Ta có: \(\left\{{}\begin{matrix}x+my=3\\mx+4y=6\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}y\left(m^2-4\right)=3m-6\\mx+4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3m-6}{m^2-4}\\mx=6-4y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3\left(m-2\right)}{\left(m+2\right)\left(m-2\right)}=\dfrac{3}{m+2}\\mx=6-4\cdot\dfrac{3}{m+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{m+2}\\mx=6-\dfrac{12}{m+2}=\dfrac{6\left(m+2\right)-12}{m+2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{m+2}\\mx=\dfrac{6m+12-12}{m+2}=\dfrac{6m}{m+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{6m}{m+2}:m=\dfrac{6m}{m+2}\cdot\dfrac{1}{m}=\dfrac{6}{m+2}\\y=\dfrac{3}{m+2}\end{matrix}\right.\)

Để phương trình có nghiệm x>1 và y>0 thì \(\left\{{}\begin{matrix}\dfrac{6}{m+2}>1\\\dfrac{3}{m+2}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{m+2}-1>0\\m+2>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{m+2}-\dfrac{m+2}{m+2}>0\\m>-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6-m-2}{m+2}>0\\m>-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4-m>0\\m>-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-m>-4\\m>-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\m>-2\end{matrix}\right.\Leftrightarrow-2< m< 4\)

Vậy: Để hệ phương trình có nghiệm x>1 và y>0 thì -2<m<4

Đáp án A

a: ĐKXĐ: x<>0; y<>0

\(\begin{cases}\frac{1}{x}+\frac{1}{y}=\frac12\\ \frac{3}{x}-\frac{4}{y}=-1\end{cases}\Rightarrow\begin{cases}\frac{1}{y}=\frac12-\frac{1}{x}\\ \frac{3}{x}-4\left(\frac12-\frac{1}{x}\right)=-1\end{cases}\)

=>\(\begin{cases}\frac{1}{y}=\frac12-\frac{1}{x}\\ \frac{3}{x}-2+\frac{4}{x}=-1\end{cases}\Rightarrow\begin{cases}\frac{7}{x}=1\\ \frac{1}{y}=\frac12-\frac{1}{x}\end{cases}\Rightarrow\begin{cases}x=7\\ \frac{1}{y}=\frac12-\frac17=\frac{5}{14}\end{cases}\)

=>x=7(nhận ) và y=14/5(nhận)

b: ĐKXĐ: 2x-y<>0 và x+y<>0

\(\begin{cases}\frac{3}{2x-y}-\frac{6}{x+y}=-1\\ \frac{1}{2x-y}-\frac{1}{x+y}=0\end{cases}\Rightarrow\begin{cases}\frac{1}{2x-y}=\frac{1}{x+y}\\ \frac{3}{x+y}-\frac{6}{x+y}=1\end{cases}\)

=>\(\begin{cases}2x-y=x+y\\ \frac{-3}{x+y}=1\end{cases}\Rightarrow\begin{cases}2x-y=x+y\\ x+y=-3\end{cases}\)

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

=>x=-2(nhận) và y=-3-(-2)=-3+2=-1(nhận)

c: ĐKXĐ: x<>-1 và y<>3

\(\begin{cases}\frac{5x}{x+1}+\frac{y}{y-3}=27\\ \frac{2x}{x+1}-\frac{3y}{y-3}=4\end{cases}\Rightarrow\begin{cases}\frac{y}{y-3}=27-\frac{5x}{x+1}\\ \frac{2x}{x+1}-3\left(27-\frac{5x}{x+1}\right)=4\end{cases}\)

=>\(\begin{cases}\frac{2x}{x+1}-81+\frac{15x}{x+1}=4\\ \frac{y}{y-3}=27-\frac{5x}{x+1}\end{cases}\Rightarrow\begin{cases}\frac{17x}{x+1}=85\\ \frac{y}{y-3}=27-\frac{5x}{x+1}\end{cases}\)

=>\(\begin{cases}\frac{x}{x+1}=5\\ \frac{y}{y-3}=27-5\cdot5=2\end{cases}\Rightarrow\begin{cases}5x+5=x\\ 2y-6=y\end{cases}\)

=>4x=-5 và y=6

=>x=-5/4(nhận) và y=6(nhận)

d: ĐKXĐ: x<>-2; y<>0

\(\begin{cases}\frac{7}{x+2}+\frac{3}{y}=2\\ \frac{4}{x+2}-\frac{1}{y}=\frac52\end{cases}\Rightarrow\begin{cases}\frac{1}{y}=\frac{4}{x+2}-\frac52\\ \frac{7}{x+2}+3\left(\frac{4}{x+2}-\frac52\right)=2\end{cases}\)

