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\(\hept{\begin{cases}-1\le x\le1\\2-\sqrt{1-x^2}\end{cases}\Rightarrow-1\le x\le1\left(^∗\right)}\)
Đặt : \(\hept{\begin{cases}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{cases}\Rightarrow\hept{\begin{cases}a^2+b^2=2\\a,b\ge0\end{cases}}}\)
A tồn tại mọi x thuộc ( * )
\(A=\frac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\frac{\sqrt{a^2-2ab+b^2}\left(a+b\right)\left(a^2+b^2-ab\right)}{2-ab}\)
\(A=\frac{\sqrt{2}\sqrt{\left(a-b\right)^2}\left(a+b\right)\left(2-ab\right)}{\left(2-ab\right)}\) . Vói đk ( \(I\)) \(A=\frac{\sqrt{2}}{2}!a-b!\left(a+b\right)\)
\(\orbr{\begin{cases}\hept{\begin{cases}a\ge b\Leftrightarrow0\le x\le1\\A=\frac{\sqrt{2}}{2}\left[\left(1+x\right)-\left(1-x\right)\right]=\frac{\sqrt{2}}{2}x\end{cases}}\\\hept{\begin{cases}a< b\Leftrightarrow-1\le x< 0\\A=\frac{-\sqrt{2}}{2}\left[\left(1+x\right)-\left(1-x\right)\right]=\frac{-\sqrt{2}}{2}x\end{cases}}\end{cases}}\)
\(\Rightarrow A=\frac{\sqrt{2}}{2}!x!\) . Với x thỏa mãn điều kiện ( * )
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
Lời giải:
Đặt \(\sqrt{(1-x)+(1-x)\sqrt{1-x^2}}=a; \sqrt{(1-x)-(1-x)\sqrt{1-x^2}}=b\)
Khi đó: \(P=a+b\geq 0\)
Ta có:
\(a^2+b^2=(1-x)+(1-x)\sqrt{1-x^2}+(1-x)-(1-x)\sqrt{1-x^2}=2(1-x)\)
Và:
\(ab=(1-x)\sqrt{(1+\sqrt{1-x^2})(1-\sqrt{1-x^2})}\)
\(=(1-x)\sqrt{1-(1-x^2)}=(1-x)\sqrt{x^2}=|x|(1-x)\)
\(\Rightarrow P^2=(a+b)^2=a^2+b^2+2ab=2(1-x)+2|x|(1-x)\)
\(=2(1-x)(1+|x|)=\frac{4036}{2017}.\frac{2018}{2017}\)
\(\Rightarrow P=\sqrt{\frac{4036.2018}{2017^2}}=\frac{\sqrt{4036.2018}}{2017}\)