Bài 1.

\(\left\{{}\begin{matrix}x-3y=5-2m\\2x+y=3\left(m+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=5-2m\\6x+3y=9m+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m+14\\x-3y=5-2m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\m+2-3y=5-2m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\-3y=-3m+3\end{matrix}\right.\)

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

\(x_0^2+y_0^2=9m\)

\(\Leftrightarrow\left(m+2\right)^2+\left(m-1\right)^2=9m\)

\(\Leftrightarrow m^2+4m+4+m^2-2m+1-9m=0\)

\(\Leftrightarrow2m^2-7m+5=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=\dfrac{5}{2}\end{matrix}\right.\) ( Vi-ét )

1: Để hệ có nghiệm duy nhất thì \(\dfrac{m}{m-1}\ne\dfrac{1}{-1}\ne-1\)

=>\(\dfrac{m+m-1}{m-1}\ne0\)

=>\(\dfrac{2m-1}{m-1}\ne0\)

=>\(m\notin\left\{\dfrac{1}{2};1\right\}\)(1)

\(\left\{{}\begin{matrix}mx+y=3\\\left(m-1\right)x-y=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}mx+\left(m-1\right)x=3+7\\mx+y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(2m-1\right)=10\\mx+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10}{2m-1}\\y=3-mx=3-\dfrac{10m}{2m-1}=\dfrac{6m-3-10m}{2m-1}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{10}{2m-1}\\y=\dfrac{-4m-3}{2m-1}\end{matrix}\right.\)

Để x và y trái dấu thì x*y<0

=>\(\dfrac{10}{2m-1}\cdot\dfrac{-4m-3}{2m-1}< 0\)

=>\(\dfrac{10\left(4m+3\right)}{\left(2m-1\right)^2}>0\)

=>4m+3>0

=>m>-3/4

Kết hợp (1), ta được: \(\left\{{}\begin{matrix}m>-\dfrac{3}{4}\\m\notin\left\{\dfrac{1}{2};1\right\}\end{matrix}\right.\)

2: Để x,y là số nguyên thì \(\left\{{}\begin{matrix}10⋮2m-1\\-4m-3⋮2m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2m-1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\\-4m+2-5⋮2m-1\end{matrix}\right.\)

=>\(2m-1\in\left\{1;-1;5;-5\right\}\)

=>\(2m\in\left\{2;0;6;-4\right\}\)

=>\(m\in\left\{1;0;3;-2\right\}\)

Kết hợp (1), ta được: \(m\in\left\{0;3;-2\right\}\)

Bài 1:

Để hpt đã cho vô nghiệm thì m = 1 (lật sách trang 25 là hiểu)

Bài 2 :

Để hpt đã cho có vô số nghiệm thì m = 1

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 \(x=\sqrt{2013}\)

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

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