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P=\(\sqrt{\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1}\)
=\(\sqrt{\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1}\)
=\(\sqrt{x-\sqrt{x}-x-\sqrt{x}+x+1}\)
=\(\sqrt{x-2\sqrt{x}+1}\)
=\(\sqrt{\left(\sqrt{x}-1\right)^2}\)
=\(\sqrt{x}-1\)
\(\hept{\begin{cases}-1\le x\le1\\2-\sqrt{1-x^2}\end{cases}\Rightarrow-1\le x\le1\left(^∗\right)}\)
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 ( * )
Điều kiện xác định \(x\ge0\)
\(A=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=x-\sqrt{x}-\left(x+\sqrt{x}\right)=-2\sqrt{x}\)
\(B=\frac{1}{3}-\sqrt{A+x+1}=\frac{1}{3}-\sqrt{x-2\sqrt{x}+1}=\frac{1}{3}-\sqrt{\left(\sqrt{x}-1\right)^2}=\frac{1}{3}-\left|\sqrt{x}-1\right|\)
\(=\frac{1}{3}-\left(1-\sqrt{x}\right)=\sqrt{x}-\frac{2}{3}\) (vì \(0\le x\le1\))
\(P=\frac{x+2}{\sqrt{x}^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(P=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
2,
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}+\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{7\sqrt{7}}{7}\)
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}+\frac{\left(\sqrt{2}+1\right)}{2-1}-\sqrt{7}\)
\(A=\sqrt{7}-\sqrt{2}+\sqrt{2}+1-\sqrt{7}=1\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(a,E=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\left(Đk:x\ge0;x\ne\pm1\right)\)(Đề như này mới đúng!)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2x-2\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{7\sqrt{x}-2-5x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{5\sqrt{x}+2\sqrt{x}-2-5x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(5\sqrt{x}-5x\right)+\left(2\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
Vậy...
\(b,\)Ta có:\(\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{-15+17-5\sqrt{x}}{\sqrt{x}+3}=\frac{\left(-15-5\sqrt{x}\right)+17}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\)
Vì \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+3\ge3\forall x\Rightarrow\frac{17}{\sqrt{x}+3}\le\frac{17}{3}\Rightarrow-5+\frac{17}{\sqrt{x}+3}\le\frac{2}{3}\Rightarrow E\le\frac{2}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Với điều kiện \(0\)\(\le x\le1\)ta có
P= \(\sqrt{\frac{\sqrt{x\left(x\sqrt{x-1}\right)}}{x+\sqrt{x}+1}-\sqrt{\frac{\sqrt{x\left(x\sqrt{x+1}\right)}}{x-\sqrt{x}+1}+x+1}}\)
sử dụng hằng đẳng thức bậc 3 : \(x^3\)- \(y^3\)và \(x^3\)+ \(y^3\)
ta có P = \(\sqrt{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1}\)= \(\sqrt{x-2\sqrt{x}+1}=\)\(\sqrt{\left(\sqrt{x}-1\right)^2}\)=\(\left|\sqrt{x}-1\right|\)
P= \(1-\sqrt{x}\)
mk nghĩ P=\(\sqrt{x}-1\) bạn ak