+2\sqrt{2-x}\)
\(\Leftrightarrow 5=(2-x)+2\sqrt{2-x}+1=(\sqrt{2-x}+1)^2\)
\(\Rightarrow \sqrt{2-x}+1=\sqrt{5}\) (do \(\sqrt{2-x}+1>0\) )
\(\Rightarrow \sqrt{2-x}=\sqrt{5}-1\)
\(\Rightarrow 2-x=6-2\sqrt{5}\)
\(\Rightarrow x=-4+2\sqrt{5}\) (thỏa mãn)
Vậy...........