Ta có : 
1 + x = 1 + (b^2 + c^2 - a^2)/2bc = [(b + c)^2 - a^2]/2bc 
1 + y = 1 + [a*2 - (b - c)^2]/[(b + c)^2 - a^2] = 4bc/[(b + c)^2 - a^2] 
Vậy : P = x + y + xy = (1 + x)(1 + y) - 1 = 2 - 1 = 1

  Ta có : 
1 + x = 1 + (b^2 + c^2 - a^2)/2bc = [(b + c)^2 - a^2]/2bc 
1 + y = 1 + [a*2 - (b - c)^2]/[(b + c)^2 - a^2] = 4bc/[(b + c)^2 - a^2] 
Vậy : P = x + y + xy = (1 + x)(1 + y) - 1 = 2 - 1 = 1

Linh_Men

a)\(M=\text{[}x^3-3xy\left(x-y\right)-y^3\text{]}-\left(x^2-2xy+y^2\right)\)

\(M=\left(x-y\right)^3-\left(x-y\right)^2\)

\(\Rightarrow M=7^3-7^2\)

\(M=294\)

2.a là x=0 , x=-1, x=-2
2.b là x=2/3 , x=-5

Đkxđ : \(x+y\ne0\)

\(x^2-2y^2=xy\Rightarrow x^2-y^2=xy+y^2\)

\(\Rightarrow\left(x-y\right)\left(x+y\right)=y\left(x+y\right)\)

\(\Rightarrow x-y=y\)

\(\Rightarrow x=2y\)

Thay x = 2y vào M có :

\(M=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)

Vậy ...

Bài 1:

\(x^2+\frac{1}{x^2}=2\Leftrightarrow (x+\frac{1}{x})^2-2.x.\frac{1}{x}=7\Leftrightarrow (x+\frac{1}{x})^2=9\)

\(\Rightarrow x+\frac{1}{x}=3\) (do \(x>0\rightarrow x+\frac{1}{x}>0\))

\(\Rightarrow (x+\frac{1}{x})^3=27\)

\(\Leftrightarrow x^3+\frac{1}{x^3}+3x.\frac{1}{x}(x+\frac{1}{x})=27\)

\(\Leftrightarrow x^3+\frac{1}{x^3}+3.3=27\Leftrightarrow x^3+\frac{1}{x^3}=18\)

Do đó:

\(x^5+\frac{1}{x^5}=(x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})-(x+\frac{1}{x})=7.18-3=123\)

Bài 2:

Ta có:

\(x^2+y^2+z^2=xy+yz+xz\)

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0\)

\(\Leftrightarrow 2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)

\(\Leftrightarrow (x^2+y^2-2xy)+(y^2+z^2-2yz)+(z^2+x^2-2xz)=0\)

\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)

Ta thấy $(x-y)^2; (y-z)^2; (z-x)^2\geq 0, \forall x,y,z\in\mathbb{R}$

Do đó để $(x-y)^2+(y-z)^2+(z-x)^2=0$ thì $(x-y)^2=(y-z)^2=(z-x)^2=0$

Hay $x=y=z$

Thay vào điều kiện thứ 2:

$\Rightarrow x^{2016}+x^{2016}+x^{2016}=3^{2017}$

$\Leftrightarrow 3.x^{2016}=3^{2017}$

$\Leftrightarrow $x=3$

$\Rightarrow y=z=x=3$

Vậy $x=y=z=3$