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a.
\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)
c.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)
Lời giải:
Ta thấy: \(f(x)=\frac{x^3}{1-3x+3x^2}\Rightarrow f(1-x)=\frac{(1-x)^3}{1-3(1-x)+3(1-x)^2}=\frac{(1-x)^3}{3x^2-3x+1}\)
\(\Rightarrow f(x)+f(1-x)=\frac{x^3}{1-3x+3x^2}+\frac{(1-x)^3}{3x^2-3x+1}=\frac{x^3+(1-x)^3}{3x^2-3x+1}=1\)
Do đó:
\(f\left(\frac{1}{2017}\right)+f\left(\frac{2016}{2017}\right)=1\)
\(f\left(\frac{2}{2017}\right)+f\left(\frac{2015}{2017}\right)=1\)
............
\(f\left(\frac{1008}{2017}\right)+f\left(\frac{1009}{2017}\right)=1\)
Cộng theo vế:
\(\Rightarrow A=f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+f\left(\frac{3}{2017}\right)+...f\left(\frac{2015}{2017}\right)+f\left(\frac{2016}{2017}\right)\)
\(=\underbrace{1+1+1...+1}_{1008}=1008\)
Ta có \(x=\sqrt{\dfrac{2-\sqrt{3}}{2}}=\sqrt{\dfrac{2\left(2-\sqrt{3}\right)}{4}}=\sqrt{\dfrac{4-2\sqrt{3}}{4}}=\dfrac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{4}}=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}=\dfrac{\sqrt{3}-1}{2}\)Ta lại có \(2x^2+2x-1=2.\left(\dfrac{\sqrt{3}-1}{2}\right)^2+2\left(\dfrac{\sqrt{3}-1}{2}\right)-1=2\left(\dfrac{3-2\sqrt{3}+1}{4}\right)+\dfrac{2\left(\sqrt{3}-1\right)}{2}-1=\dfrac{4-2\sqrt{3}}{2}+\sqrt{3}-1-1=\dfrac{2\left(2-\sqrt{3}\right)}{2}+\sqrt{3}-2=2-\sqrt{3}+\sqrt{3}-2=0\)(1)
⇒\(x^3\left(2x^2+2x-1\right)=0\Rightarrow2x^5+2x^4-x^3=0\Rightarrow2x^5+2x^4-x^3-1=-1\Rightarrow\left(2x^5+2x^4-x^3-1\right)^{2016}=\left(-1\right)^{2016}=1\)(2)
Từ (1)⇒\(x\left(2x^2+2x-1\right)=0\Rightarrow2x^3+2x^2-x=0\Rightarrow2x^3+2x^2-x-3=-3\Rightarrow\left(2x^3+2x^2-x-3\right)^{2017}=\left(-3\right)^{2017}\)(3)
Từ (1)⇒\(x^2\left(2x^2+2x-1\right)=0\Rightarrow2x^4+2x^3-x^2=0\Rightarrow2x^4+2x^3-x^2-3=-3\)(4)
Từ (2),(3),(4)⇒\(\left(2x^5+2x^4-x^3-1\right)^{2016}+\dfrac{\left(2x^3+2x^2-x-3\right)^{2017}}{2x^4+2x^3-x^2-3}=1+\dfrac{\left(-3\right)^{2017}}{-3}=1+\left(-3\right)^{2016}=3^{2016}+1\Rightarrow P=3^{2016}+1\)
ta có:
\(P=\dfrac{\sqrt{\left(x-2016\right).2017}}{\sqrt{2017}\left(x+1\right)}+\dfrac{\sqrt{\left(x-2017\right)2016}}{\sqrt{2016}\left(x-1\right)}\)
Áp dụng BĐT cauchy:\(\sqrt{\left(x-2016\right)2017}\le\dfrac{1}{2}\left(x-2016+2017\right)=\dfrac{1}{2}\left(x+1\right)\)
\(\sqrt{\left(x-2017\right)2016}\le\dfrac{1}{2}\left(x-2017+2016\right)=\dfrac{1}{2}\left(x-1\right)\)
do đó \(P\le\dfrac{x+1}{2\sqrt{2017}\left(x+1\right)}+\dfrac{x-1}{2\sqrt{2016}\left(x-1\right)}=\dfrac{1}{2\sqrt{2017}}+\dfrac{1}{2\sqrt{2016}}\)
đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-2016=2017\\x-2017=2016\end{matrix}\right.\)\(\Rightarrow x=4033\)
\(\left(\dfrac{x+1}{\sqrt{x}+1}+\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}}{x+2\sqrt{x}+1}=\left(\dfrac{x\sqrt{x}+\sqrt{x}}{x+\sqrt{x}}+\dfrac{1}{x+\sqrt{x}}-\dfrac{\sqrt{x}+1}{x}\right):\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}=\dfrac{x\sqrt{x}+\sqrt{x}}{x+\sqrt{x}}:\dfrac{\sqrt{x}}{\cdot\left(\sqrt{x}+1\right)^2}=\dfrac{\sqrt{x}\left(x+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x+1}\right)}=\dfrac{x+1}{\sqrt{x}+1}:\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x+1}\right)}{\left(\sqrt{x}+1\right)\sqrt{x}}=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{x\sqrt{x}+x+\sqrt{x}+1}{\sqrt{x}}\)
\(=x+\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\ge2017+\sqrt{2017}\Leftrightarrow x+\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2016+\sqrt{2017}\Leftrightarrow x+\sqrt{x}+\dfrac{1}{\sqrt{x}}-2016-\sqrt{2017}\ge0\)
Bài toán sắp hoàn thành rồi đấy cậu giải tiếp nhé! =))
Lời giải:
Trong TH này ta thêm điều kiện $x$ là số nguyên dương.
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x(x+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(=1-\frac{1}{x+1}=\frac{x}{x+1}\)
Vậy \(\frac{x}{x+1}=\frac{\sqrt{2017-x}+2016}{\sqrt{2016-x}+2017}\)
\(\Rightarrow x\sqrt{2016-x}+2017x=(x+1)\sqrt{2017-x}+2016(x+1)\)
\(\Leftrightarrow x\sqrt{2016-x}=(x+1)\sqrt{2017-x}+2016-x\)
\(\Leftrightarrow x(\sqrt{2017-x}-\sqrt{2016-x})+\sqrt{2017-x}+2016-x=0\)
\(\Leftrightarrow \frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}+\sqrt{2017-x}+(2016-x)=0\)
Hiển nhiên ta thấy:
\(\frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}>0\)
\(\sqrt{2017-x}\geq 0\)
\(2016-x\geq 0\)
Do đó pt trên vô nghiệm
Tức là không tìm đc $x$ thỏa mãn.