\(\left(1+y^2\right)\left(1+y^2\right)\)
\(=\left(1^2+y^2\right)\left[x^2+\left(-1\right)^2\right]\)
\(\ge\left[1.x+\left(-1\right).y\right]^2\)
\(=\left(x-y\right)^2\)
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21 tháng 1 2021
Ta có: \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2\left(xy+1\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy+2\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x+xy-y+2\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+xy-y-x^2-x-xy+y-2=0\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2x-2=0\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\\left(y+1\right)^2=\left(y-1\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y^2+2y+1=y^2-3y+2+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y^2+2y+1-y^2+3y-2-2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\3y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
21 tháng 1 2021
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2\left(xy+1\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy+2\\y^2+y-xy-x=y^2-2y+xy-2x-2xy\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-2x=2\\x+3y=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\-1+3y=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\3y=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy hpt trên có nghiệm duy nhất (x;y) = (-1; \(\dfrac{1}{3}\))
Chúc bn học tốt!
20 tháng 1
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2\left(xy+1\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy+2\\y^2+y-xy-x=y^2-2y+xy-2x-2xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2+xy-x-y=x^2+xy+x-y+2\\y^2+y-xy-x=y^2-xy-2y-2x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x=2\\y-x+2y+2x=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-1\\x+3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\3y=-x=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
15 tháng 12 2021
\(ĐK:x,y\in R\)
Từ 2 PT \(\Leftrightarrow\sqrt{\left(x+1\right)^2+\left(y-1\right)^2}=\sqrt{\left(x-5\right)^2+\left(y+1\right)^2}\)
\(\Leftrightarrow x^2+2x+y^2-2y+2=x^2-10x+y^2+2y+26\\ \Leftrightarrow12x-4y-24=0\\ \Leftrightarrow3x-y-6=0\\ \Leftrightarrow y=3x-6\)
Thay vào \(PT\left(1\right)\Leftrightarrow\sqrt{\left(x-1\right)^2+\left(3x-8\right)^2}=\sqrt{\left(x+1\right)^2+\left(3x-7\right)^2}\)
\(\Leftrightarrow10x^2-50x+65=10x^2-40x+50\\ \Leftrightarrow10x=15\Leftrightarrow x=\dfrac{3}{2}\Leftrightarrow y=-\dfrac{3}{2}\)
Vậy hệ có nghiệm \(\left(x;y\right)=\left(\dfrac{3}{2};-\dfrac{3}{2}\right)\)
7 tháng 6 2018
\(2\left(x^2+1\right)=\left(1+1\right)\left(x^2+1\right)>=\left(x+1\right)^2\)(bđt bunhiacopxki)(1)
\(2\left(y^2+1\right)=\left(1+1\right)\left(y^2+1\right)>=\left(y+1\right)^2\)(bđt bunhiacopxki)(2)
\(\left(x^2+1\right)\left(y^2+1\right)>=\left(xy+1\right)^2\)(3)
từ (1) (2) và (3)\(\Rightarrow\left(2\left(x^2+1\right)\left(y^2+1\right)\right)^2>=\left(\left(x+1\right)\left(y+1\right)\left(xy+1\right)\right)^2\)
\(\Rightarrow2\left(x^2+1\right)\left(y^2+1\right)>=\left(x+1\right)\left(y+1\right)\left(xy+1\right)\)
dấu = xảy ra khi x=y=1
21 tháng 11 2017
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
Mượn chỗ nhok chút !
ta có pt
<=>\(5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=2\left(x^2-x+1\right)+2\left(x+1\right)\)
đặt \(\sqrt{x+1}=a;\sqrt{x^2-x+1}=b\)
Ta có PT <=> \(5ab=2a^2+2b^2\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
đến đây thì dex rồi ^_^
Mượn chỗ nhok chút !
Áp dụng bđt svacxơ, ta có
\(M\ge\frac{\left(x^3+y^3+z^3\right)^2}{2\left(x^3+y^3+z^3\right)}=\frac{x^3+y^3+z^3}{2}\)
Áp dụng bài toán \(a^2+b^2+c^2\ge ab+bc+ca\) (dễ dàng chứng minh ) , ta có
\(x^3+y^3+z^3\ge xy\sqrt{xy}+yz\sqrt{yz}+zx\sqrt{zx}=1\)
=> \(M\ge\frac{1}{2}\)
dấu = xảy ra <=> x=y=z=\(\frac{1}{\sqrt[3]{3}}\)