\(\sqrt{x+1}=a\ge0;\sqrt{1-x}=b\ge0\) -1\lex\le 1

Suy ra: \(a^2+b^2=2;\frac{a^2-b^2}{2}=x\).

PT \(\Leftrightarrow\frac{\left(a^2-b^2\right)^2}{4}+a+b-\sqrt{2\left(a^2+b^2\right)}=0\)

\(\Leftrightarrow\left(a-b\right)^2\left[\frac{\left(a+b\right)^2}{4}-\frac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\right]=0\)

Nếu \(a=b\Rightarrow\sqrt{x+1}=\sqrt{1-x}\Leftrightarrow x=0\)

Ngoặc to chịu.

DK \(-1\le x\le1\)

Dat \(\sqrt{x+1}=a\ge0,\sqrt{1-x}=b\ge0\)

ta co \(a^2+b^2=x+1+1-x=2\) 

va \(1-x^2=\left(1-x\right)\left(1+x\right)\)

ta co hpt

\(\hept{\begin{cases}a^2+b^2=2\\a+b-1=a^2b^2\end{cases}}\)

Dat \(a+b=S\ge0,ab=P\ge0\left(S^2\ge4P\right)\)

lai co he moi 

\(\hept{\begin{cases}S^2-2P=2\\S-1=P^2\end{cases}}\)

den day de roi thay S=P^2 +1 vao phuong trinh 1 roi tinh tiep nha

xài delta rùi giải thôi bạn

Xét \(\Delta=\left(m+5\right)^2-4\left(m+4\right)=m^2+6m+9=\left(m+3\right)^2\)

\(\Rightarrow x=\frac{m+5\pm\sqrt{\left(m+3\right)^2}}{2}=\frac{m+5\pm m+3}{2}\)

:P

Trả lời:

\(P=\left(\frac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\frac{x}{x-2\sqrt{x}}\right)\div\frac{1-\sqrt{x}}{2-\sqrt{x}}\left(ĐK:x>0,x\ne1,x\ne4\right)\)

\(P=\left[\frac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}-\frac{x}{\sqrt{x}.\left(\sqrt{x}-2\right)}\right]\div\frac{-\left(\sqrt{x}-1\right)}{-\left(\sqrt{x}-2\right)}\)

\(P=\left[\frac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\sqrt{x}-2}\right]\div\frac{\sqrt{x}-1}{\sqrt{x}-2}\)

\(P=\left[\frac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}\right]\times\frac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(P=\left[\frac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}\right]\times\frac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(P=\left[\frac{-2\sqrt{x}+2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}\right]\times\frac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(P=\frac{-2.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}\times\frac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(P=\frac{-2}{\sqrt{x}+1}\)

Vậy \(P=\frac{-2}{\sqrt{x}+1}\)với \(x>0,x\ne1,x\ne4\)