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Cho f(x)=\(\frac{1+\sqrt{1+x}}{x+1}+\frac{1+\sqrt{1-x}}{x-1}\) và a=\(\frac{\sqrt{3}}{2}\).Tính f(a)
\(a=\dfrac{4}{\sqrt{3}+\dfrac{1}{\sqrt{3}}}\)
\(=4:\dfrac{4\sqrt{3}}{3}\)
\(=\sqrt{3}\)
\(f\left(x\right)=\dfrac{\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}}{\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1}}\)
\(=\dfrac{\left(\sqrt{3}+1+\sqrt{3}-1+2\cdot\sqrt{2}\right)}{2}\)
\(=\sqrt{3}+\sqrt{2}\)
\(a=\sqrt{2}+\sqrt{7-2\sqrt{5}-1}+1\)
\(=\sqrt{2}+\sqrt{5}-1+1=\sqrt{2}+\sqrt{5}\)
f(x)=x^4(x+2)-14x^2(x+2)+9(x+2)+1
=(x+2)(x^4-14x^2+9)+1
\(=\left(\sqrt{2}+\sqrt{5}+2\right)\left[\left(7+2\sqrt{10}\right)^2-14\left(7+2\sqrt{10}\right)+1\right]\)+1
\(=\left(\sqrt{2}+\sqrt{5}+2\right)\left(89+28\sqrt{10}-84-28\sqrt{10}+1\right)\)+1
=6(căn 2+căn 5+1)+1
Ta có: \(f\left(x\right)=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)= \(\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}{\left(\sqrt{x+1}-\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}\)=\(\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{x+1-\left(x-1\right)}\)
= \(\frac{x+1+x-1+2\sqrt{\left(x-1\right)\left(x+1\right)}}{2}\)= \(\frac{2x+2\sqrt{x^2-1}}{2}\)=\(x+\sqrt{x^2-1}\)
Với a= \(\sqrt{3}\)=> \(f\left(\sqrt{3}\right)=\sqrt{3}+\sqrt{\left(\sqrt{3}\right)^2-1}\)=\(\sqrt{3}+\sqrt{2}\)