\(a,x\left(y-z\right)+y\left(z-x\right)+z\left(x-y\right)\\ =xy-xz+yz-xy+xz-yz\\ =\left(xy-xy\right)+\left(xz-xz\right)+\left(yz-yz\right)\\ =0+0+0\\ =0\left(dpcm\right)\)

\(b,x\left(y+z-yz\right)-y\left(z+x-zx\right)+z\left(y-x\right)\\ =xy+xz-xyz-yz-xy+xyz+yz-xz\\ =\left(xy-xy\right)+\left(xz-xz\right)+\left(xyz-xyz\right)+\left(yz-yz\right)\\ =0+0+0+0\\ =0\left(dpcm\right)\)

a,\(\dfrac{x-y}{xy}+\dfrac{y-z}{yz}+\dfrac{z-x}{zx}\)

=\(\dfrac{\left(x-y\right).z}{xyz}+\dfrac{\left(y-z\right).x}{xyz}+\dfrac{\left(z-x\right).y}{xyz}\)

=\(\dfrac{xz-yz}{xyz}+\dfrac{xy-xz}{xyz}+\dfrac{yz-xy}{xyz}\)

=\(\dfrac{xz-yz+xy-xz+yz-xy}{xyz}\)

=\(\dfrac{0}{xyz}\)=0

Vậy biểu thức trên ko phụ thuộc vào x,y,z

b,\(\dfrac{1}{\left(x-y\right).\left(y-z\right)}-\dfrac{1}{\left(x-z\right).\left(y-z\right)}-\dfrac{1}{\left(x-y\right).\left(x-z\right)}\)

=\(\dfrac{1.\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}-\dfrac{\left(x-y\right).1}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}-\dfrac{1\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

=\(\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)=\(\dfrac{0}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)=0

Vậy biểu thức trên ko phụ thuộc vào x,y,z

ai giúp mình với

...

Ta có:
         \(\dfrac{x-y}{1+xy}\)+\(\dfrac{y-z}{1+yz}\)+\(\dfrac{z-x}{1+xz}\) = \(\dfrac{x-y}{1+xy}\)+\(\dfrac{-\left(x-y\right)-\left(z-x\right)}{1+yz}\)+\(\dfrac{z-x}{1+xz}\)

         =\(\dfrac{x-y}{1+xy}\)\(-\dfrac{x-y}{1+yz}\) \(-\dfrac{z-x}{1+yz}\)+\(\dfrac{z-x}{1+xz}\) 

         = \(\left(x-y\right)\)\(\left(\dfrac{\left(1+yz\right)-\left(1+xy\right)}{\left(1+yz\right)\left(1+xy\right)}\right)\)+(\(z-x\))\(\left(\dfrac{\left(1+yz\right)-\left(1+zx\right)}{\left(1+yz\right)\left(1+zx\right)}\right)\)

         =\(\left(x-y\right)\)\(\dfrac{y\left(z-x\right)}{\left(1+yz\right)\left(1+xy\right)}\)+(\(z-x\))\(\dfrac{-z\left(x-y\right)}{\left(1+yz\right)\left(1+zx\right)}\)

         =\(\left(\dfrac{\left(x-y\right)\left(z-x\right)}{1+yz}\right)\)\(\left(\dfrac{y\left(1+xz\right)-z\left(1+xy\right)}{\left(1+xz\right)\left(1+xy\right)}\right)\)

       =đpcm

Áp dụng BĐT AM-GM ta có:

\(\frac{\left(y+z\right)\sqrt{yz}}{x}\ge\frac{2\sqrt{yz}\cdot\sqrt{yz}}{x}=\frac{2\sqrt{\left(yz\right)^2}}{x}=\frac{2yz}{x}\)

Tương tự cho 2 BĐT còn lại ta cũng có

\(\frac{\left(x+y\right)\sqrt{xy}}{z}\ge\frac{2xy}{z};\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xz}{y}\)

\(\Leftrightarrow\frac{\left(y+z\right)\sqrt{yz}}{x}+\frac{\left(x+y\right)\sqrt{xy}}{z}+\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\)

Cần chứng minh \(\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\ge2\left(x+y+z\right)\)

\(\Leftrightarrow\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge x+y+z\)

Áp dụng BĐT AM-GM:

\(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy}{z}\cdot\frac{yz}{x}}=2\sqrt{y^2}=2y\)

Tương tự rồi cộng theo vế ta có ĐPCM

Khi \(x=y=z\)

Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)

\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)

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

(vì \(2013=3.671=3\left(xy+yz+zx\right)\))

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

\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)

\(=\dfrac{1}{x+y+z}\)

ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)

\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)

\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))

Vậy ta có đpcm.

