đk của x,y,z là x,y,z\(\ge\sqrt{2014}\) nhé, xin lỗi chép sót đề

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-2014}=a\left(a\ge0\right)\\\sqrt{y^2-2014}=b\left(b\ge0\right)\\\sqrt{z^2-2014}=c\left(c\ge0\right)\end{matrix}\right.\)

\(\Rightarrow ab+bc+ca=2014\)

Ta có: \(\sqrt{x^2-2014}=a\)

\(\Leftrightarrow x^2-2014=a^2\)

\(\Rightarrow x^2=a^2+2014=a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\)

Tương tự, ta có:

\(y^2=\left(b+c\right)\left(b+a\right)\)

\(z^2=\left(c+a\right)\left(c+b\right)\)

Xét \(A=xyz\left(\dfrac{\sqrt{x^2-2014}}{x^2}+\dfrac{\sqrt{y^2-2014}}{y^2}+\dfrac{\sqrt{z^2-2014}}{z^2}\right)\)

\(=\sqrt{\left(a+b\right)\left(a+c\right)}\times\sqrt{\left(b+c\right)\left(b+c\right)}\times\sqrt{\left(c+a\right)\left(c+b\right)}\)

\(\times\left[\dfrac{a}{\left(a+b\right)\left(a+c\right)}+\dfrac{b}{\left(b+c\right)\left(b+a\right)}+\dfrac{c}{\left(c+a\right)\left(c+b\right)}\right]\)

\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\times\dfrac{a\left(b+c\right)\times b\left(c+a\right)\times c\left(b+a\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)

\(=2\left(ab+bc+ac\right)=4028\)

câu này mik vừa làm sáng ngày ne

ta đặt \(\sqrt{x^2-2014}=a;\sqrt{y^2-2014}=b;\sqrt{z^2-2014}=c\)

ta có \(ab+bc+ca=2014\Rightarrow ab+bc+ca+a^2=x^2-2014+2014=x^2\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)=x^2\)

tương tự ta có \(\left(b+c\right)\left(b+a\right)=y^2;\left(c+a\right)\left(c+b\right)=z^2\)

nhân cả 3 vào ta có \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=xyz\)

=> \(\hept{\begin{cases}\left(a+b\right)z^2=xyz\\\left(b+c\right)x^2=xyz\\\left(c+a\right)y^2=xyz\end{cases}\Rightarrow\hept{\begin{cases}a+b=\frac{xy}{z}\\b+c=\frac{yz}{x}\\c+a=\frac{zx}{y}\end{cases}}}\)

cậu nhân tung A ra rồi thay \(\frac{xy}{z};\frac{yz}{x};\frac{zx}{y}\) như vừa tính vào thì cậu sẽ ra kết quả là A=4028

Ta có \(\left(x+y+z\right)^2-x^2-y^2-z^2=a^2-b\Rightarrow2\left(xy+yz+zx\right)=2048\Rightarrow xy+yz+zx=2014\)

với xy+yz+zx=2014, thay vào, ta có A=\(\sum x\sqrt{\dfrac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}=\sum x\sqrt{\dfrac{\left(y+z\right)^2\left(y+x\right)\left(z+x\right)}{\left(x+z\right)\left(x+y\right)}}=\sum x\left(y+z\right)=2\left(xy+yz+zx\right)=2048\)

\(\left(\sqrt{x^2+2014}-x\right)\left(x+\sqrt{x^2+2014}\right)\left(y+\sqrt{y^2+2014}\right)\)

nhân lên

1) x+ căn x^2+2014=2014/ y- căn y^2+2014= 2014(y+căn  y^2+ 2014)/-2014=-y-(căn y^2+2014)

tương tự , đuwa bên x+ căn... qua=> 1 pt y+ căn//..... =??

sau đó kết hợp 2 cái này là ra

`(x+3)^2014 = (x+3)^2012`

`(x+3)^2014 -(x+3)^2012 =0`

`(x+3)^2012 [(x+3)^2 -1]=0`

TH1 :`(x+3)^2012 =0 => x+3 =0 => x=-3`

TH2 :`(x+3)^2 -1 =0 => (x+3)^2 =1 => [(x+3=1),(x+3=-1):}`

`=> [(x=-2),(x=-4):}`

`(x+3)^2014 = (x+3)^2012`

`=> (x+3)^2014 - (x+3)^2012 = 0`

`=> (x+3)^2 * (x+3)^2012 - (x+3)^2012 = 0`

`=> (x+3)^2012 * [ (x+3)^2 - 1] =0`

`=>`\(\left[{}\begin{matrix}\left(x+3\right)^{2012}=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)

Vậy, `x = {-3; -2; -4}.`

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:

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

\(f\left(1\right)=\left(1^2+1-1\right)^{2014}+\left(1^2-1-1\right)^{2014}-2=1+1-2=0\)

Nên \(f\left(x\right)⋮\left(x-1\right)\)

\(f\left(-1\right)=\left[\left(-1\right)^2+\left(-1\right)-1\right]^{2014}.\left[\left(-1\right)^2-\left(-1\right)-1\right]^{2014}-2=1+1-2=0\)

Nên \(f\left(x\right)⋮\left(x+1\right)\)

Vậy \(f\left(x\right)⋮\left[\left(x-1\right)\left(x+1\right)\right]\Rightarrow f\left(x\right)⋮\left(x^2-1\right)\)