b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)

\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)

\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)

\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)

\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)

caau a) binh phuong len ra no x=y tuong tu

c)

ĐK $y \geqslant 0$

Hệ đã cho tương đương với

$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$

Trừ từng vế $2$ phương trình ta được

$x^2+2+2\sqrt{y(x^2+2)}-3y=0$

$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$

$\Leftrightarrow x^2+2=y$

a/

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}ab=-6\\a+b=1\end{matrix}\right.\) với

Theo Viet đảo, a và b là nghiệm của:

\(t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x-3=0\\y=-2x-2\end{matrix}\right.\) (bấm casio)

b/

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

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

\(\Rightarrow2x^3=\left(x+y\right)\left(x^2+y^2-xy\right)\)

\(\Leftrightarrow2x^3=x^3+y^3\)

\(\Leftrightarrow x^3=y^3\Rightarrow x=y\)

Thay vào pt đầu:

\(2x^2=4\Rightarrow x^2=2\Rightarrow x=y=\pm\sqrt{2}\)

\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\)  \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)

TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)

Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)

TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)

2 câu dưới hình như em hỏi rồi?

1)Điều kiện: \(x + y > 0\)\((1) \Leftrightarrow (x + y)^2 - 2xy + \dfrac{2xy}{x + y} - 1 = 0 \\ \Leftrightarrow (x + y)^3 - 2xy(x + y) + 2xy -(x + y) = 0 \\ \Leftrightarrow (x+y)[(x+y)^2- 1]-2xy(x+y-1)=0 \\ \Leftrightarrow (x+y)(x+y+1)(x+y-1)-2xy(x+y-1)=0 \\ \Leftrightarrow (x + y - 1)[(x+y)(x + y + 1)-2xy] = 0 \\ \Leftrightarrow \left[ \begin{matrix}x + y = 1 \,\, (3) \\ x^2+y^2+x+y=0 \,\, (4) \end{matrix} \right.\)(4) vô nghiệm vì x + y > 0

Thế (3) vào (2) , giải được nghiệm của hệ :\((x =1 ; y = 0)\)và \((x = -2 ; y = 3)\)

\((1)\Leftrightarrow (x-2y)+(2x^3-4x^2y)+(xy^2-2y^3)=0\)\(\Leftrightarrow (x-2y)(1+2x^2+y^2)=0\)

\(\Leftrightarrow x=2y\)(vì \(1+2x^2+y^2>0, \forall x,y\))

Thay vào phương trình (2) giải dễ dàng.

Điều kiện:\(9y^2+(2y+3)(y-x)\geq 0;xy\geq 0;-1\leq x\leq 1\)

Từ phương trình thứ nhất có \(x\geq 0\Rightarrow y\geq 0\)

Xét \(\left\{\begin{matrix} x=0\\ y=0 \end{matrix}\right.\) thỏa mãn hệ

Xét x,y không đồng thời bằng 0, ta có

\(\sqrt{9y^2+(2y+3)(y-x)}-3x+4\sqrt{xy}-4x=0\)

\(\Leftrightarrow \frac{9y^2+(2y+3)(y-x)-9x^2}{\sqrt{9y^2+(2y-3)(y-x)+3x}}+\frac{4(xy-x^2)}{\sqrt{xy}+x}=0\)

\(\Leftrightarrow (y-x)\left [ \frac{11y+9x+3}{\sqrt{11y^2+(2y-3)(y-x)+3x}}+\frac{4x}{\sqrt{xy}+x} \right ]=0\Leftrightarrow y=x\)

Tới đây thay vào phương trình (2) giải dễ dàng.

1/PT (1) cho ta nhân tử x - y - 1:)

\(\left\{{}\begin{matrix}\left(17-3x\right)\sqrt{5-x}+\left(3y-14\right)\sqrt{4-y}=0\left(1\right)\\2\sqrt{2x+y+5}+3\sqrt{3x+2y+11}=x^2+6x+13\left(2\right)\end{matrix}\right.\)

ĐK: \(x\le5;y\le4\); \(2x+y+5\ge0;3x+2y+11\ge0\)

PT (1) \(\Leftrightarrow\left(17-3x\right)\left(\sqrt{5-x}-\sqrt{4-y}\right)-3\left(x-y-1\right)\sqrt{4-y}=0\)

\(\Leftrightarrow\left(3x-17\right)\left(\frac{x-y-1}{\sqrt{5-x}+\sqrt{4-y}}\right)-3\left(x-y-1\right)\sqrt{4-y}=0\)

\(\Leftrightarrow\left(x-y-1\right)\left(\frac{3x-17}{\sqrt{5-x}+\sqrt{4-y}}-3\sqrt{4-y}\right)=0\)

Dễ thấy cái ngoặc to < 0

Do đó x= y + 1

Thay xuống PT (2):\(y^2+8y+20=2\sqrt{3y+7}+3\sqrt{5y+14}\)\(\left(y+1\right)\left(y+2\right)=y^2+3y+2\)

ĐK: \(y\ge-\frac{7}{3}\) (để các căn thức được thỏa mãn)

PT (2) \(\Leftrightarrow y^2+3y+2+2\left(y+3-\sqrt{3y+7}\right)+3\left(y+4-\sqrt{5y+14}\right)=0\)

\(\Leftrightarrow\left(y^2+3y+2\right)\left(1+\frac{2}{y+3+\sqrt{3y+7}}+\frac{3}{y+4+\sqrt{5y+14}}\right)=0\)

Cái ngoặc to > 0 =>...

