\(1,ĐK:x,y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}2x^2y^2=y^3+1\\2x^2y^2=x^3+1\end{matrix}\right.\\ \Leftrightarrow x^3+1=y^3+1\\ \Leftrightarrow x^3=y^3\Leftrightarrow x=y\)

Thay vào PT 1

\(\Leftrightarrow2x^4=x^3+1\\ \Leftrightarrow2x^4-x^3-1=0\\ \Leftrightarrow2x^4-2x^3+x-1=0\\ \Leftrightarrow\left(2x^3+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^3=-\dfrac{1}{2}\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=\sqrt[3]{-\dfrac{1}{2}}\\x=y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(\sqrt[3]{-\dfrac{1}{2}};\sqrt[3]{-\dfrac{1}{2}}\right);\left(1;1\right)\)

\(2,ĐK:x,y\ge1\\ HPT\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)+\sqrt{y-1}=\dfrac{1}{2}\\2\left(y-1\right)+\sqrt{x-1}=\dfrac{1}{2}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}2a^2+b=\dfrac{1}{2}\\2b^2+a=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow2\left(a-b\right)\left(a+b\right)-\left(a-b\right)=0\\ \Leftrightarrow\left(a-b\right)\left(2a+2b-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\2a+2b=1\end{matrix}\right.\)

Với \(a=b\Leftrightarrow x-1=y-1\Leftrightarrow x=y\)

Thay vào \(PT\left(1\right)\Leftrightarrow2x+\sqrt{x-1}=\dfrac{5}{2}\Leftrightarrow2\sqrt{x-1}=5-4x\)

\(\Leftrightarrow4x-4=25-40x+16x^2\\ \Leftrightarrow16x^2-44x+29=0\\ \Leftrightarrow\left[{}\begin{matrix}x=y=\dfrac{11+\sqrt{5}}{8}\left(tm\right)\\x=y=\dfrac{11-\sqrt{5}}{8}\left(tm\right)\end{matrix}\right.\)

Với \(2a+2b=1\Leftrightarrow b=\dfrac{1}{2}-a\Leftrightarrow\sqrt{y-1}=\dfrac{1}{2}-\sqrt{x-1}\)

Thay vào \(PT\left(1\right)\Leftrightarrow2x+\dfrac{1}{2}-\sqrt{x-1}=\dfrac{5}{2}\Leftrightarrow2x-2=\sqrt{x-1}\)

\(\Leftrightarrow4x^2-8x+4=x-1\\ \Leftrightarrow4x^2-9x+5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\Rightarrow y=1\left(tm\right)\\x=1\Rightarrow y=\dfrac{5}{4}\left(tm\right)\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(\dfrac{11+\sqrt{5}}{8};\dfrac{11+\sqrt{5}}{8}\right);\left(\dfrac{11-\sqrt{5}}{8};\dfrac{11-\sqrt{5}}{8}\right);\left(\dfrac{5}{4};1\right);\left(1;\dfrac{5}{4}\right)\)

9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)

11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\y=\dfrac{x+4}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)

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

14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)

15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))

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

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

5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)

Thay từng TH rồi làm nha bạn

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

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

thay nhá

Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)

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

+) Với y = x - 1 thay vào pt (2):

\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))

Anh quy đồng lên đê, chắc cần vài con trâu đó:))

+) Với y = 2x + 3...

a, \(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2\right)=13\\\left(x+y\right)\left(x^2-y^2\right)=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-y\right)\left(x^2+y^2\right)=26\\\left(x-y\right)\left(x+y\right)^2=25\end{matrix}\right.\)

Trừ vế theo vế \(pt\left(1\right)\) cho \(pt\left(2\right)\) ta được:

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

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

\(\Leftrightarrow x-y=1\)

Khi đó hệ trở thành:

\(\left\{{}\begin{matrix}x^2+y^2=13\\\left(x+y\right)^2=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=13\\13+2xy=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=13\\2xy=12\end{matrix}\right.\)

Cộng vế theo vế 2 phương trình:

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

\(\Leftrightarrow x+y=\pm5\)

TH1: \(x+y=5\)

Ta có hệ: \(\left\{{}\begin{matrix}x-y=1\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

TH2: \(x+y=-5\)

Ta có hệ: \(\left\{{}\begin{matrix}x-y=1\\x+y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

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

ĐK: \(y\ne0\)

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

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

Đặt \(\dfrac{1}{y}=t\), hệ trở thành:

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

\(\Rightarrow\left(x-t\right)\left(x+t+1\right)=0\)

\(\Leftrightarrow...\)

1. \(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\8x-4y=-44\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\11x=-33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=-3\end{matrix}\right.\)

2. \(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=0\\4x+2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)

3.\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+5y=18\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=12\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

(1) + rút y từ pt (2) thay vào pt (1), ta được pt bậc hai 1 ẩn x, dễ rồi, tìm x rồi suy ra y

(2) + (3)

+ pt nào có nhân tử chung thì đặt nhân tử chung (thật ra chỉ có pt (2) của câu 2 là có nhân từ chung)

+ trong hệ, thấy biểu thức nào giống nhau thì đặt cho nó 1 ẩn phụ

VD hệ phương trình 3: đặt a= x+y ; b= căn (x+1)

+ khi đó ta nhận được một hệ phương trình bậc nhất hai ẩn, giải hpt đó rồi suy ra x và y

hello bạn

a.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)

b.

ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)

Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:

\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)

\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)

\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)

Thay xuống pt dưới:

\(6y+y=14\Rightarrow y=2\)

\(\Rightarrow x=4\)