\(x=9-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}+\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\)

\(=9-\frac{2}{\sqrt{9-4\sqrt{5}}}+\frac{2}{\sqrt{9+4\sqrt{5}}}\)

\(=9-\frac{2}{\sqrt{\left(\sqrt{5}-2\right)^2}}+\frac{2}{\sqrt{\left(\sqrt{5}+2\right)^2}}\)

\(=9-\frac{2}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+2}\)

\(=9-\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

\(=9-\frac{8}{5-4}\)

= 1

\(f\left(x\right)=\left(1^4-3+1\right)^{2016}=1\)

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 !!!

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

a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)

\(\frac{1}{3}:2x=\frac{-21}{4}\)

\(2x=\frac{-4}{63}\)

\(x=\frac{2}{63}\)

b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)

Vậy.........

Trả lời nhanh giúp mình với mình cần gấp lắm

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

1) Đk \(x\ge1\)

Ta có \(\hept{\begin{cases}\sqrt{x-1}\ge0\\\sqrt{x+3}\ge\sqrt{1+3}=2\\\sqrt{\left(x-1\right)\left(x^2-3x+5\right)}\ge0\end{cases}\Rightarrow VT\ge2}\)

Mà \(4-2x\le2\Rightarrow VT\ge VP\)

Dấu = xảy ra <=> x=1 

Vậy ...^^

Ma tai sao \(\sqrt{x-1}+\sqrt{x+3}\ge2\)