\(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)
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* Có BĐT : \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\) với $x,y>0$ ( Chứng minh bằng xét hiệu )

Ta có BĐT : \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\Rightarrow\dfrac{x+y}{x^2+y^2}\le\dfrac{2\left(x+y\right)}{\left(x+y\right)^2}=\dfrac{2}{x+y}\)

Chứng minh tương tự khi đó :

\(P\le\dfrac{2}{x+y}+\dfrac{2}{y+z}+\dfrac{2}{z+x}\)

\(\Rightarrow2P\le\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}=2.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=4032\)

\(\Rightarrow P\le2016\)

7 tháng 5 2018

nhân cả 2 vế với 2 rồi bunhia

6 tháng 4 2018

câu c là \(\dfrac{1}{2}\)(x+y+z) nhé, mih chép nhầm

29 tháng 12 2017

Nhân ra thôi

30 tháng 12 2017

\(A=\left(xy+yz+xz\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-xyz\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\\ =y+x+\dfrac{xy}{z}+y+z+\dfrac{yz}{x}+x+z+\dfrac{xz}{y}-\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\\ =2\left(x+y+z\right)=2.2018=4036\)

13 tháng 11 2017

1) \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}=0\)

\(\Leftrightarrow\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=0\)

\(\Leftrightarrow\dfrac{3}{x-3}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{x+3}=0\)

\(\Leftrightarrow\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+2.x.3+3^2}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{x+3}{x-3}=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy x=-3

13 tháng 11 2017

bạn ơi x ko thể bằng -3 đc vì

\(\dfrac{x}{x+3}=\dfrac{-3}{-3+3}=\dfrac{-3}{0}\) là sai

12 tháng 7 2017

\(P=\dfrac{1}{x^2+x}+\dfrac{1}{y^2+y}+\dfrac{1}{z^2+z}\)

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

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

Áp dụng BĐT \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) và BĐT Cauchy Shwarz dạng Engel, ta có:

\(P\ge\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{4}\left(\dfrac{1}{x}+1+\dfrac{1}{y}+1+\dfrac{1}{z}+1\right)\)

\(=\dfrac{3}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{4}\)

\(\ge\dfrac{3}{4}\left[\dfrac{\left(1+1+1\right)^2}{x+y+z}\right]-\dfrac{3}{4}=\dfrac{3}{4}\left(\dfrac{9}{3}-1\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi x = y = z = 1.

Min P = 1,5 <=> x = y = z = 1.

13 tháng 7 2017

T xài phương pháp chuẩn hóa thử, lên C3 có gặp mấy bài này chém dễ dàng, có sai thì đừng ném đá nha :vv.

Ta chứng minh BĐT sau:

\(\dfrac{1}{x^2+x}\ge-0,75x+1,25\) \(\forall x\in\left(0;1\right)\) ( Để ra cái BĐT này t dùng casio, ra cái này là ra hết bài :D )

Thật vậy: \(\dfrac{1}{x^2+x}+0,75x-1,25\ge0\)

\(\Rightarrow\dfrac{1+0,75x\left(x^2+x\right)-1,25\left(x^2+x\right)}{x^2+x}\ge0\)

\(\Rightarrow1+0,75x^3+0,75x^2-1,25x^2+1,25x\ge0\)

\(\Rightarrow0,75\left(x-1\right)^2\left(x+\dfrac{4}{3}\right)\ge0\) \(\forall x\in\left(0;1\right)\) (BĐT này luôn đúng)

Tương tự: \(\dfrac{1}{y^2+y}\ge-0,75y+1,25\)

\(\dfrac{1}{z^2+z}\ge-0,75z+1,25\)

Cộng vế theo vế các BĐT vừa chứng minh, ta được: \(P\ge-0,75\left(x+y+z\right)+1,25.3\)

\(P\ge1\)

Vậy Min P =1 khi x=y=z =1

17 tháng 7 2017

\(=>P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\sqrt{3}\)

CHÚC BẠN HỌC TỐT..........

3 tháng 7 2018

\(1.\) Giả sử : \(a\ge b\ge c\Rightarrow a+b\ge a+c\ge b+c\)

Ta có : \(\dfrac{c}{a+b}\le\dfrac{c}{b+c};\dfrac{b}{a+c}\le\dfrac{b}{b+c};\dfrac{a}{b+c}=\dfrac{a}{b+c}\)

\(\Rightarrow\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\le\dfrac{b+c}{b+c}+\dfrac{a}{b+c}=1+\dfrac{a}{b+c}< 1+1=2\left(đpcm\right)\)

\(2.\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

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

\(\Leftrightarrow\left(x+y+z\right)\left(xy+yz+xz\right)=xyz\)

\(\Leftrightarrow x^2y+x^2z+xy^2+y^2z+xyz+xyz+yz^2+xz^2=0\)

\(\Leftrightarrow xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)y\left(x+z\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz+xz\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=-z\\x=-z\end{matrix}\right.\)

+) Với : \(x=-y\) , ta có :

Đpcm \(\Leftrightarrow-\dfrac{1}{y^{2011}}+\dfrac{1}{y^{2011}}+\dfrac{1}{z^{2011}}=\dfrac{1}{-y^{2011}+y^{2011}+z^{2011}}\)

\(\Leftrightarrow\dfrac{1}{z^{2011}}=\dfrac{1}{z^{2011}}\left(luôn-đúng\right)\)

Tương tự với 2 TH còn lại .

\(\RightarrowĐCPM\)

12 tháng 12 2017

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)

\(\Rightarrow ayz+bxz+cxy=0\)

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

\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

12 tháng 12 2017

Có:

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)

\(\Rightarrow ayz+bxz+cxy=0\)

Có:

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

\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1^2\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(ayz+bxz+cxy=0\right)\)

22 tháng 8 2017

1)

\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)

\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)

22 tháng 8 2017

dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)

\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)

\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)