Lời giải:

Đặt \((x,y,z)=(a^2,b^2,c^2)\). Bài toán tương đương với:

\(\frac{bc(b+c)}{a}+\frac{ac(a+c)}{b}+\frac{ab(a+b)}{c}\geq 2(a^2+b^2+c^2)\)

Biến đổi ta thấy:

\(\text{VT}=a^2\left ( \frac{b}{c}+\frac{c}{b} \right )+b^2\left ( \frac{a}{c}+\frac{c}{a} \right )+c^2\left ( \frac{a}{b}+\frac{b}{a} \right )\)

Áp dụng BĐT AM-GM:

\(\left\{\begin{matrix} \frac{a}{b}+\frac{b}{a}\geq 2\\ \frac{a}{c}+\frac{c}{a}\geq 2\\ \frac{b}{c}+\frac{c}{b}\geq 2\end{matrix}\right.\Rightarrow \text{VT}\geq 2(a^2+b^2+c^2)=\text{VP}\)

Do đó ta có đpcm

Dấu bằng xảy ra khi \(a=b=c\Leftrightarrow x=y=z>0\)

Áp dụng BĐT AM-GM ta có:

\(\dfrac{\left(y+z\right)\sqrt{yz}}{x}\ge\dfrac{2\sqrt{yz}\cdot\sqrt{yz}}{x}=\dfrac{2yz}{x}\)

Tương tự cho 2 BĐT còn lại thì được:

\(\dfrac{2xy}{z}+\dfrac{2yz}{x}+\dfrac{2xz}{y}\ge2\left(x+y+z\right)\)

\(\Leftrightarrow\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\ge x+y+z\)

Tiếp tục dùng AM-GM:

\(\dfrac{xy}{z}+\dfrac{yz}{x}\ge2\sqrt{y^2}=2y\)

Tương tự rồi cộng theo vế có:

\(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\ge x+y+z\) (đúng)

Hay ta có ĐPCM. Khi \(x=y=z\)

Đề này à: \(\dfrac{\left(y+z\right)\sqrt{yz}}{x}+\dfrac{\left(z+x\right)\sqrt{zx}}{y}+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge2\left(x+y+z\right)\)

Dùng máy tính kiểm tra. (đề sai không?)

Thế x=1, y=2, z=3

VT = 17,12576389

VP = 12

\(=\dfrac{x}{xy}-\dfrac{y}{xy}+\dfrac{y}{yz}-\dfrac{z}{yz}+\dfrac{z}{zx}-\dfrac{x}{zx}\)

\(=\dfrac{1}{y}-\dfrac{1}{x}+\dfrac{1}{z}-\dfrac{1}{y}+\dfrac{1}{x}-\dfrac{1}{z}\)

= 0

=> KO PHỤ THUỘC

* Chứng minh biểu thức sau phụ thuộc vào x , y , z

\(\dfrac{x-y}{xy}+\dfrac{y-z}{yz}+\dfrac{z-x}{zx}\)

= \(\dfrac{(x-y)z+(y-z)x+(z-x)y}{xyz} \)

= \(\dfrac{xz-yz+xy-xz+zy-xy}{xyz}\)

= \(\dfrac{0}{xyz}\)

= 0

Vậy \(\dfrac{x-y}{xy} + \dfrac{y-z}{yz} + \dfrac{z-x}{zx} \) phụ thuộc vào x , y ,z

SORY I'M I GRADE 6

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mày hỏi vả bài kiểm tra à thằng điên