Tính:
\(\dfrac{x\left(y^2-z\right)+y\left(x-xy\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\) . \(\dfrac{2\left(x^3+y^3+z^3-3xyz\right)}{xy^2-xz\left(2y-z\right)}\)
Giúp mình với!!! Mình cần gấp nha!!!
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23 tháng 9 2017
\(\dfrac{x^3+y^3+z^3-3xyz}{xy^2+xz\left(2y+z\right)}.\dfrac{x\left(x+y\right)+y\left(x-xy\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)}{xy^2+2xyz+x^2z}.\dfrac{x^2+xy-xy-xy^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\\ =\dfrac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2xy^2+4xyz+2x^2z}.\dfrac{x^2-xy^2}{\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2-xy\right)}{2xy^2+4xy+2x^2z}\)
@@ ko ra nữa
14 tháng 1 2021
Áp dụng bất đẳng thức AM - GM:
\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).
Áp dụng bất đẳng thức AM - GM ta có:
\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).
Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).
Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)
\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).
Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)
Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)
\(\Rightarrow P\ge\dfrac{15}{2}\).
Vậy...
HN
17 tháng 1 2021
Áp dụng bất đẳng thức AM - GM:
P≥33√(xy+1)(yz+1)(zx+1)xyz.
Áp dụng bất đẳng thức AM - GM ta có:
xy+1=xy+14+14+14+14≥55√xy44.
Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.
Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412
⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.
Mà xyz≤(x+y+z)327=18
Nên (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258
⇒P≥152.
9 tháng 12 2018
\(\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\dfrac{y^2-xz}{\left(y+z\right)\left(x+y\right)}+\dfrac{z^2-xy}{\left(x+z\right)\left(z+y\right)}\)
\(=\dfrac{\left(x^2-yz\right)\left(y+z\right)+\left(y^2-xz\right)\left(x+z\right)+\left(z^2-xy\right)\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\left\{{}\begin{matrix}\left(x^2-yz\right)\left(y+z\right)=x^2y+x^2z-y^2z-yz^2\\\left(y^2-xz\right)\left(x+z\right)=y^2x+y^2z-x^2z-xz^2\\\left(z^2-xy\right)\left(x+y\right)=z^2x+z^2y-x^2y-xy^2\end{matrix}\right.\)
Đa thức trên bằng 0
\(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-x\right)\left(y-z\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\dfrac{-x^2}{\left(x-y\right)\left(z-x\right)}+\dfrac{-y^2}{\left(x-y\right)\left(y-z\right)}+\dfrac{-z^2}{\left(z-x\right)\left(y-z\right)}\)
\(=\dfrac{-x^2\left(y-z\right)-y^2\left(z-x\right)-z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
Xét: \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z+y^2z-xy^2+z^2\left(x-y\right)\)
\(\)\(=xy\left(x-y\right)-z\left(x^2-y^2\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-xz-yz+z^2\right)\)
\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]\)
\(=\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
Thêm dấu - đằng trc nữa suy ra bt có giá trị bằng 1 :P
21 tháng 11 2017
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
MH
22 tháng 11 2023
Ta có:
\(x^2+1=x^2+xy+yz+zx\)
\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
Tương tự:
\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)
\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
TH1: x,y,z <0
\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)
TH2: x,y,z>0
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)
22 tháng 11 2023
Ta có \(1+z^2=xy+yz+zx+z^2\)
\(=y\left(x+z\right)+z\left(x+z\right)\)
\(=\left(x+z\right)\left(y+z\right)\)
CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)
Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)
Tương tự như thế, ta được
\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.