Cho x,y,z,t \(\in R\) và \(1\le x,y,z,t\le2\)
Tìm giá trị lớn nhất của \(A=\dfrac{\left(x+y\right)\left(y+t\right)}{\left(x+z\right)\left(z+y\right)}+\dfrac{\left(y+t\right)\left(t+z\right)}{\left(z+x\right)\left(x+t\right)}\)
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CW
13 tháng 3 2017
Quy đồng tính bình thường.
\(A=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+2\left(\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{z-x}\right)\)\(=\dfrac{2x^2+2y^2+2z^2-2xy-2yz-2xz}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\dfrac{2yz+2xz+2xy-2x^2-2y^2-2z^2}{ }\)
=0
TT
20 tháng 12 2017
a,
\(-\dfrac{x}{\left(x-y\right)\left(z-x\right)}-\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(z-x\right)\left(y-z\right)}\)
\(\dfrac{-x\left(y-z\right)-y\left(z-x\right)-z\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(\dfrac{-xy+xz-yz+xy-zx+yz}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
= 0
OY
29 tháng 1 2018
\(P=\dfrac{\left(x+y\right)\left(y+z\right)}{z+x}+\dfrac{\left(y+z\right)\left(z+x\right)}{x+y}+\dfrac{\left(z+x\right)\left(x+y\right)}{y+z}\)
Áp dụng BĐT Cauchy ta có:
\(\left\{{}\begin{matrix}x+y\ge2\sqrt{xy}\\z+y\ge2\sqrt{yz}\\x+z\ge2\sqrt{xz}\end{matrix}\right.\)
\(\Rightarrow\dfrac{\left(x+y\right)\left(y+z\right)}{z+x}\ge\dfrac{2\sqrt{xy}.2\sqrt{yz}}{2\sqrt{xz}}\)
\(\Leftrightarrow\dfrac{\left(x+y\right)\left(y+z\right)}{z+x}\ge2y\) (1)
Chứng minh tương tự ta có:
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\left(y+z\right)\left(z+x\right)}{x+y}\ge2z\left(2\right)\\\dfrac{\left(y+x\right)\left(z+x\right)}{z+y}\ge2x\left(3\right)\end{matrix}\right.\)
Từ (1),(2),(3)
\(\Rightarrow P\ge2x+2y+2z\)
\(\Rightarrow P\ge2.3\)
\(\Rightarrow P\ge6\)
Dấu "=" xảy ra khi
\(x=y=z\)
Vậy Min P là 6 khi \(x=y=z\)
29 tháng 1 2018
Otasaka Yu: Cosi nhưng đừng là ở dưới đó.... (it's same some mô típ i've read and seen Manga and Anime Japan ( ͡° ͜ʖ ͡°))
\(\dfrac{\left(x+y\right)\left(y+z\right)}{x+z}+\dfrac{\left(y+z\right)\left(x+z\right)}{x+y}\ge2\sqrt{\left(y+z\right)^2}=2\left(y+z\right)\)
Tương tự rồi cộng theo vế:
\(2P\ge2\left(x+y+z\right)\Leftrightarrow P\ge x+y+z=3\)
\("=" <=> x=y=z=1\)
It's A jOke. DoN't TriGgeRed my dude !
