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Ta có : \(\left(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\right)^2\)
\(=\frac{1}{\frac{9}{4}+\sqrt{5}}+\frac{1}{\frac{9}{4}-\sqrt{5}}-2.\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}.\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)
\(=\frac{\frac{9}{4}-\sqrt{5}+\frac{9}{4}+\sqrt{5}}{\frac{1}{16}}-2.\frac{1}{\frac{1}{4}}\)
\(=72-8=64\)
Mà \(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}< \frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)
\(\Rightarrow\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}< 0\)
Do đó : \(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}=-8\)
Khi đó : \(x=9-8=1\)
Với x =1 ta có ;
\(f\left(1\right)=\left(1^4-3.1+1\right)^{2016}=\left(-1\right)^{2016}=1\)
Chúc bạn học tốt !!!
\(x=9-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}+\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\)
\(=9-\frac{2}{\sqrt{9-4\sqrt{5}}}+\frac{2}{\sqrt{9+4\sqrt{5}}}\)
\(=9-\frac{2}{\sqrt{\left(\sqrt{5}-2\right)^2}}+\frac{2}{\sqrt{\left(\sqrt{5}+2\right)^2}}\)
\(=9-\frac{2}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+2}\)
\(=9-\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=9-\frac{8}{5-4}\)
= 1
\(f\left(x\right)=\left(1^4-3+1\right)^{2016}=1\)
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
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.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)
Có: \(\left(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\right)^2\)
\(=\frac{1}{\frac{9}{4}+\sqrt{5}}+\frac{1}{\frac{9}{4}-\sqrt{5}}-2\cdot\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\cdot\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)
\(=\frac{\frac{9}{4}-\sqrt{5}+\frac{9}{4}+\sqrt{5}}{\frac{1}{16}}-2\cdot\frac{1}{\frac{1}{4}}\)
\(=72-8=64\)
Mà; \(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}< \frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)
\(\Rightarrow\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}< 0\)
Do đó: \(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}=-8\)
Khi đó: \(x=9-8=1\)
Với \(x=1\), ta có:
\(f\left(1\right)=\left(1^4-3\cdot1+1\right)^{2016}=\left(-1\right)^{2016}=1\)
cảm ơn bạn nhieuf nha