@Akai Haruma, Nguyen, Nguyễn Thị Ngọc Thơsvtkvtm

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

pt đã cho <=>\(\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)-2\left(x+y\right)-\left(x+y+2\sqrt{xy}\right)+2\sqrt{xy}+4\left(\sqrt{x}+\sqrt{y}\right)-4=0\)

<=>\(\left(\sqrt{x}+\sqrt{y}\right)\left(x+y\right)-\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)-2\left(x+y\right)+2\sqrt{xy}-\left(\sqrt{x}+\sqrt{y}-2\right)^2=0\)

<=>\(\left(\sqrt{x}+\sqrt{y}-2\right)\left(x+y-\sqrt{xy}-\sqrt{x}-\sqrt{y}+2\right)=0\)

<=>\(\orbr{\begin{cases}\sqrt{x}+\sqrt{y}=2\\x+y-\sqrt{xy}-\sqrt{x}-\sqrt{y}+2=0\end{cases}}\)

th2: nhân cả hai vế với 2 ta được

\(\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+2>0\)

=>th2 vô nghiệm

do đó M=\(\sqrt{xy}\)

áp dụng bdt cô si ta có \(\sqrt{x}+\sqrt{y}>=2\sqrt{\sqrt{xy}}\)

<=>1>=\(\sqrt{\sqrt{xy}}\)(do \(\sqrt{x}+\sqrt{y}=2\))

<=>\(\sqrt{xy}< =1\)

<=>M<=1

Ta có:

\(x^2+1=x^2+xy+yz+zx\)

           \(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)

Tương tự:

\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)

\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

TH1: x,y,z <0

\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)

TH2: x,y,z>0

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)

Ta có \(1+z^2=xy+yz+zx+z^2\)

\(=y\left(x+z\right)+z\left(x+z\right)\)

\(=\left(x+z\right)\left(y+z\right)\)

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

Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)

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

 Tương tự như thế, ta được

\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

 Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.

\(\sqrt{2000}\)=\(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

\(\Rightarrow2000=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+y^2\right)\left(1+x^2\right)}\)

                  =\(x^2y^2+1+x^2+y^2+x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

                 \(\Rightarrow x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2000-1=1999\)

ma \(S^2=x^2\left(1+y^2\right)+y^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

           =\(x^2+x^2y^2+y^2+x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

          =\(x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\) =\(1999\Rightarrow S=\sqrt{1999}\)

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

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

\(P=\sqrt{x^2+y^2-\sqrt{2\left(x^2+1\right)\left(y^2+1\right)}+2}+xy\)

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

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

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

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

\(\sqrt{x\left(1-y\right)\left(1-z\right)}=\sqrt{x\left(yz-y-z+1\right)}=\sqrt{x\left(yz-y-z+x+y+z+2\sqrt{xyz}\right)}\)

\(=\sqrt{x\left(yz+x+2\sqrt{xyz}\right)}=\sqrt{x^2+2x\sqrt{xyz}+xyz}=\sqrt{\left(x+\sqrt{xyz}\right)^2}\)

\(=x+\sqrt{xyz}\)

Tương tự: \(\sqrt{y\left(1-x\right)\left(1-z\right)}=y+\sqrt{xyz}\) ; \(\sqrt{z\left(1-x\right)\left(1-y\right)}=z+\sqrt{xyz}\)

\(\Rightarrow VT=x+y+z+3\sqrt{xyz}=1-2\sqrt{xyz}+3\sqrt{xyz}=1+\sqrt{xyz}\) (đpcm)