a, dk \(x\ge0.x\ne1\)

\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)

 =\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)

phan b,c ban tu lam not nhe dai lam mk ko lam dau  mk co vc ban rui

Ta có :

 Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)

=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)

=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)

=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)

=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)

=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)

=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)

=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)

           =\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)

           =\(4-2\sqrt{4-3}\)

           =\(4-2\)

           =\(2\)

=>\(A=\sqrt{2}\)

Ehem mk nghi pa nhan binh Phương 2 ve len

@Akai Haruma, Nguyen, Nguyễn Thị Ngọc Thơsvtkvtm

Bạn tham khảo tại đây:

Câu hỏi của Vũ Sơn Tùng - Toán lớp 9 | Học trực tuyến

\(\Rightarrow\left|a\right|\le1\),\(\left|b\right|\le1\),\(\left|c\right|\le1\)

\(\Rightarrow1-a\ge0\)tương tự 1-b,1-c............

\(\Rightarrow\left(1\right)\ge0\)

dấu = khi a=1b=0c=0 và hoán vị

Đang nổi cơn làm biếng mà nhìn thấy hệ còn buồn ngủ hơn:

a/ \(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-2x\right)\left(2x-y\right)=6\\x^2-2x-2\left(2x-y\right)=1\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2-2x=a\\2x-y=b\end{matrix}\right.\) ta được:

\(\left\{{}\begin{matrix}ab=6\\a-2b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=6\\a=2b+1\end{matrix}\right.\)

\(\Rightarrow b\left(2b+1\right)=6\Leftrightarrow2b^2+b-6=0\)

\(\Leftrightarrow...\)

b/ ĐKXĐ: ...

\(\Leftrightarrow\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}+\sqrt{2-\frac{1}{y}}-\sqrt{2-\frac{1}{x}}=0\)

\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{\frac{1}{x}-\frac{1}{y}}{\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}}=0\)

\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{y-x}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}=0\)

\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}=0\)

\(\Leftrightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\frac{1}{\sqrt{xy}}+\frac{\sqrt{y}+\sqrt{x}}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}\right)=0\)

\(\Leftrightarrow\sqrt{y}=\sqrt{x}\Rightarrow x=y\)

Thay vào pt đầu:

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

\(\Leftrightarrow\frac{1}{x}+2-\frac{1}{x}+2\sqrt{\frac{2}{x}-\frac{1}{x^2}}=4\)

\(\Leftrightarrow\sqrt{\frac{2}{x}-\frac{1}{x^2}}=1\)

\(\Leftrightarrow\frac{2}{x}-\frac{1}{x^2}=1\)

\(\Leftrightarrow\left(\frac{1}{x}-1\right)^2=0\)

c/ \(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\x^2+y^2+x+y+xy=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\x^2+y^2=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\\left(x+y\right)^2-2xy=10\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\) ta được:

\(\left\{{}\begin{matrix}a+b=7\\a^2-2b=10\end{matrix}\right.\) \(\Rightarrow a^2+2a-24=0\Rightarrow\left[{}\begin{matrix}a=4;b=3\\a=-6;b=13\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)

Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\)

Tương tự  \(1+y^2=\left(x+y\right)\left(y+z\right)\)

\(1+z^2=\left(x+z\right)\left(y+z\right)\)

Thay vào A ta được

\(P=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)

=2(xy+xz+yz)=2

\(b,VT=VP\)

\(\Leftrightarrow\frac{x}{xy+yz+zx+x^2}+\frac{y}{xy+yz+zx+y^2}+\frac{z}{xy+yz+zx+z^2}\)

                                                                                                                                                                                                                                                                                    \(=\frac{2xyz}{\sqrt{\left(xy+yz+zx+x^2\right)\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}}\)

\(\Leftrightarrow\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(x+y\right)\left(y+z\right)}+\frac{z}{\left(x+z\right)\left(y+z\right)}\)

                                                                                \(=\frac{2xyz}{\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)\left(z+x\right)\left(y+z\right)}}\)

\(\Leftrightarrow\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(\Leftrightarrow xy+xz+xy+yz+xz+yz=2xyz\)

\(\Leftrightarrow2=2xyz\)

\(\Leftrightarrow xyz=1\)

Đù =)))