\(1;\left\{{}\begin{matrix}mx+2y=7\\2x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7-mx}{2}\\2x+\dfrac{3\left(7-mx\right)}{2}=5\left(1\right)\end{matrix}\right.\) 

\(hệ\) \(pt\) \(có\) \(nghiệm\) \(duy\) \(nhất\Leftrightarrow\left(1\right)có\) \(ngo\) \(duy\) \(nhất\)

\(\left(1\right)\Leftrightarrow\dfrac{4x+3\left(7-mx\right)}{2}=5\Leftrightarrow4x+21-3mx=10\Leftrightarrow x\left(4-3m\right)=-11\)

\(với:m\ne\dfrac{4}{3}\) \(thì\) \(hpt\) \(có\) \(ngo\) \(duy-nhất\left(x;y\right)=\left\{\dfrac{-11}{4-3m};\dfrac{7-m\left(\dfrac{-11}{4-3m}\right)}{2}\right\}\)

\(2,\left\{{}\begin{matrix}2x-y=m\\-4x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x-m\\-4x+2\left(2x-m\right)=4\left(1\right)\end{matrix}\right.\)

hệ pt vô nghiệm khi (1) vô nghiệm

(1)\(\Leftrightarrow-4x+4x-2m=4\Leftrightarrow m=-2\Rightarrow m=-2\)

thì hệ pt có vô số nghiệm

\(\Rightarrow m\ne-2\) thì hpt vô nghiệm

Để hệ có nghiệm duy nhất thì \(\dfrac{m}{2m}\ne\dfrac{1}{3}\)

=>\(\dfrac{1}{2}\ne\dfrac{1}{3}\)(luôn đúng)

\(\left\{{}\begin{matrix}mx+y=5\\2mx+3y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2mx+2y=10\\2mx+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y=4\\mx+y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-4\\mx=5-y=5-\left(-4\right)=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-4\\x=\dfrac{9}{m}\end{matrix}\right.\)

\(\left(2m-1\right)\cdot x+\left(m+1\right)\cdot y=m\)

=>\(\dfrac{9}{m}\left(2m-1\right)+\left(m+1\right)\cdot\left(-4\right)=m\)

=>\(\dfrac{9\left(2m-1\right)}{m}=m+4m+4=5m+4\)

=>m(5m+4)=18m-9

=>\(5m^2-14m+9=0\)

=>(m-1)(5m-9)=0

=>\(\left[{}\begin{matrix}m=1\\m=\dfrac{9}{5}\end{matrix}\right.\)

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 )

để hệ (1) lsao hả bạn

- Với \(m=0\) hệ có nghiệm (vô số nghiệm)

- Với \(m=\dfrac{1}{2}\) hệ có nghiệm

Hệ phương trình \(\left\{{}\begin{matrix}ax+by=c\\a'x+b'y=c'\end{matrix}\right.\) vô nghiệm khi \(\dfrac{a}{a'}=\dfrac{b}{b'}\ne\dfrac{c}{c'}\)

- Với \(m\ne\left\{\dfrac{1}{2};0\right\}\) , xét điều kiện: \(\dfrac{a}{a'}=\dfrac{b}{b'}\)

Hay \(\dfrac{m^2}{m}=\dfrac{2-m}{2m-1}\Leftrightarrow m=\dfrac{2-m}{2m-1}\)

\(\Rightarrow m^2=1\Rightarrow\left[{}\begin{matrix}m=1\\m=-1\end{matrix}\right.\)

+ Với \(m=1\Rightarrow\dfrac{m^2}{m}=\dfrac{2-m}{2m-1}=1\ne\dfrac{m^3+4}{m^5-2}=-5\)  thỏa mãn hệ vô nghiệm

+ Với \(m=-1\) \(\Rightarrow\dfrac{m^2}{m}=\dfrac{2-m}{2m-1}=-1=\dfrac{m^3+4}{m^5-2}=-1\) ko thỏa mãn

Vậy \(m=1\) thì hệ vô nghiệm

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 \(x=\sqrt{2013}\)

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