\(\left\{{}\begin{matrix}\sqrt{y}\left(\sqrt{x+3}+\sqrt{x}\right)=3\left(1\right)\\\sqrt{x}+\sqrt{y}=x+1\left(2\right)\end{matrix}\right.\)

ĐKXĐ: x;y≥0

Ta thấy: \(x+3\ne x\Rightarrow\sqrt{x+3}-\sqrt{x}\ne0\)

=>(1)<=>\(\frac{3\sqrt{y}}{\sqrt{x+3}-\sqrt{x}}=3\Leftrightarrow\sqrt{x}+\sqrt{y}=\sqrt{x+3}\). Thay vào (2):

\(\sqrt{x+3}=x+1\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+x-2=0\end{matrix}\right.\Leftrightarrow x=1\) (tm đkxđ)

Thay vào (1) => y=1 (tm đkxđ)

Vậy hệ có nghiệm (x;y)\(\in\){(1;1)}

\(\sqrt{x+y}+\sqrt{x-y}=1+\sqrt{\left(x-y\right)\left(x+y\right)}.\\ \left(\sqrt{x+y}-1\right)\left(\sqrt{x-y}-1\right)=0.\)

Chắc bạn cũng biết phải làm gì :))

Câu 1:

\(ĐK:x\ge2\)

Áp dụng BĐT cauchy ta có:

\(\left(x+1\right)+4\ge2\sqrt{4\left(x+1\right)}=4\sqrt{x+1}\\ \Leftrightarrow2\sqrt{x+1}\le\dfrac{x+5}{2}\)

Ta có \(\left(x-2\right)+1\ge2\sqrt{x-2}\Leftrightarrow\sqrt{x-2}\le\dfrac{x-1}{2}\)

\(\Leftrightarrow P\le\dfrac{x+5}{2}+\dfrac{x-1}{2}-x+2013=x+2-x+2013=2015\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x-2=1\end{matrix}\right.\Leftrightarrow x=3\)

Câu 2:

\(HPT\Leftrightarrow\left\{{}\begin{matrix}10\sqrt{x}+15y^3=140\\4y^3-10\sqrt{x}=12\end{matrix}\right.\left(x\ge0\right)\\ \Leftrightarrow19y^3=152\\ \Leftrightarrow y^3=8\Leftrightarrow y=2\\ \Leftrightarrow2\sqrt{x}+24=28\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

Vậy \(\left(x;y\right)=\left(4;2\right)\)

Câu 3:

\(HPT\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\my+2m+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=\dfrac{3-2m}{m+1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{m+1}\\x=\dfrac{3-2m}{m+1}\end{matrix}\right.\\ \Leftrightarrow xy=\dfrac{5\left(3-2m\right)}{\left(m+1\right)^2}\)

Đặt \(xy=t\)

\(\Leftrightarrow m^2t+2mt+t=15-10m\\ \Leftrightarrow m^2t+2m\left(t+5\right)+t-15=0\)

PT có nghiệm nên \(\Delta'=\left(t+5\right)^2-t\left(t-15\right)\ge0\)

\(\Leftrightarrow10t+25+15t\ge0\Leftrightarrow t\ge-1\)

Vậy \(xy_{min}=-1\Leftrightarrow\dfrac{5\left(2m-3\right)}{\left(m+1\right)^2}=1\Leftrightarrow m^2-8m+16=0\Leftrightarrow m=4\)

Câu 4: \(a^2+b^2=4a+bc+540\)

c đâu ra vậy?

Câu 5:

Thay \(x=3\Leftrightarrow P\left(2\right)+2P\left(2\right)=3^2\Leftrightarrow P\left(2\right)=3\)

Thay \(x=\sqrt{2013}\)

\(\Leftrightarrow P\left(\sqrt{2013}-1\right)+2P\left(2\right)=\left(\sqrt{2013}\right)^2=2013\\ \Leftrightarrow P\left(\sqrt{2013}-1\right)+6=2013\\ \Leftrightarrow P\left(\sqrt{2013}-1\right)=2007\)

Bài 2: 

a) Thay m=3 vào hệ pt, ta được:

\(\left\{{}\begin{matrix}x-2y=7\\2x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=14\\2x+y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-5y=5\\x-2y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=7+2y=5\end{matrix}\right.\)

Vậy: Khi m=3 thì hệ phương trình có nghiệm duy nhất là (x,y)=(5;-1)

nhanh đy mằ =(

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge2\\y\ge-\dfrac{1}{2}\end{matrix}\right.\)

Ta có \(\left\{{}\begin{matrix}x+2y-1-2\sqrt{2xy+x-4y-2}=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)+\left(2y+1\right)-2\sqrt{\left(x-2\right)\left(2y+1\right)}=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x-2}-\sqrt{2y+1}\right)^2=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\4\sqrt{2y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\2y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)