AH
Akai Haruma
Giáo viên
6 tháng 6 2018
Lời giải:
Ta có:
\(P=\frac{x^3}{(x-y)(x-z)}+\frac{y^3}{(y-x)(y-z)}+\frac{z^3}{(z-y)(z-x)}\)
\(=\frac{x^3(y-z)+y^3(x-z)+z^3(y-x)}{(x-y)(y-z)(z-x)}\)
\(=\frac{xz(x^2-z^2)+xy(y^2-x^2)+zy(z^2-y^2)}{(x-y)(y-z)(z-x)}\)
\(=\frac{xz(x-z)(x+z)+xy(y-x)(y+x)+zy(z-y)(z+y)}{(x-y)(y-z)(z-x)}\)
\(=\frac{xz(x-z)(2008-y)+xy(y-x)(2008-z)+zy(z-y)(2008-x)}{(x-y)(y-z)(z-x)}\)
\(=\frac{2008[xz(x-z)+xy(y-x)+zy(z-y)-xyz(x-z+y-x+z-y)}{(x-y)(y-z)(z-x)}\)
\(=\frac{2008[xz(x-z)+xy(y-x)+zy(z-y)]}{xz(x-z)+xy(y-x)+zy(z-y)}=2008\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 8 2020
\(P=3x^2+3z^2+10y^2+10t^2+8xy+8zt+4zx+2yz+2xt\)
\(P\le5x^2+5z^2+10y^2+10t^2+8xy+8zt+2yz+2xt\)
\(P\le10+5y^2+5t^2+8xy+8zt+2yz+2xt\)
\(\left\{{}\begin{matrix}8xy=\left(2+2\sqrt{5}\right)\left[2.x.\frac{\left(\sqrt{5}-1\right)}{2}y\right]\le\left(2+2\sqrt{5}\right)\left[x^2+\left(\frac{3-\sqrt{5}}{2}\right)y^2\right]\\8zt\le\left(2+2\sqrt{5}\right)\left[z^2+\left(\frac{3-\sqrt{5}}{2}\right)t^2\right]\\2yz\le\left(\frac{\sqrt{5}+1}{2}\right)\left[z^2+\left(\frac{3-\sqrt{5}}{2}\right)y^2\right]\\2xt\le\left(\frac{\sqrt{5}+1}{2}\right)\left(x^2+\left(\frac{3-\sqrt{5}}{2}\right)t^2\right)\end{matrix}\right.\)
\(\Rightarrow P\le10+\frac{5}{2}\left(\sqrt{5}+1\right)\left(x^2+y^2+z^2+t^2\right)\le15+5\sqrt{5}\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x=z=\sqrt{\frac{5-\sqrt{5}}{10}}\\y=t=\sqrt{\frac{5+\sqrt{5}}{10}}\end{matrix}\right.\)
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)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
AH
Akai Haruma
Giáo viên
2 tháng 3 2019
Lời giải:
Xét hiệu:
\(\frac{x^4}{(x^2+y^2)(x+y)}+\frac{y^4}{(y^2+z^2)(y+z)}+\frac{z^4}{(z^2+x^2)(z+x)}-\left(\frac{y^4}{(x^2+y^2)(x+y)}+\frac{z^4}{(y^2+z^2)(y+z)}+\frac{x^4}{(z^2+x^2)(z+x)}\right)\)
\(=\frac{x^4-y^4}{(x^2+y^2)(x+y)}+\frac{y^4-z^4}{(y^2+z^2)(y+z)}+\frac{z^4-x^4}{(z^2+x^2)(z+x)}\)
\(=x-y+y-z+z-x=0\)
\(\Rightarrow \frac{x^4}{(x^2+y^2)(x+y)}+\frac{y^4}{(y^2+z^2)(y+z)}+\frac{z^4}{(z^2+x^2)(z+x)}=\frac{y^4}{(x^2+y^2)(x+y)}+\frac{z^4}{(y^2+z^2)(y+z)}+\frac{x^4}{(z^2+x^2)(z+x)}\)
Do đó:
\(2F=\frac{x^4+y^4}{(x^2+y^2)(x+y)}+\frac{y^4+z^4}{(y^2+z^2)(y+z)}+\frac{z^4+x^4}{(z^2+x^2)(z+x)}\)
\(\geq \frac{\frac{(x^2+y^2)^2}{2}}{(x^2+y^2)(x+y)}+\frac{\frac{(y^2+z^2)^2}{2}}{(y^2+z^2)(y+z)}+\frac{\frac{(z^2+x^2)^2}{2}}{(z^2+x^2)(z+x)}\) (áp dụng BĐT Cauchy)
hay \(2F\geq \frac{x^2+y^2}{2(x+y)}+\frac{y^2+z^2}{2(y+z)}+\frac{z^2+x^2}{2(z+x)}\)
Mà cũng theo BĐT Cauchy thì:
\(\frac{x^2+y^2}{2(x+y)}+\frac{y^2+z^2}{2(y+z)}+\frac{z^2+x^2}{2(z+x)}\geq \frac{\frac{(x+y)^2}{2}}{2(x+y)}+\frac{\frac{(y+z)^2}{2}}{2(y+z)}+\frac{\frac{(z+x)^2}{2}}{2(x+z)}=\frac{x+y+z}{2}=\frac{1}{2}\)
\(\Rightarrow 2F\geq \frac{1}{2}\Rightarrow F\geq \frac{1}{4}\)
Vậy \(F_{\min}=\frac{1}{4}\Leftrightarrow x=y=z=\frac{1}{3}\)