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\(x^2+y^2=3\frac{1}{3}xy\)hay \(x^2+y^2=\frac{10}{3}xy\)
\(\Rightarrow x^2+2xy+y^2=\frac{16}{3}xy\)\(\Rightarrow\left(x+y\right)^2=\frac{16}{3}xy\)
tương tự : \(\left(x-y\right)^2=\frac{4}{3}xy\)
\(\Rightarrow\frac{\left(x-y\right)^2}{\left(x+y\right)^2}=\frac{1}{4}\Rightarrow\orbr{\begin{cases}\frac{x-y}{x+y}=\frac{1}{2}\\\frac{x-y}{x+y}=\frac{-1}{2}\end{cases}}\)
vì x > y > 0 nên x - y > 0 \(\Rightarrow\frac{x-y}{x+y}>0\)
Vậy \(\frac{x-y}{x+y}=\frac{1}{2}\)
Xét\(x^2+2xy+y^2=\frac{10}{3}xy+2xy=\frac{16}{3}xy\)
\(x^2-2xy+y^2=\frac{10}{3}xy-2xy=\frac{4}{3}xy\)
Từ đó ta được:
\(\frac{\left(x-y\right)^2}{\left(x+y\right)^2}=\frac{\left(\frac{4}{3}xy\right)}{\left(\frac{16}{3}xy\right)}=\frac{1}{4}\)
\(\Rightarrow\sqrt{\frac{\left(x-y\right)^2}{\left(x+y\right)^2}}=\frac{1}{2}\Rightarrow\left|\frac{x-y}{x+y}\right|=\frac{1}{2}\)
Hihi
đến đây bạn tự làm nốt nha
^-^ Học tốt
\(\frac{16}{\sqrt{x-6}}+\frac{4}{\sqrt{y-2}}+\frac{256}{\sqrt{z-1750}}+\sqrt{x-6}+\sqrt{y-2}+\sqrt{z-1750}=44\) (Điều kiện xác định : \(x>6;y>2;z>1750\))
\(\Leftrightarrow\left(\sqrt{x-6}+\frac{16}{\sqrt{x-6}}-8\right)+\left(\sqrt{y-2}+\frac{4}{\sqrt{y-2}}-4\right)+\left(\sqrt{z-1750}+\frac{256}{\sqrt{z-1750}}-32\right)=0\)
\(\Leftrightarrow\frac{\left(x-6\right)-8\sqrt{x-6}+16}{\sqrt{x-6}}+\frac{\left(y-2\right)-4\sqrt{y-2}+4}{\sqrt{y-2}}+\frac{\left(z-1750\right)-32\sqrt{z-1750}+256}{\sqrt{z-1750}}=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-6}-4\right)^2}{\sqrt{x-6}}+\frac{\left(\sqrt{y-2}-2\right)^2}{\sqrt{y-2}}+\frac{\left(\sqrt{z-1750}-16\right)^2}{\sqrt{z-1750}}=0\)
Vì \(\frac{\left(\sqrt{x-6}-4\right)^2}{\sqrt{x-6}}\ge0\) , \(\frac{\left(\sqrt{y-2}-2\right)^2}{\sqrt{y-2}}\ge0\) , \(\frac{\left(\sqrt{z-1750}-16\right)^2}{\sqrt{z-1750}}\ge0\) với mọi x>6 , y>2 , z>1750 nên phương trình trên tương đương với :
\(\begin{cases}\frac{\left(\sqrt{x-6}-4\right)^2}{\sqrt{x-6}}=0\\\frac{\left(\sqrt{y-2}-2\right)^2}{\sqrt{y-2}}=0\\\frac{\left(\sqrt{z-1750}-16\right)^2}{\sqrt{z-1750}}=0\end{cases}\) \(\Leftrightarrow\begin{cases}\left(\sqrt{x-6}-4\right)^2=0\\\left(\sqrt{y-2}-2\right)^2=0\\\left(\sqrt{z-1750}-16\right)^2=0\end{cases}\) \(\Leftrightarrow\begin{cases}x=22\\y=6\\z=2006\end{cases}\) (TMĐK)
Vậy (x;y;z) = (22;6;2006)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{x}{1+y+xz}=\frac{x\left(x^2+y+\frac{z}{x}\right)}{\left(1+y+xz\right)\left(x^2+y+\frac{z}{x}\right)}\le\frac{x^3+xy+z}{\left(x+y+z\right)^2}\)
\(\le\frac{x+y+z}{\left(x+y+z\right)}=\frac{1}{x+y+z}\)
Tương tự ta cũng có: \(\frac{y}{1+z+xy}\le\frac{1}{x+y+z};\frac{z}{1+x+yz}\le\frac{1}{x+y+z}\)
Cộng theo vế ta có: \(\frac{x}{1+y+xz}+\frac{y}{1+z+xy}+\frac{z}{1+x+yz}\le\frac{1+1+1}{x+y+z}=\frac{3}{x+y+z}\)
a)\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
\(\Leftrightarrow2c+2\sqrt{\left(a+c\right)\left(b+c\right)}=0\)
\(\Leftrightarrow\sqrt{\left(a+c\right)\left(b+c\right)}=-c\)
\(\Leftrightarrow\begin{cases}c< 0\\ab+bc+ca+c^2=c^2\end{cases}\)\(\Leftrightarrow ab+bc+ca=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}=0\)
Đpcm
a) \(x^4-3x^3-x+3=x^4-x^3-2x^3+2x^2-2x^2+2x-3x+3\)
\(=x^3\left(x-1\right)-2x^2\left(x-1\right)-2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x^3-2x^2-2x-3\right)\left(x-1\right)=\left(x^3+x^2+x-3x^2-3x-3\right)\left(x-1\right)\)
\(=\left(x\left(x^2+x+1\right)-3\left(x^2+x+1\right)\right)\left(x+1\right)=\left(x^2+x+1\right)\left(x-3\right)\left(x-1\right)\)
b) \(x^2y^2\left(y-x\right)+y^2z^2\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=-x^2y^2\left(x-y\right)+y^2z^2\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=\left(y^2z^2-x^2y^2\right)\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z^2-x^2\right)\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z+x\right)\left(z-x\right)\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=\left(y^2\left(z+x\right)\left(x-y\right)-z^2x^2\right)\left(z-x\right)\)
c) câu này đề có sai o bn
hình như đề là : \(4x^2+4x^2y-8y^2\) mới đúng chứ ?? ?
! Bài khó quá à!
Thank you very much!!!!