ai giups tui di

Ta có\(x\sqrt{\frac{\left(2015+y^2\right)\left(2015+z^2\right)}{2015+x^2}}=x\sqrt{\frac{\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}{xy+yz+zx+x^2}}\)

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

Tương tự:\(y\sqrt{\frac{\left(2015+x^2\right)\left(2015+z^2\right)}{2015+y^2}}=yx+yz\)

               \(z\sqrt{\frac{\left(2015+x^2\right)\left(2015+y^2\right)}{2015+z^2}}=zx+zy\)

Ta có :\(P=xy+xz+yx+yz+zx+zy=2\left(xy+yz+zx\right)=4030\)

=>P không phải là số chính phương

\(\left(x+\sqrt{x^2+2015}\right)\left(y+\sqrt{y^2+2015}\right)=2015\)

\(\Leftrightarrow\frac{2015}{\sqrt{x^2+2015}-x}\left(y+\sqrt{y^2+2015}\right)=2015\)

\(\Leftrightarrow\sqrt{x^2+2015}-x=y+\sqrt{y^2+2015}\left(1\right)\)

Tương tự : \(x+\sqrt{x^2+2015}=\sqrt{y^2+2015}-y\left(2\right)\)

(1)+(2):

\(x+\sqrt{x^2+2015}+y+\sqrt{y^2+2015}=\sqrt{x^2+2015}+\sqrt{y^2+2015}-x-y\)

\(\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)

Bài 1

Từ giả thiết, bình phương 2 vế, ta được:

\(x^2y^2+\left(x^2+1\right)\left(y^2+1\right)+2xy\sqrt{x^2+1}\sqrt{y^2+1}=2015\)

\(\Leftrightarrow2x^2y^2+x^2+y^2+2xy\sqrt{x^2+1}\sqrt{y^2+1}=2014.\)

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

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

\(=2014\)

\(\Rightarrow A=\sqrt{2014}.\)

Bài 2:

Đặt \(\sqrt{2015}=a>0\)

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

Do \(\sqrt{y^2+a}-y>\sqrt{y^2}-y=\left|y\right|-y\ge0\) nên ta nhân cả 2 vế với \(\sqrt{y^2+a}-y\)

\(\left(1\right)\Leftrightarrow\left(x+\sqrt{x^2+a}\right)\left[\left(y^2+a\right)-y^2\right]=a.\left(\sqrt{y^2+a}-y\right)\)

\(\Leftrightarrow\sqrt{x^2+a}+x=\sqrt{y^2+a}-y\)

Tương tự ta có: \(\sqrt{y^2+a}+y=\sqrt{x^2+a}-x\)

Cộng theo vế 2 phương trình trên, ta được \(x+y=-\left(x+y\right)\Leftrightarrow x+y=0\)

Bài 3

Áp dụng bất đẳng thức Côsi

\(x\sqrt{x}+y\sqrt{y}+z\sqrt{z}\ge3\sqrt[3]{x\sqrt{x}.y\sqrt{y}.z\sqrt{z}}=3\sqrt{xyz}\)

Dấu bằng xảy ra khi và chỉ khi \(x=y=z\)

Thay vào tính được \(A=2.2.2=8\text{ }\left(x=y=z\ne0\right).\)

Em mới hoc lớp 7

Ta có:\(\left(x+\sqrt{x^2+2015}\right)\left(y-\sqrt{y^2+2015}\right)\left(y+\sqrt{y^2+2015}\right)=2015\left(y-\sqrt{y^2+2015}\right)\)

\(\Leftrightarrow-2015\left(x+\sqrt{x^2+2015}\right)=2015\left(y-\sqrt{y^2+2015}\right)\)

\(\Leftrightarrow x+\sqrt{x^2+2015}=\sqrt{y^2+2015}-y\)                                                 (1)

 Lại có:\(\left(x+\sqrt{x^2+2015}\right)\left(x-\sqrt{x^2+2015}\right)\left(y+\sqrt{y^2+2015}\right)=2015\left(x-\sqrt{x^2+2015}\right)\)

\(\Leftrightarrow-2015\left(y+\sqrt{y^2+2015}\right)=2015\left(x-\sqrt{x^2+2015}\right)\)

\(\Leftrightarrow y+\sqrt{y^2+2015}=\sqrt{x^2+2015}-x\)                                               (2)

Cộng theo vế \(\left(1\right)\) và \(\left(2\right)\) ta có:\(x+\sqrt{x^2+2015}+y+\sqrt{y^2+2015}=\sqrt{y^2+2015}+\sqrt{x^2+2015}-x-y\)

\(\Leftrightarrow2x+2y=0\Leftrightarrow x+y=0\)

Từ \(\left(x+\sqrt{x^2+2015}\right)\left(y+\sqrt{y^2+2015}\right)=2015\)

\(\Leftrightarrow2015\left(x+\sqrt{x^2+2015}\right)=2015\left(\sqrt{y^2+2015}-y\right)\)

\(\Leftrightarrow x+\sqrt{x^2+2015}=\sqrt{y^2+2015}-y\left(1\right)\)

Tương tự cũng có: \(y+\sqrt{y^2+2015}=\sqrt{x^2+2015}-x\left(2\right)\)

\(\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)