a) Thay m=1 vào hệ phương trình, ta được:

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=5\\x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x+10=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=5\end{matrix}\right.\)

Vậy: Khi m=1 thì hệ phương trình có nghiệm duy nhất là (x,y)=(-8;5)

b) Ta có: \(\left\{{}\begin{matrix}x+2y=m+1\\2x+3y=m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=2m+2\\2x+3y=m-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=m+4\\x+2\cdot\left(m+4\right)=m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2m+8=m+1\\y=m+4\end{matrix}\right.\)

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

Để hệ phương trình có nghiệm (x,y) thỏa mãn x>3 và y<5 thì \(\left\{{}\begin{matrix}-m-7>3\\m+4< 5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-m>10\\m< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< -10\\m< 1\end{matrix}\right.\Leftrightarrow m< -10\)

Vậy: Để hệ phương trình có nghiệm (x,y) thỏa mãn x>3 và y<5 thì m<-10

a/ Theo Viet đảo, x và y là nghiệm của pt:

\(t^2-5t+5=0\Rightarrow t=\frac{5\pm\sqrt{5}}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\frac{5+\sqrt{5}}{2}\\y=\frac{5-\sqrt{5}}{2}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=\frac{5-\sqrt{5}}{2}\\y=\frac{5+\sqrt{5}}{2}\end{matrix}\right.\)

b/ Đặt \(Y=-y\Rightarrow\left\{{}\begin{matrix}x+Y=1\\xY=-6\end{matrix}\right.\)

Theo Viet đảo, x và Y là nghiệm của: \(t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\Y=-2\Rightarrow y=2\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-2\\Y=3\Rightarrow y=-3\end{matrix}\right.\)

ý 2

Do cắt trục tung tại điểm có tung độ bằng -4--->b=-4(1)

Cắt trục hoành tại điểm có hoành độ bằng 2

-->x=2,y=0

-->2a+b=0 hay 2a=-b(2)

Thay (1) vào (2) ta dc

2x=4

-->x=2

Vậy a=2,b=-4

Từ pt (1) \(\Rightarrow x=8+\left|y-5\right|\ge8\Rightarrow x+1>0\)

- Nếu \(y\ge5\Rightarrow3\left|y+3\right|\ge24>21\Rightarrow\) vô nghiệm

- Nếu \(-5\le y\le5\) hệ trở thành:

\(\left\{{}\begin{matrix}x-\left(5-y\right)=8\\x+1+3\left(y+5\right)=21\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=13\\x+3y=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=17\\y=-4\end{matrix}\right.\)

- Nếu \(y< -5\) hệ trở thành:

\(\left\{{}\begin{matrix}x-\left(5-y\right)=8\\x+1+3\left(-y-5\right)=21\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=13\\x-3y=35\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{37}{2}\\y=\dfrac{-11}{2}\end{matrix}\right.\)

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

(2)\(\Leftrightarrow x^5-x^2+y^5-y^2=0\)

\(\Leftrightarrow x^2\left(x^3-1\right)+y^2\left(y^3-1\right)=0\)

\(\Leftrightarrow x^2\left(-y\right)^3+y^2\left(-x\right)^3=0\)

\(\Leftrightarrow x^2y^3+y^2x^3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=1\\y=0\Rightarrow x=1\\x=-y\left(loại\right)\end{matrix}\right.\)

a, Ta có ( I ) : \(\left\{{}\begin{matrix}x+y=5\\xy=5\end{matrix}\right.\)

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

=> \(\left\{{}\begin{matrix}x=5-y\\5y-y^2-5=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=5-y\\y^2-5y+5=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=5-y\\y^2-2.\frac{5}{2}y+\left(\frac{5}{2}\right)^2-1,25=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=5-y\\\left(y-2,5\right)^2=1,25\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=5-y\\\left[{}\begin{matrix}y-2,5=\frac{\sqrt{5}}{2}\\y-2,5=-\frac{\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=5-\frac{\sqrt{5}}{2}-2,5=\frac{5-\sqrt{5}}{2}\\x=5-2,5+\frac{\sqrt{5}}{2}=\frac{15-\sqrt{5}}{2}\end{matrix}\right.\\\left[{}\begin{matrix}y=\frac{\sqrt{5}}{2}+2,5\\y=2,5-\frac{\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy hệ phương trình có 2 nghiệm là : \(\left(x,y\right)=\left(\frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2}\right),\left(\frac{15-\sqrt{5}}{2},\frac{5-\sqrt{5}}{2}\right)\) .

\(\left\{{}\begin{matrix}4x+3x=-6\\\dfrac{x+3y}{3}-\dfrac{y-2}{5}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7x=-6\\\dfrac{5\left(x+3y\right)-3\left(y-2\right)}{15}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\5x+15y-3y+6=15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\12y=9-5x=9+5\cdot\dfrac{6}{7}=9+\dfrac{30}{7}=\dfrac{93}{7}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\y=\dfrac{93}{7\cdot12}=\dfrac{93}{84}=\dfrac{31}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=5\\\left(x+y\right)^2-2xy+x+y=8\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=5\\a^2+a-2b=8\end{matrix}\right.\) \(\Rightarrow a^2+a-2\left(5-a\right)=8\)

\(\Leftrightarrow a^2+3a-18=0\Rightarrow\left[{}\begin{matrix}a=3\Rightarrow b=2\\a=-6\Rightarrow b=11\left(l\right)\end{matrix}\right.\)

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

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=5\\\dfrac{1}{xy}=6\end{matrix}\right.\left(x,y\ne0\right)\)\(\Leftrightarrow\left(I\right)\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=5\\\dfrac{1}{x}\cdot\dfrac{1}{y}=6\end{matrix}\right.\)

Đặt \(\dfrac{1}{x}=a,\dfrac{1}{y}=b\left(a,b>0\right)\)

Hệ (I) \(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\ab=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\ab=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\b\left(5-b\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\-\left(b^2-5b\right)=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\b^2-5b+6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\\left(b-3\right)\left(b-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=5-b\\b-3=0\end{matrix}\right.\\\left\{{}\begin{matrix}a=5-b\\b-2=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=5-b\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=5-b\\b=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\end{matrix}\right.\)

Trả lại biến cũ

\(\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=2\\\dfrac{1}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,5\\y=\dfrac{1}{3}\end{matrix}\right.\left(TM\right)\)và ngược lại

Vậy HPT có các cặp nghiệm là \(\left(0,5;\dfrac{1}{3}\right);\left(\dfrac{1}{3};0,5\right)\)

P/S: Bạn kiểm tra kết quả lại giúp mình nhé