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

\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{x^2+y^2+z^2-xy-yz-xz}\)

\(=x+y+z\)

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

Đặt \(N=x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right).z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2-3xy\right)\)

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

Vậy \(M=\frac{N}{x^2+y^2+z^2-xy-yz-zx}=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{x^2+y^2+z^2-xy-yz-zx}=x+y+z=2016\)

(*) bn ghi sai đề 1 chỗ nhé:ở mẫu thức của M phải là  \(x^2+y^2+z^2-xy-yz-zx\) nhé!

Ta có: 

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)

\(=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(x+y\right).z-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yx-3xz-3yz-3xy\right)\)

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

=> \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz\)

Đáp án:

$P=\pm 36$

Giải thích các bước giải:

Ta có:

x2+y2+z2=16xy−yz+zx=−10⇒(x2+y2+z2)−2.(xy−yz+zx)=16−2.(−10)⇔x2+y2+z2−2xy+2yz−2zx=36⇔(x2−2xy+y2)+z2+2yz−2zx=36⇔(x−y)2+2z(y−x)+z2=36⇔(x−y)2−2.(x−y).z+z2=36⇔(x−y−z)2=36⇔x−y−z=±6P=x3−y3−z3−3xyz=(x3−3x2y+3xy2−y3)−z3+3x2y−3xy2−3xyz=(x−y)3−z3+3x2y−3xy2−3xyz=[(x−y)−z].[(x−y)2+(x−y).z+z2]+3xy(x−y−z)=(x−y−z).(x2−2xy+y2+xz−yz+z2+3xy)=(x−y−z).(x2+y2+z2+xy−yz+zx)Trường hợp 1: x−y−z=6⇒P=6.(16+(−10))=36Trường hợp 2: x−y−z=−6⇒P=(−6).(16+(−10))=−36x2+y2+z2=16xy−yz+zx=−10⇒(x2+y2+z2)−2.(xy−yz+zx)=16−2.(−10)⇔x2+y2+z2−2xy+2yz−2zx=36⇔(x2−2xy+y2)+z2+2yz−2zx=36⇔(x−y)2+2z(y−x)+z2=36⇔(x−y)2−2.(x−y).z+z2=36⇔(x−y−z)2=36⇔x−y−z=±6P=x3−y3−z3−3xyz=(x3−3x2y+3xy2−y3)−z3+3x2y−3xy2−3xyz=(x−y)3−z3+3x2y−3xy2−3xyz=[(x−y)−z].[(x−y)2+(x−y).z+z2]+3xy(x−y−z)=(x−y−z).(x2−2xy+y2+xz−yz+z2+3xy)=(x−y−z).(x2+y2+z2+xy−yz+zx)Trường hợp 1: x−y−z=6⇒P=6.(16+(−10))=36Trường hợp 2: x−y−z=−6⇒P=(−6).(16+(−10))=−36

Vậy $P=\pm 36$.

MÌNH CHỈ BIẾT LÀM B7 THÔI NHA

P= 811^3+ 812^3+815^3+3.811.812.(-815)=  31694

K ĐÚNG HỘ TỚ NHA

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Lời giải:
a)

\(\frac{x^4-3x^2+1}{x^4-x^2-2x-1}=\frac{(x^4-2x^2+1)-x^2}{(x^4-x)-(x^2+x+1)}=\frac{(x^2-1)^2-x^2}{x(x^3-1)-(x^2+x+1)}\)

\(=\frac{(x^2-1-x)(x^2-1+x)}{x(x-1)(x^2+x+1)-(x^2+x+1)}=\frac{(x^2-1-x)(x^2-1+x)}{(x^2+x+1)(x^2-x-1)}=\frac{x^2+x-1}{x^2+x+1}\)

\(=\frac{x^2+x+1-2}{x^2+x+1}=1-\frac{2}{x^2+x+1}\)

b)

Xét tử số:

\(x^3+y^3+z^3-3xyz=(x+y)^3-3xy(x+y)+z^3-3xyz\)

\(=[(x+y)^3+z^3]-3xy(x+y+z)\)

\(=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)\)

\(=(x+y+z)[(x+y)^2-(x+y)z+z^2-3xy]\)

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

Do đó:

\(\frac{x^3+y^3+z^3-3xyz}{x^2+y^2+z^2-xy-yz-xz}=\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}{x^2+y^2+z^2-xy-yz-xz}=x+y+z\)

-3xyz hayu -xyz vậy bạn?

mình biến đổi bước xy+yz+zx=3xyz roi nhe 1/x+1/y+1/z=3

Ta có: \(\frac{x^3}{x^2+z}=\frac{x^3+xz}{x^2+z}-\frac{xz}{x^2+z}\ge x-\frac{xz}{2x\sqrt{z}}=x-\frac{\sqrt{z}}{2}\)

Lại có: \(\sqrt{z}\le\frac{z+1}{2}\)

\(\Rightarrow\frac{x^3}{x^2+z}\ge x-\frac{z+1}{4}\)

Tương tự cộng vào ta có: 

\(VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\)

Lại có: \(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)

\(\Rightarrow x+y+z\ge3\)

\(\ge VT\ge\frac{3}{4}.3-\frac{3}{4}=1,5\)

Dấu = xảy ra khi x=y=z=1