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)

\(\left\{{}\begin{matrix}x+y=4\\2x+3y=m\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x+3y=12\\2x+3y=m\end{matrix}\right.\)

trừ 2 vế của pt cho nhau ta tìm được

\(\left\{{}\begin{matrix}x=12-m\\y=m-8\end{matrix}\right.\)

để \(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m< 12\\m< 8\end{matrix}\right.\Rightarrow}m< 8}\)

Vì \(\frac31<>\frac{-2}{-1}\left(3<>2\right)\)

nên hệ luôn có nghiệm duy nhất

\(\begin{cases}3x-2y=2m^2-3\\ x-y=3\end{cases}\Rightarrow\begin{cases}3x-2y=2m^2-3\\ 2x-2y=6\end{cases}\)

=>\(\begin{cases}3x-2y-2x+2y=2m^2-3-6\\ x-y=3\end{cases}\Rightarrow\begin{cases}x=2m^2-9\\ y=x-3=2m^2-9-3=2m^2-12\end{cases}\)

\(x^2-y^2=-15\)

=>(x-y)(x+y)=-15

=>\(\left(2m^2-9-2m^2+12\right)\left(2m^2-9+2m^2-12\right)=-15\)

=>\(3\left(4m^2-21\right)=-15\)

=>\(4m^2-21=-5\)

=>\(4m^2=16\)

=>\(m^2=4\)

=>m=2 hoặc m=-2

Bài 2:

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

=>-4x-2y=3 và 8x+2y=-2

=>x=1/4; y=-2

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)

=>y=6 và x-2=5/4

=>x=13/4; y=6

c: =>x+y=24 và 3x+y=78

=>-2x=-54 và x+y=24

=>x=27; y=-3

d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)

=>y+2=1 và x-1=25

=>x=26; y=-1

Để hệ có nghiệm duy nhất thì \(\frac11<>\frac{-1}{-3}\) (luôn đúng)

=>Hệ luôn có nghiệm duy nhất

\(\begin{cases}x-y=4m+8\\ x-3y=6-2m^2\end{cases}\Rightarrow\begin{cases}x-y-x+3y=4m+8-6+2m^2\\ x-y=4m+8\end{cases}\)

=>\(\begin{cases}2y=2m^2+4m+2\\ x=y+4m+8\end{cases}\Rightarrow\begin{cases}y=m^2+2m+1=\left(m+1\right)^2\\ x=m^2+2m+1+4m+8=m^2+6m+9=\left(m+3\right)^2\end{cases}\)

\(\sqrt{x}+\sqrt{y}=8\)

=>\(\sqrt{\left(m+1\right)^2}+\sqrt{\left(m+3\right)^2}=8\)

=>|m+1|+|m+3|=8

=>m+1+m+3=8(Do m>0)

=>2m+4=8

=>2m=4

=>m=2(nhận)

Để hệ có nghiệm duy nhất thì \(\frac12<>\frac{3}{-1}\)

=>0,5<>-3(đúng)

=>Hệ luôn có nghiệm duy nhất

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

=>\(\begin{cases}2x-6y=10-4m\\ 2x+y=3\left(m+1\right)\end{cases}\Rightarrow\begin{cases}2x+y-2x+6y=3\left(m+1\right)+4m-10\\ x-3y=5-2m\end{cases}\)

=>\(\begin{cases}7y=3m+3+4m-10\\ x=3y+5-2m\end{cases}\Rightarrow\begin{cases}y=m-1\\ x=3\left(m-1\right)+5-2m=3m-3+5-2m=m+2\end{cases}\)

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

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

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

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

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

=>(m-1)(2m-5)=0

=>m=1 hoặc m=5/2

Ta có: \(\left\{{}\begin{matrix}3x+y=2m+9\\x+y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x+5-x=2m+9\\y=5-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=2m+4\\y=5-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\y=5-m-2\end{matrix}\right.\)

Gọi A=xy+x-1, ta có: \(A=\left(m+2\right)\left(5-m-2\right)+m+2-1\)

\(A=\left(m+2\right)\left(3-m\right)+m+1\)

\(A=-m^2+m+6+m+1\)

\(A=-m^2+2m+7=-\left(m-1\right)^2+8\)

\(A_{max}=7\Leftrightarrow m=1\) Khi đó x=3, y=2

thay m=2 vào HPT ta có
\(\left\{{}\begin{matrix}x+2y=2+1\\2x+y=2.2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=3\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+4y=6\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3y=2\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
vậy ..........
 

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