=>\(\begin{cases}\frac{1}{y}=\frac{4}{x+2}-\frac52\\ \frac{19}{x+2}-\frac{15}{2}=2\end{cases}\Rightarrow\begin{cases}\frac{19}{x+2}=\frac{19}{2}\\ \frac{1}{y}=\frac{4}{x+2}-\frac52\end{cases}\)

=>\(\begin{cases}x+2=2\\ \frac{1}{y}=\frac42-\frac52=\frac{-1}{2}\end{cases}\Rightarrow\begin{cases}x=0\\ y=-2\end{cases}\) (nhận)

Đặt \(x+\dfrac{1}{x}=a;y+\dfrac{1}{y}=b\left(\left|a\right|\ge2;\left|b\right|\ge2\right)\)

\(\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\x^3+y^3+\dfrac{1}{x^3}+\dfrac{1}{y^3}=15m-25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x^3+\dfrac{1}{x^3}\right)+\left(y^3+\dfrac{1}{y^3}\right)=15m-25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)^3-3\left(y+\dfrac{1}{y}\right)=15m-25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3-3\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)=15m-25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\a^3+b^3=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\\left(a+b\right)^3-3ab\left(a+b\right)=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\125-15ab=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\ab=9-m\end{matrix}\right.\)

\(\Rightarrow a,b\) là nghiệm của phương trình \(t^2-5t+9-m=0\left(1\right)\)

a, Nếu \(m=3\), phương trình \(\left(1\right)\) trở thành

\(t^2-5t+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y+\dfrac{1}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\y^2-3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3\pm\sqrt{5}}{2}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=3\\y+\dfrac{1}{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3\pm\sqrt{5}}{2}\\y=1\end{matrix}\right.\)

Vậy ...

b, \(\left(1\right)\Leftrightarrow t=\dfrac{5\pm\sqrt{4m-11}}{2}\left(m\ge\dfrac{11}{4}\right)\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5\pm\sqrt{4m-11}}{2}\\b=\dfrac{5\mp\sqrt{4m-11}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{5\pm\sqrt{4m-11}}{2}\\y+\dfrac{1}{y}=\dfrac{5\mp\sqrt{4m-11}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-\left(5\pm\sqrt{4m-11}\right)+2=0\left(2\right)\\2y^2-\left(5\mp\sqrt{4m-11}\right)+2=0\end{matrix}\right.\)

Yêu cầu bài toán thỏa mãn khi phương trình \(\left(2\right)\) có nghiệm dương

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(5\pm\sqrt{4m-11}\right)^2-16\ge0\\\dfrac{5\pm\sqrt{4m-11}}{2}>0\\1>0\end{matrix}\right.\)

\(\Leftrightarrow...\)

Ta có x + y + z = 0 

<=> (x + y + z)² = 0

<=> \(x^2+y^2+z^2+2xy+2yz+2zx=0\)

\(\Leftrightarrow xy+yz+zx=-3\) (vì x² + y² + z² = 6)

\(\Leftrightarrow x\left(y+z\right)+yz=-3\)

\(\Leftrightarrow-x^2+yz=-3\Leftrightarrow yz=x^2-3\) (vì x + y + z = 0)

Khi đó \(x^3+y^3+z^3=x^3+(y+z).(y^2+z^2-yz)\)

\(=x^3-x.[6-x^2-(x^2-3)]\)

\(=x^3-x.(9-2x^2)=3x^3-9x=6\)

Ta được \(\Leftrightarrow x^3-3x-2=0\Leftrightarrow(x^3+1)-3(x+1)=0\)

\(\Leftrightarrow(x+1)(x^2-x-2)=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(x-2\right)=0\)

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

Với x = -1 ta có hệ \(\left\{{}\begin{matrix}y+z=1\\y^2+z^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-z\\(1-z)^2+z^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-z\\z^2-z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-z\\\left[{}\begin{matrix}z=-1\\z=2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2\\z=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\z=2\end{matrix}\right.\end{matrix}\right.\)

Với x = 2 ta có hệ : \(\left\{{}\begin{matrix}y+z=-2\\y^2+z^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2-z\\(-2-z)^2+z^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2-z\\z^2+2z+1=0\end{matrix}\right.\)

