\(x^2-25=y\left(y+6\right)\)

\(\Leftrightarrow x^2-25=y^2+6y\)

\(\Leftrightarrow x^2-25-y^2-6y=0\)

\(\Leftrightarrow x^2-\left(y^2+6y+9\right)-16=0\)

\(\Leftrightarrow x^2-\left(y+3\right)^2=16\)

\(\Leftrightarrow\left(x+y+3\right)\left(x-y-3\right)=16\)

\(\Leftrightarrow\left(x+y+3\right);\left(x-y-3\right)\in\left\{-1;1;-2;2;-4;4;-8;8;-16;16\right\}\)

Ta giải các hệ phương trình sau :

1) \(\left\{{}\begin{matrix}x+y+3=-1\\x-y-3=-16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-4\\x-y=-15\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=-11\left(loại\right)\\x-y=-15\end{matrix}\right.\)

2) \(\left\{{}\begin{matrix}x+y+3=1\\x-y-3=16\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-2\\x-y=19\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=17\left(loại\right)\\x-y=19\end{matrix}\right.\)

3) \(\left\{{}\begin{matrix}x+y+3=2\\x-y-3=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\x-y=11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=10\\x-y=11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-6\end{matrix}\right.\)

4) \(\left\{{}\begin{matrix}x+y+3=-2\\x-y-3=-8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-5\\x-y=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=-10\\x-y=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=0\end{matrix}\right.\)

5) \(\left\{{}\begin{matrix}x+y+3=-4\\x-y-3=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-7\\x-y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=-6\\x-y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

6) \(\left\{{}\begin{matrix}x+y+3=4\\x-y-3=4\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\x-y=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=8\\x-y=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-3\end{matrix}\right.\)

7) \(\left\{{}\begin{matrix}x+y+3=-8\\x-y-3=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-11\\x-y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=-10\\x-y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=-6\end{matrix}\right.\)

8) \(\left\{{}\begin{matrix}x+y+3=8\\x-y-3=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=5\\x-y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=10\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=0\end{matrix}\right.\)

9) \(\left\{{}\begin{matrix}x+y+3=-16\\x-y-3=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-19\\x-y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=-17\left(loại\right)\\x-y=2\end{matrix}\right.\)

10) \(\left\{{}\begin{matrix}x+y+3=16\\x-y-3=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=15\\x-y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=19\left(loại\right)\\x-y=4\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(5;-6\right);\left(-5;0\right);\left(-3;-2\right);\left(4;-3\right);\left(-5;-6\right);\left(5;0\right)\right\}\)

Ta có : 

\(2x^4-2x^2y+y^2=16\)

\(\Leftrightarrow\left(x^4-2x^2y+y^2\right)+x^4=16\)

\(\Leftrightarrow\left(x^2-y\right)^2+\left(x^2\right)^2=16\)

Vì \(x,y\) nguyên mà \(16=0^2+\left(2^2\right)^2=0^2+\left[\left(-2\right)^2\right]^2\)

Nên ta sẽ tìm được 2 cặp nghiệm nguyên của hai phương trình này.

Đáp số : 2.

pt <=>\(x^4+\left(x^4-2x^2y+y^2\right)=16\)

\(\Leftrightarrow x^4+\left(x^2-y\right)^2=16\)

\(\Leftrightarrow x^4=16-\left(x^2-y\right)^2\le16\)

\(\Leftrightarrow0\le x^2\le4\) (*)

Do \(x\in Z\) \(\Rightarrow x^2\in N\) và \(x^2\) là số chính phương

=> \(x^2\in\left\{0;1;4\right\}\) \(\Leftrightarrow x\in\left\{0;-1;1;-2;2\right\}\)

Tại x=0 thay vào pt ta được: \(y^2=16\) \(\Leftrightarrow y=\pm4\) => Tìm được 2 cặp

Tại x²=1 thay vào pt tìm được \(\left[{}\begin{matrix}y=1+\sqrt{15}\\y=1-\sqrt{15}\end{matrix}\right.\) không thỏa mãn y nguyên => Loại

