Ta có:

\(M=\frac{x\left(yz-x^2\right)+y\left(zx-y^2\right)+z\left(xy-z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{xyz-x^3+xyz-y^3+xyz-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{3xyz-x^3-y^3-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)

\(-M=\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)

Xét đẳng thức phụ:

\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=\left[\left(a +b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-ab\right]=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-abc-ac\right)\)

\(=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)

Thay vào -M ta có:

\(-M=\frac{\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{1}{2}\left(x+y+z\right)\Rightarrow M=-\frac{1}{2}\left(x+y+z\right)\)

Giờ thay: \(x=2014^{2015}-20142015;y=20142015-2015^{2014};z=2015^{2014}-2014^{2015}\)

Ta có:

\(M=-\frac{1}{2}\left(2014^{2015}-20142015+20142015-2015^{2014}+2015^{2014}-2014^{2015}\right)=0\)

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

Từ giả thiết ta có ngay \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

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

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

Suy ra x + y = 0 hoặc y + z = 0 hoặc z + x = 0

Tới đây bạn tự làm nhé :)

AD BĐT cô - si cho ba số không âm x² ; y² ; z² , ta có :

x² + y² ≥ 2√x²y² = 2xy ( dấu bằng xảy ra khi x = y )

Tương tự : y² + z² ≥ 2yz ( dấu ... khi y = x )

z² + x² ≥ 2zx ( ... z = x )

⇒ 2 ( x² + y² + z² ) ≥ 2 ( xy + yz + zx )

⇔ x² + y² + z² ≥ xy + yz + zx

Dấu = xảy ra khi x = y = z

⇒ x²⁰¹⁵ + y²⁰¹⁵ + z²⁰¹⁵ = 3x²⁰¹⁵ = 3²⁰¹⁶

⇔ 3²⁰¹⁵. x = 3²⁰¹⁵. 3 ⇒ x = 3

⇒ x = y = z = 3

đề thi casio lớp 9 đó
 

mk ms có lớp 8 nên ngồi cắn bút k pt làm

kho....................wa..................troi.......................thi.....................ret.................lanh................wa..................tich............................ung.........................ho..............minh......................cho....................do....................lanh

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

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

Ta co: \(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)

\(\Rightarrow\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}=\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=y+z\)

Thê vào ta được

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