Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
câu này mik vừa làm sáng ngày ne
ta đặt \(\sqrt{x^2-2014}=a;\sqrt{y^2-2014}=b;\sqrt{z^2-2014}=c\)
ta có \(ab+bc+ca=2014\Rightarrow ab+bc+ca+a^2=x^2-2014+2014=x^2\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)=x^2\)
tương tự ta có \(\left(b+c\right)\left(b+a\right)=y^2;\left(c+a\right)\left(c+b\right)=z^2\)
nhân cả 3 vào ta có \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=xyz\)
=> \(\hept{\begin{cases}\left(a+b\right)z^2=xyz\\\left(b+c\right)x^2=xyz\\\left(c+a\right)y^2=xyz\end{cases}\Rightarrow\hept{\begin{cases}a+b=\frac{xy}{z}\\b+c=\frac{yz}{x}\\c+a=\frac{zx}{y}\end{cases}}}\)
cậu nhân tung A ra rồi thay \(\frac{xy}{z};\frac{yz}{x};\frac{zx}{y}\) như vừa tính vào thì cậu sẽ ra kết quả là A=4028
Ta có :
\(\left(1.x+9.\frac{1}{y}\right)^2\le\left(1^2+9^2\right)\left(x^2+\frac{1}{y^2}\right)\Rightarrow\sqrt{x^2+\frac{1}{y^2}}\ge\frac{1}{\sqrt{82}}\left(x+\frac{9}{y}\right)\)
tương tự : \(\sqrt{y^2+\frac{1}{z^2}}\ge\frac{1}{\sqrt{82}}.\left(y+\frac{9}{z}\right)\); \(\sqrt{z^2+\frac{1}{x^2}}\ge\frac{1}{\sqrt{82}}.\left(z+\frac{9}{x}\right)\)
\(\Rightarrow\sqrt{x^2+\frac{1}{y^2}}+\sqrt{y^2+\frac{1}{z^2}}+\sqrt{z^2+\frac{1}{x^2}}\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\right)\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{81}{x+y+z}\right)\)
\(=\frac{1}{\sqrt{82}}\left[\left(x+y+z+\frac{1}{x+y+z}\right)+\frac{80}{x+y+z}\right]\ge\sqrt{82}\)
2a)với a,b,c là các số thực ta có
\(a^2-ab+b^2=\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2\ge\frac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\sqrt{a^2-ab+b^2}\ge\sqrt{\frac{1}{4}\left(a+b\right)^2}=\frac{1}{2}\left|a+b\right|\)
tương tự \(\sqrt{b^2-bc+c^2}\ge\frac{1}{2}\left|b+c\right|\)
tương tự \(\sqrt{c^2-ca+a^2}\ge\frac{1}{2}\left|a+c\right|\)
cộng từng vế mỗi BĐT ta được \(\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\ge\frac{2\left(a+b+c\right)}{2}=a+b+c\)
dấu "=" xảy ra khi và chỉ khi a=b=c
Bài 2:Áp dụng BĐT AM-GM ta có:
\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}\)
\(\frac{1}{y}+\frac{1}{z}\ge2\sqrt{\frac{1}{yz}}\)
\(\frac{1}{x}+\frac{1}{z}\ge2\sqrt{\frac{1}{xz}}\)
CỘng theo vế 3 BĐT trên có:
\(2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge2\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\)
Khi x=y=z
Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)
\(..........................\)
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
Cộng theo vế ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{100}{10}=10\)
Đặt \(\sqrt{x-2014}=a;\sqrt{y-2015}=b;\sqrt{z=2016}=c\)(với a,b,c>0). Khi đó pt trở thành:
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2=0\Leftrightarrow a=b=c=2\)
\(\Rightarrow x=2018;y=2019;z=2020\)
\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)
\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}-\left(\frac{1}{x-2014+y-2015+z-2016}\right)=\frac{3}{4}\)
\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}+0=\frac{3}{4}\)
\(\frac{\sqrt{x}-\sqrt{2014}}{x-2014}+\frac{\sqrt{y}-\sqrt{2015}}{y-2015}+\frac{\sqrt{z}-\sqrt{2016}}{z-2016}=\frac{3}{4}\)
\(x=2018,y=2019,z=2020\)
Đặt \(\sqrt{x-2013}=a\left(a>0\right)\)
\(\sqrt{y-2014}=b\left(b>0\right)\)
\(\sqrt{z-2015}=c\left(c>0\right)\)
Có \(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
<=> \(\frac{a-1}{a^2}-\frac{1}{4}+\frac{b-1}{b^2}-\frac{1}{4}+\frac{c-1}{c^2}-\frac{1}{4}=0\)
<=> \(\frac{4a-4-a^2}{4.a^2}+\frac{4b-4-b^2}{4b^2}+\frac{4c-4+c^2}{4c^2}=0\)
<=>\(\frac{-\left(a^2-4a+4\right)}{4a^2}-\frac{b^2-4b+4}{4b^2}-\frac{c^2-4c+4}{4c^2}=0\)
<=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}=0\).
Có \(\frac{\left(a-2\right)^2}{4a^2}\ge0\forall a>0\)
\(\frac{\left(b-2\right)^2}{4b^2}\ge0\forall b>0\)
\(\frac{\left(c-2\right)^2}{4c^2}\ge0\forall c>0\)
=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}\ge0\) với moi a,b,c >0
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}\sqrt{x-2013}=2\\\sqrt{y-2014}=2\\\sqrt{z-2015}=2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x-2013=4\\y-2014=4\\z-2015=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)(t/m)
Vậy \(\left(x,y,z\right)\in\left\{\left(2017,2018,2019\right)\right\}\)
Bài 2 : đã cm bên kia
Bài 1: :|
we had điều này:
\(2=\frac{2014}{x}+\frac{2014}{y}+\frac{2014}{z}\)
\(\Leftrightarrow\frac{x-2014}{x}+\frac{y-2014}{y}+\frac{z-204}{z}=1\)
Xòng! bunyakovsky
P/s : Bệnh lười kinh niên tái phát nên ít khi ol sorry :<