a/ Bạn tự giải

b/ ĐKXĐ:...

Cộng vế với vế: \(\frac{x-y}{y+12}=3\Rightarrow x-y=3y+36\Rightarrow x=4y+36\)

Thay vào pt đầu: \(\frac{4y+36}{y}-\frac{y}{y+12}=1\)
Đặt \(\frac{y+12}{y}=a\Rightarrow4a-\frac{1}{a}=1\Rightarrow4a^2-a-1=0\)

\(\Rightarrow a=\frac{1\pm\sqrt{17}}{8}\) \(\Rightarrow\frac{y+12}{y}=\frac{1\pm\sqrt{17}}{8}\)

\(\Rightarrow\left[{}\begin{matrix}y+12=y\left(\frac{1+\sqrt{17}}{8}\right)\\y+12=y\left(\frac{1-\sqrt{17}}{8}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(\frac{-7+\sqrt{17}}{8}\right)y=12\\\left(\frac{-7-\sqrt{17}}{8}\right)y=12\end{matrix}\right.\) \(\Rightarrow y=...\)

Chắc bạn ghi sai đề, nghiệm quá xấu

3/ \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+y^2=5\\3x^2-9y=3\end{matrix}\right.\) \(\Rightarrow y^2+9y=2\Rightarrow y^2+9y-2=0\Rightarrow y=...\)

4/ ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{3x-1}-3\sqrt{2y+1}=3\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)

\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)

\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)

1. Đề này là 18 chứ không phải 15 nhé

\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+x+\sqrt{y^2+x+y+1}+y=18\left(1\right)\\\sqrt{x^2+x+y+1}-x+\sqrt{y^2+x+y+1}-y=2\left(2\right)\end{matrix}\right.\)

Lấy (1) + (2) và (1) - (2) ta được hệ mới

\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+\sqrt{y^2+x+y+1}=10\\x+y=8\end{matrix}\right.\)

\(\Rightarrow x=8-y\)

\(\Rightarrow\sqrt{x^2+9}+\sqrt{y^2+9}=10\)\(\Leftrightarrow\sqrt{x^2+9}=10-\sqrt{y^2+9}\)

\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\x^2+9=100-20\sqrt{y^2+9}+y^2+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\x^2=100-20\sqrt{y^2+9}+y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\\left(8-y\right)^2=100-20\sqrt{y^2+9}+y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\9y^2-72y+144=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)

2. Dễ thấy x = y = 0 không phải là nghiệm của phương trình

HPT\(\Leftrightarrow\left\{{}\begin{matrix}1-\dfrac{12}{y+3x}=\dfrac{2}{\sqrt{x}}\left(1\right)\\1+\dfrac{12}{y+3x}=\dfrac{6}{\sqrt{y}}\left(2\right)\end{matrix}\right.\)

Lấy (1) + (2) ; (1) - (2) ta được

\(\left\{{}\begin{matrix}1=\dfrac{1}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}\left(3\right)\\\dfrac{12}{y+3x}=\dfrac{3}{\sqrt{y}}-\dfrac{1}{\sqrt{x}}\left(4\right)\end{matrix}\right.\)

Lấy ( 3) nhân (4)

\(\dfrac{12}{y+3x}=\dfrac{9}{y}-\dfrac{1}{x}=\dfrac{9x-y}{xy}\)

\(\Leftrightarrow27x^2-6xy-y^2=0\Leftrightarrow\left(9x+y\right)\left(3x-y\right)=0\)

\(\Rightarrow y=3x\)

đến đây thì dễ rồi

3. Đây là hệ đối xứng loại I

\(\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=8\\\left(x+y\right)+2xy=2\end{matrix}\right.\)

Đặt S = a + b ; P = ab (\(S^2\ge4P\) )

xong giải ra thôi mà

a, #Góp ý từ nhiều người nhưng họ không giải nên t làm giùm

ĐK: \(x\le3\)

\(\left\{{}\begin{matrix}x^2+y^2+1=2\left(xy-x+y\right)\left(1\right)\\x^3+3y^2+5x-12=\left(12-y\right)\sqrt{3-x}\left(2\right)\end{matrix}\right.\)

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

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

\(\left(2\right)\)\(\Leftrightarrow x^3+3\left(x+1\right)^2+5x-12=\left[12-\left(x+1\right)\right]\sqrt{3-x}\)

\(\Leftrightarrow x^3+3x^2+11x-9=\left(11-x\right)\sqrt{3-x}\)

\(\Leftrightarrow x^3+3x^2+8x=\left(11-x\right)\sqrt{3-x}+3\left(3-x\right)\)

\(\Leftrightarrow x^3+3x^2+8x=\left(3-x\right)\sqrt{3-x}+8\sqrt{3-x}+3\left(3-x\right)\)

\(\Leftrightarrow x^3+3x^2+8x=\sqrt{\left(3-x\right)^3}+3\sqrt{\left(3-x\right)^2}+8\sqrt{3-x}\)

\(\Leftrightarrow x=\sqrt{3-x}\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x-3=0\end{matrix}\right.\) \(\Rightarrow x=\frac{-1+\sqrt{13}}{2}\left(tm\right)\Rightarrow y=\frac{1+\sqrt{13}}{2}\)

Vậy...

Akai Haruma, No choice teen, Arakawa Whiter, Phạm Hoàng Lê Nguyên, Vũ Minh Tuấn, tth, HISINOMA KINIMADO, Nguyễn Việt Lâm

Mn giúp e vs ạ! thanks!

Với những bài như thế này thì nên tag @Nguyễn Thị Ngọc Thơ vô

\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)

Vì \(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)

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

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

\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)

Làm nốt nha

\(\left\{{}\begin{matrix}\sqrt{x-1}+\sqrt{y-1}=2\\\frac{1}{x}+\frac{1}{y}=1\end{matrix}\right.\left(x;y\ge1\right)\)

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

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

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

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

Làm nốt

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

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

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

⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)

\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))

mk lm 1 bài còn lại bn lm tương tự nha :

a) điều kiện xác định : \(x\ge0;y\ge1\)

đặc \(a=\sqrt{x};b=\sqrt{y-1}\)

\(\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}a+2b=5\\4a-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)

ta có : \(a=1\Rightarrow\sqrt{x}=1\Leftrightarrow x=1\left(tmđk\right)\) ; \(b=2\Rightarrow\sqrt{y-1}=2\Leftrightarrow y=5\left(tmđk\right)\)

vậy phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(1;5\right)\)

b) bn đặc : \(a=\dfrac{1}{x};b=\dfrac{1}{y+12}\)

c) bn đặc : \(a=\dfrac{x}{x+1};b=\dfrac{y}{y+1}\)

nhớ điều kiện nha