Tại \(x^2=4\)thay vào pt tìm được \(y=4\) => Tìm đc 2 cặp

Vậy tìm đc 4 cặp tm

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

\(\Leftrightarrow4x^2-4xy+4y^2+8\left(x+y\right)=0\)

\(\Leftrightarrow\left(2x-y\right)^2+4\left(2x-y\right)+4+3y^2+12y+12=-16\)

\(\Leftrightarrow\left(2x-y+2\right)^2+3\left(y+2\right)^2=-16\)

Dễ thấy VT \(\ge0\) ; VP < 0 nên phương trình vô nghiệm 

\(x^2+y^2-2\left(x+y\right)=xy\)

\(\Rightarrow x^2-2x+1+y^2-2y+1=2+xy\)

\(\Rightarrow\left(x-1\right)^2+\left(y-1\right)^2=2+xy\)

Ta lại có : \(\left(x-1\right)^2+\left(y-1\right)^2\ge2\left(x-1\right)\left(y-1\right)\) (Bất đẳng thức Cauchy)

a.

\(\Leftrightarrow2x^2-4x+4y^2=4xy+4\)

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(x^2-4x+4\right)=8\)

\(\Leftrightarrow\left(x-2y\right)^2+\left(x-2\right)^2=8\) (1)

Do \(\left(x-2y\right)^2\ge0;\forall x;y\)

\(\Rightarrow\left(x-2\right)^2\le8\)

\(\Rightarrow\left(x-2\right)^2=\left\{0;1;4\right\}\)

TH1: \(\left(x-2\right)^2\Rightarrow x=2\) thế vào (1)

\(\Rightarrow\left(2-2y\right)^2=8\Rightarrow\left(1-y\right)^2=2\) (ko tồn tại y nguyên t/m do 2 ko phải SCP)

TH2: \(\left(x-2\right)^2=1\Rightarrow\left(x-2y\right)^2=8-1=7\), mà 7 ko phải SCP nên pt ko có nghiệm nguyên

TH3: \(\left(x-2\right)^2=4\Rightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\) thế vào (1):

- Với \(x=0\Rightarrow\left(-2y\right)^2+4=8\Rightarrow y^2=1\Rightarrow y=\pm1\)

- Với \(x=2\Rightarrow\left(2-2y\right)^2+4=8\Rightarrow\left(1-y\right)^2=1\Rightarrow\left[{}\begin{matrix}y=0\\y=2\end{matrix}\right.\)

Vậy pt có các cặp nghiệm là: 

\(\left(x;y\right)=\left(0;1\right);\left(0;-1\right);\left(2;0\right);\left(2;2\right)\)

b.

\(\Leftrightarrow2x^2+4y^2+4xy-4x=14\)

\(\Leftrightarrow\left(x^2+4xy+4y^2\right)+\left(x^2-4x+4\right)=18\)

\(\Leftrightarrow\left(x+2y\right)^2+\left(x-2\right)^2=18\) (1)

Lý luận tương tự câu a ta được 

\(\left(x-2\right)^2\le18\Rightarrow\left(x-2\right)^2=\left\{0;1;4;9;16\right\}\)

Với \(\left(x-2\right)^2=\left\{0;1;4;16\right\}\) thì \(18-\left(x-2\right)^2\) ko phải SCP nên ko có giá trị nguyên x;y thỏa mãn

Với \(\left(x-2\right)^2=9\Rightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\) thế vào (1)

- Với \(x=5\Rightarrow\left(5+2y\right)^2+9=18\Rightarrow\left(5+2y\right)^2=9\)

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

- Với \(x=-1\Rightarrow\left(-1+2y\right)^2=9\Rightarrow\left[{}\begin{matrix}-1+2y=3\\-1+2y=-3\end{matrix}\right.\)

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

Vậy \(\left(x;y\right)=\left(5;-1\right);\left(5;-4\right);\left(-1;3\right);\left(-1;-3\right)\)

5x²+2y+y²-4x-40=0

△=(-4)²-4.5.(2y+y²-40)

△=16-40y-20y²+800

△=-(784+40y+20y²)

△=-(32y+8y+16y²+4y²+16+4+764)

△=-[(4y+4)²+(2y+2)²+764]<0

=>PHƯƠNG TRÌNH VÔ NGHIỆM.

