Rút gọn biểu thức : A=\(\frac{\sqrt{1-\sqrt{1-x^2}}\left(\sqrt{\left(x+1\right)^3}+\sqrt{\left(1-x\right)^3}\right)}{2-\sqrt{1-x^2}}\) với -1\(\le\) x \(\le\) 1
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27 tháng 11 2019
\(\hept{\begin{cases}-1\le x\le1\\2-\sqrt{1-x^2}\end{cases}\Rightarrow-1\le x\le1\left(^∗\right)}\)
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 ( * )
Y
16 tháng 6 2023
\(A=3\left(x+2\sqrt{x}\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3x+6\sqrt{x}-\left(x-1\right)\)
\(=3x+6\sqrt{x}-x+1\)
\(=2x+6\sqrt{x}+1\)
\(B=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-2\left(\sqrt{x}-1\right)^2\)
\(=x+3\sqrt{x}+\sqrt{x}+3-2\left(x-2\sqrt{x}+1\right)\)
\(=x+4\sqrt{x}+3-2x+4\sqrt{x}-2\)
\(=-x+8\sqrt{x}+1\)
\(C=3x-3\sqrt{x}-2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3x-3\sqrt{x}-2+\left(\sqrt{x^2}-1\right)\)
\(=3x-3\sqrt{x}-2+x-1\)
\(=4x-3\sqrt{x}-3\)
\(D=\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
\(=x-9-\left(2x-3\sqrt{x}-2\right)\)
\(=x-9-2x+3\sqrt{x}+2\)
\(=-x+3\sqrt{x}-7\)
\(E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)
\(=\sqrt{x^2}-2^2-2\left(2x+4\sqrt{x}-\sqrt{x}-2\right)\)
\(=x-4-2\left(2x+3\sqrt{x}-2\right)\)
\(=x-4-4x-6\sqrt{x}+4\)
\(=-3-6\sqrt{x}\)
26 tháng 8 2020
a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+\left|x-3\right|\)
\(=x+3-\left(x-3\right)\)
\(=x+3-x+3\)
\(=6\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)
\(=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
\(=x+2-\left(-x\right)\)
\(=x+2+x\)
\(=2x+2=2\left(x+1\right)\)
c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)
\(=\frac{\sqrt{\left(x-1\right)^2}}{x-1}\)
\(=\frac{\left|x-1\right|}{x-1}\)
\(=\frac{x-1}{x-1}=1\)
d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)
\(=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}\)
\(=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}\)
\(=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}\)
\(=\left|x-2\right|-1\)
\(=-\left(x-2\right)-1\)
\(=-x+2-1\)
\(=-x+1=-\left(x-1\right)\)
\(\left\{\begin{matrix}-1\le x\le1\\2-\sqrt{1-x^2}\end{matrix}\right.\) \(\Rightarrow-1\le x\le1\) (*)
Đặt \(\left\{\begin{matrix}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{matrix}\right.\Rightarrow\left\{\begin{matrix}a^2+b^2=2\\a,b\ge0\end{matrix}\right.\)
A tồn tại mọi x thuộc (*)
\(A=\frac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\frac{\sqrt{\frac{a^2-2ab+b^2}{2}}\left(a+b\right)\left(a^2+b^2-ab\right)}{2-ab}\)
\(A=\frac{\frac{\sqrt{2}}{2}\sqrt{\left(a-b\right)^2}\left(a+b\right)\left(2-ab\right)}{\left(2-ab\right)}\) Vơi đk (I)\(A=\frac{\sqrt{2}}{2}!a-b!\left(a+b\right)\)
\(\left[\begin{matrix}\left\{\begin{matrix}a\ge b\Leftrightarrow0\le x\le1\\A=\frac{\sqrt{2}}{2}\left[\left(1+x\right)-\left(1-x\right)\right]=\frac{\sqrt{2}}{2}x\end{matrix}\right.\\\left\{\begin{matrix}a< b\Leftrightarrow-1\le x< 0\\A=\frac{-\sqrt{2}}{2}\left[\left(1+x\right)-\left(1-x\right)\right]=\frac{-\sqrt{2}}{2}x\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow A=\frac{\sqrt{2}}{2}!x!\) Với x thỏa mãn đk (*)
Kết luận viết thế này cho đẹp
\(A=\left\{\begin{matrix}!x!\le1\\\frac{\sqrt{2}}{2}!x!\end{matrix}\right.\)