P/s: Is that true? Ko đúng thì chịu thua-_- Mất nửa tiếng đồng hồ để gõ bài này đấy:(

2/ĐK: \(x\ge-y;y\ge0\)

PT (1) \(\Leftrightarrow x\left(x+y\right)+\sqrt{x+y}=2y^2+\sqrt{2y}\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+y\left(x-y\right)+\sqrt{x+y}-\sqrt{2y}=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+2y+\frac{1}{\sqrt{x+y}+\sqrt{2y}}\right)=0\)

Cái ngoặc to \(\ge y+\frac{1}{\sqrt{x+y}+\sqrt{2y}}>0\).

Do đó x = y \(\ge0\)

Thay xuống pt dưới: \(x^3-5x^2+14x-4=6\sqrt[3]{x^2-x+1}\)

Lập phương hai vế lên ra pt bậc 6, tuy nhiên cứ yên tâm, nghiệm rất đẹp: x = 1:)

Em đưa kết quả luôn: \(\left(x-1\right)\left(x^2-4x+7\right)\left(x^6-10x^5+56x^4-160x^3+272x^2-64x+40\right)=0\)

P/s: khúc cuối em ko còn cách nào khác nên đành lập phương:((

1,ĐK: \(x,y\ne-2\)

HPT<=> \(\left\{{}\begin{matrix}x\left(x+2\right)+y\left(y+2\right)=\left(x+2\right)\left(y+2\right)\left(1\right)\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x^2\left(x+2\right)^2+2xy\left(x+2\right)\left(y+2\right)+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)

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

<=>\(2xy=0\) (do x+2 và y+2 \(\ne0\))

<=> \(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Tại x=0 thay vào (1) có: \(y\left(y+2\right)=2\left(y+2\right)\) <=> y= \(\pm2\) => y=2 (vì y khác -2)

Tại y=0 thay vào (1) có: \(x\left(x+2\right)=2\left(x+2\right)\) => x=2

Vậy HPT có 2 nghiệm duy nhất (2,0),(0,2)

2, ĐK: \(y\ne-1\)

HPT <=> \(\left\{{}\begin{matrix}x^2=2\left(x+3\right)\left(y+1\right)\left(1\right)\\\frac{3x^2}{y+1}=4-x\end{matrix}\right.\)

=> \(\frac{6\left(3+x\right)\left(y+1\right)}{y+1}=4-x\)

<=> 6(x+3)=4-x

<=> \(14=-7x\)

<=> \(x=-2\) thay vào (1) có \(4=2\left(y+1\right)\)

<=>y=1\(\)( tm)

Vậy hpt có một nghiệm duy nhất (-2,1)

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

PT (1) <=> \(\left(x-y\right)\left(x+y\right)+\left(x-y\right)=0\)

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

<=>\(\left[{}\begin{matrix}x=y\\y=-x-1\end{matrix}\right.\)

Tại x=y thay vào (2) có \(y^2-y=y+3\) <=> \(y^2-2y-3=0\) <=> (y-3)(y+1)=0 <=> \(\left[{}\begin{matrix}y=3\\y=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Tại y=-1-x thay vào (2) có: \(x^2-x=-1-x+3\) <=> \(x^2=2\) <=> \(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}y=-1-\sqrt{2}\\y=-1+\sqrt{2}\end{matrix}\right.\)

Vậy hpt có 4 nghiệm (3,3),(-1,-1), ( \(\sqrt{2},-1-\sqrt{2}\)),( \(-\sqrt{2},-1+\sqrt{2}\))

4,\(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=\frac{9}{2}\left(1\right)\\xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}=5\left(2\right)\end{matrix}\right.\)(đk:\(x\ne0,y\ne0\))

<=> \(\left\{{}\begin{matrix}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=\frac{9}{2}\\\left(y+\frac{1}{y}\right)\left(x+\frac{1}{x}\right)=5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=u\\y+\frac{1}{y}=v\end{matrix}\right.\)

Có \(\left\{{}\begin{matrix}u+v=\frac{9}{2}\\uv=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\v\left(\frac{9}{2}-v\right)=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left(v-\frac{5}{2}\right)\left(v-2\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left[{}\begin{matrix}v=\frac{5}{2}\\v=2\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\\\left[{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

Tại \(\left\{{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=2\\y+\frac{1}{y}=\frac{5}{2}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)\left(y-\frac{1}{2}\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=1\\\left[{}\begin{matrix}y=2\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

Tại \(\left\{{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=\frac{5}{2}\\y+\frac{1}{y}=2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left(x-2\right)\left(x-\frac{1}{2}\right)=0\\\left(y-1\right)^2=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\\y=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}x=\frac{1}{2}\\y=1\end{matrix}\right.\end{matrix}\right.\)

Vậy hpt có 4 nghiệm (1,2),( \(1,\frac{1}{2}\)) ,( 2,1),(\(\frac{1}{2},1\)).

10.

\(\left\{{}\begin{matrix}2x^2-3xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-2xy-xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(2x-y+1\right)=0\\x^2+x+1=y^2\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\3x\left(x+1\right)=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=1\\\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2x+1\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x=-1\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

@Nguyễn Việt Lâm @Lê Thị Thục Hiền @Akai Haruma @Trần Thanh Phương