 Ta có \(x^2+y^2=1\Leftrightarrow\left(x+y\right)^2=2xy+1\)

 Từ đó \(P=\dfrac{\left(x+y\right)^2}{x+y+1}\). Đặt \(x+y=t\left(t\ge0\right)\). Vì \(x+y\le\sqrt{2\left(x^2+y^2\right)}=2\) nên \(t\le\sqrt{2}\). ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{\sqrt{2}}\). Ta cần tìm GTLN của \(P\left(t\right)=\dfrac{t^2}{t+1}\) với \(0\le t\le\sqrt{2}\). 

 Giả sử có \(0\le t_1\le t_2\le\sqrt{2}\). Ta có BDT luôn đúng \(\left(t_2-t_1\right)\left(t_2+t_1+t_2t_1\right)\ge0\) \(\Leftrightarrow t_2^2-t_1^2+t_2^2t_1-t_2t_1^2\ge0\) \(\Leftrightarrow t_1^2\left(t_2+1\right)\le t_2^2\left(t_1+1\right)\) \(\Leftrightarrow\dfrac{t_1^2}{t_1+1}\le\dfrac{t_2^2}{t_2+1}\) \(\Leftrightarrow P\left(t_1\right)\le P\left(t_2\right)\).  Như vậy với \(0\le t_1\le t_2\le\sqrt{2}\) thì \(P\left(t_1\right)\le P\left(t_2\right)\). Do đó P là hàm đồng biến. Vậy GTLN của P đạt được khi \(t=\sqrt{2}\) hay \(x=y=\dfrac{1}{\sqrt{2}}\), khi đó \(P=2\sqrt{2}-2\)

Lời giải:
$P=\frac{2xy+1}{x+y+1}=\frac{2xy+x^2+y^2}{x+y+1}=\frac{(x+y)^2}{x+y+1}$

$=\frac{a^2}{a+1}$ với $x+y=a$

Áp dụng BĐT AM-GM:

$1=x^2+y^2\geq \frac{(x+y)^2}{2}=\frac{a^2}{2}$

$\Rightarrow a^2\leq 2\Rightarrow a\leq \sqrt{2}$

$P=\frac{a^2}{a+1}=\frac{a}{1+\frac{1}{a}}$
Vì $a\leq \sqrt{2}\Rightarrow 1+\frac{1}{a}\geq 1+\frac{1}{\sqrt{2}}=\frac{2+\sqrt{2}}{2}$

$\Rightarrow P\leq \frac{\sqrt{2}}{\frac{2+\sqrt{2}}{2}}=-2+2\sqrt{2}$

Vậy $P_{\max}=-2+2\sqrt{2}$ khi $x=y=\frac{1}{\sqrt{2}}$

a: Thay m=-2 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}x-2y=-2+1=-1\\-2x+y=3\cdot\left(-2\right)-1=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y=-2\\-2x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3y=-9\\x-2y=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x=2y-1=2\cdot3-1=5\end{matrix}\right.\)

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

=>\(m^2\ne1\)

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

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

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

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

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

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

=>\(\left\{{}\begin{matrix}y=\dfrac{m-1}{m+1}\\x=m+1-m\cdot\dfrac{m-1}{m+1}=\left(m+1\right)-\dfrac{m^2-m}{m+1}\end{matrix}\right.\)

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

\(x^2-y^2=4\)

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

=>\(\dfrac{9m^2+6m+1-m^2+2m+1}{\left(m+1\right)^2}=4\)

=>\(8m^2+8m+2=4\left(m+1\right)^2\)

=>\(8m^2+8m+2-4m^2-8m-4=0\)

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

=>\(m^2=\dfrac{1}{2}\)

=>\(m=\pm\dfrac{1}{\sqrt{2}}\)