Lời giải:

Ta có: \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)

Mà theo BĐT Cauchy-Schwarz: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\)

Do đó: \(3\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3\)

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Ta có: \(\text{VT}=x-\frac{xz}{x^2+z}+y-\frac{xy}{y^2+x}+z-\frac{yz}{z^2+y}\)

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

\(\geq x+y+z-\frac{1}{2}\left(\frac{xy}{\sqrt{xy^2}}+\frac{yz}{\sqrt{z^2y}}+\frac{xz}{\sqrt{x^2z}}\right)\) (AM-GM)

\(=x+y+z-\frac{1}{2}(\sqrt{x}+\sqrt{y}+\sqrt{z})\)

Tiếp tục AM-GM: \(\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \frac{x+1}{2}+\frac{y+1}{2}+\frac{z+1}{2}=\frac{x+y+z+3}{2}\)

Suy ra:

\(\text{VT}\geq x+y+z-\frac{1}{2}.\frac{x+y+z+3}{2}=\frac{3}{4}(x+y+z)-\frac{3}{4}\)

\(\geq \frac{9}{4}-\frac{3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Ta có đpcm

Dấu bằng xảy ra khi $x=y=z=1$

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

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

\(=x+y+z\)

thay 1 vào tử, thấy: 
căn(5-x) = căn 4= 2; 
căn bậc 3(x^2+7)=căn bậc 3 của 8=2 
=> thêm bớt 2. 
Bài làm: 
lim {[căn(5-x)-2]-[căn bậc 3(x^2-7)-2]}/(x^2-1) 
tương đương: lim [căn(5-x)-2]/(x^2-1) - lim [căn bậc 3(x^2-7)-2]/(x^2-1) 
Tính lim từng số hạng như thường.

Bạn trả lời rõ dùm mình với

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

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

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

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

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

\(\dfrac{x^3+y^3+z^3-3xyz}{xy^2+xz\left(2y+z\right)}.\dfrac{x\left(x+y\right)+y\left(x-xy\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)}{xy^2+2xyz+x^2z}.\dfrac{x^2+xy-xy-xy^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\\ =\dfrac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2xy^2+4xyz+2x^2z}.\dfrac{x^2-xy^2}{\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2-xy\right)}{2xy^2+4xy+2x^2z}\)

@@ ko ra nữa

a)

\(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)

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

b) 

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

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

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

Lời giải:

Áp dụng hằng đẳng thức dạng:

\(a^3+b^3=(a+b)^3-3ab(a+b)=(a+b)(a^2-ab+b^2)\) ta có:

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

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

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

\(=(x+y+z)(x^2+y^2+2xy-zx-zy+z^2-3xy)\)

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

Ta có đpcm.

Lời giải:

Áp dụng hằng đẳng thức dạng:

\(a^3+b^3=(a+b)^3-3ab(a+b)=(a+b)(a^2-ab+b^2)\) ta có:

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

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

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

\(=(x+y+z)(x^2+y^2+2xy-zx-zy+z^2-3xy)\)

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

Ta có đpcm.