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

Vậy (x;y;z) = (2;-1;-1) ; (-1 ; 2 ; -1) ; (-1 ; -1 ; 2)

em cảm ơn ạ

Bài 1:

ĐKXĐ: \(-x^2+2x+3\ge0\)

=>\(x^2-2x-3\le0\)

=>(x-3)(x+1)<=0

=>-1<=x<=3

Đặt a=x+1

=>x=a-1

Phương trình ban đầu sẽ trở thành:

\((a-1)^3 + (a-1) + 6 = 2a\sqrt{3 + 2(a-1) - (a-1)^2}\)

=>\(a^3 - 3a^2 + 4a + 4 = 2a\sqrt{4a - a^2} \quad (1)\)

\(2a\sqrt{4a-a^2}\le a^2+\left(\sqrt{4a-a^2}\right)^2=a^2+4a-a^2=4a\)

Dấu '=' xảy ra khi \(a=\sqrt{4a-a^2}\)

=>\(4a-a^2=a^2\)

=>\(2a^2-4a=0\)

=>2a(a-2)=0

=>a=0(loại) hoặc a=2(nhận)

\(\left(a-2\right)^2\cdot\left(a+1\right)>=0\forall a\) >0

=>\(\left(a^2-4a+4\right)\left(a+1\right)>=0\)

=>\(a^3+a^2-4a^2-4a+4a+4>=0\)

=>\(a^3-3a^2+4>=0\)

=>\(a^3-3a^2+4+4a\ge4a\)

Dấu '=' xảy ra khi a=2

Do đó, ta có: \(VT=a^3-3a^2+4a+4\ge4a\ge2a\sqrt{4a - a^2}=VP\)

Dấu '=' xảy ra khi a=2

=>x=a-1=2-1=1(nhận)

Để hệ có nghiệm duy nhất thì \(\frac{m}{4}<>\frac{1}{m}\)

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

=>m∉{2;-2}

\(\begin{cases}mx+y=3\\ 4x+my=6\end{cases}\Rightarrow\begin{cases}y=3-mx\\ 4x+m\left(3-mx\right)=6\end{cases}\)

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

=>\(\begin{cases}y=3-mx\\ x=\frac{6-3m}{4-m^2}=\frac{3\left(2-m\right)}{\left(2-m\right)\left(2+m\right)}=\frac{3}{2+m}\end{cases}\Rightarrow\begin{cases}x=\frac{3}{2+m}\\ y=3-mx=3-\frac{3m}{m+2}=\frac{3m+6-3m}{m+2}=\frac{6}{m+2}\end{cases}\)

x>1 và y>0

=>\(\begin{cases}\frac{3}{m+2}>1\\ \frac{6}{m+2}>0\end{cases}\Rightarrow\begin{cases}\frac{3-m-2}{m+2}>0\\ m+2>0\end{cases}\Rightarrow\begin{cases}\frac{1-m}{m+2}>0\\ m+2>0\end{cases}\)

=>\(\begin{cases}1-m>0\\ m+2>0\end{cases}\Rightarrow\begin{cases}m<1\\ m>-2\end{cases}\Rightarrow-2

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

=>\(m^2\ne-3\)(luôn đúng)

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

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

=>\(\left\{{}\begin{matrix}y=mx-2\\x=\dfrac{2m+5}{m^2+3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+5}{m^2+3}\\y=\dfrac{2m^2+5m}{m^2+3}-2=\dfrac{2m^2+5m-2m^2-6}{m^2+3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2m+5}{m^2+3}\\y=\dfrac{5m-6}{m^2+3}\end{matrix}\right.\)

\(x+y=\dfrac{3}{m^2+3}\)

=>\(\dfrac{2m+5+5m-6}{m^2+3}=\dfrac{3}{m^2+3}\)

=>\(7m-1=3\)

=>7m=4

=>m=4/7(nhận)

\(a,\text{Thay }x=-2;y=3\\ HPT\Leftrightarrow\left\{{}\begin{matrix}3m-2=4\\3-2n=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=2\\n=3\end{matrix}\right.\\ b,HPT\Leftrightarrow\left\{{}\begin{matrix}x=4-my\\n\left(4-my\right)+y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4-my\\4n-mny+y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=4-my\\y\left(mn-1\right)=4n+3\end{matrix}\right.\)

HPT có vô số nghiệm \(\Leftrightarrow\left\{{}\begin{matrix}mn-1=0\\4n+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=-\dfrac{4}{3}\\n=-\dfrac{3}{4}\end{matrix}\right.\)