GHPT sau:
\(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x+2}}+\dfrac{1}{\sqrt{y-1}}=\dfrac{2}{\sqrt{x+y}}\\x^2+y^2+4xy-4x+2y-5=0\end{matrix}\right.\)
Giải HPT:
1. \(\left\{{}\begin{matrix}x^2+y^2+\dfrac{2xy}{x+y}=1\\\sqrt{x+y}=x^2-y\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x+2}}+\dfrac{1}{\sqrt{y-1}}=\dfrac{2}{\sqrt{x+y}}\\x^2+y^2+4xy-4x+2y-5=0\end{matrix}\right.\)
c. ĐKXĐ: ...
\(x^2+y^2+2xy-2xy+\dfrac{2xy}{x+y}-1=0\)
\(\Leftrightarrow\left(x+y\right)^2-1-2xy\left(1-\dfrac{1}{x+y}\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x+y+1\right)-\dfrac{2xy\left(x+y-1\right)}{x+y}=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x+y+1-\dfrac{2xy}{x+y}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=1\\x^2+y^2+x+y=0\left(vô-nghiệm\right)\end{matrix}\right.\)
Thế \(y=1-x\) xuống pt dưới:
\(\sqrt{x+1-x}=x^2-\left(1-x\right)\)
\(\Leftrightarrow x^2+x-2=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=0\\x=-2\Rightarrow y=3\end{matrix}\right.\)
d.
ĐKXĐ: \(x>-2;y>1;x+y>0\)
\(\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}+\sqrt{\dfrac{x+y}{y-1}}=2\\2\left(x+y\right)^2=\left(x+2\right)^2+\left(y-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}+\sqrt{\dfrac{x+y}{y-1}}=2\\\left(\dfrac{x+2}{x+y}\right)^2+\left(\dfrac{y-1}{x+y}\right)^2=2\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}=a>0\\\sqrt{\dfrac{x+y}{y-1}}=b>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2\\\dfrac{1}{a^4}+\dfrac{1}{b^4}=2\end{matrix}\right.\)
Ta có: \(\dfrac{1}{a^4}+\dfrac{1}{b^4}\ge\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^4\ge\dfrac{1}{8}\left(\dfrac{4}{a+b}\right)^4=\dfrac{1}{8}.\left(\dfrac{4}{2}\right)^4=2\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+y}{x+2}=1\\\dfrac{x+y}{y-1}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
Giải hệ
a) \(\left\{{}\begin{matrix}x^2+y^2-2y-6+2\sqrt{2y+3}=0\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2y+2y+x=4xy\\\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{x}{y}=3\end{matrix}\right.\)
Ghpt:
a) \(\left\{{}\begin{matrix}x^2+2y^2=2x-2xy+1\\3x^2+2xy-y^2=2x-y+5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}4xy+4x^2+4y^2+\dfrac{3}{\left(x+y\right)^2}=7\\2x+\dfrac{1}{x+y}=3\end{matrix}\right.\)
1.\(\left\{{}\begin{matrix}2x^2=y+\dfrac{1}{y^2}\\2y^2=x+\dfrac{1}{x^2}\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}2x+\sqrt{y-1}=\dfrac{5}{2}\\2y+\sqrt{x-1}=\dfrac{5}{2}\end{matrix}\right.\)
\(1,ĐK:x,y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}2x^2y^2=y^3+1\\2x^2y^2=x^3+1\end{matrix}\right.\\ \Leftrightarrow x^3+1=y^3+1\\ \Leftrightarrow x^3=y^3\Leftrightarrow x=y\)
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 \(\left(x;y\right)=\left(\dfrac{11+\sqrt{5}}{8};\dfrac{11+\sqrt{5}}{8}\right);\left(\dfrac{11-\sqrt{5}}{8};\dfrac{11-\sqrt{5}}{8}\right);\left(\dfrac{5}{4};1\right);\left(1;\dfrac{5}{4}\right)\)
GHPT sau: \(\left\{{}\begin{matrix}\dfrac{25}{9}+\sqrt{9x^2-4}=\dfrac{1}{9}\left(\dfrac{2}{x}+\dfrac{18x}{y^2-2y+2}+25y\right)\\7x^3+y^3+3xy\left(x-y\right)-12x^2+6x=1\end{matrix}\right.\)
\(Ghpt:\left\{{}\begin{matrix}\sqrt{3x}\left(1+\dfrac{1}{x+y}\right)=2\\\sqrt{7y}\left(1-\dfrac{1}{x+y}\right)=4\sqrt{2}\end{matrix}\right.\)
ĐKXĐ: \(x;y\ge0\)
Với \(x=0\) hoặc \(y=0\) đều ko là nghiệm
Với \(x;y>0\) hệ tương đương:
\(\left\{{}\begin{matrix}1+\dfrac{1}{x+y}=\dfrac{2}{\sqrt{3x}}\\1-\dfrac{1}{x+y}=\dfrac{4\sqrt{2}}{\sqrt{7y}}\end{matrix}\right.\)
Lần lượt cộng vế với vế và trừ vế cho vế ta được:
\(\left\{{}\begin{matrix}1=\dfrac{1}{\sqrt{3x}}+\dfrac{2\sqrt{2}}{\sqrt{7y}}\\\dfrac{1}{x+y}=\dfrac{1}{\sqrt{3x}}-\dfrac{2\sqrt{2}}{\sqrt{7y}}\end{matrix}\right.\)
Nhân vế với vế:
\(\dfrac{1}{x+y}=\dfrac{1}{3x}-\dfrac{8}{7y}\)
\(\Leftrightarrow\dfrac{y}{3}-\dfrac{8x}{7}=1\)
\(\Rightarrow y=\dfrac{24x+21}{7}\)
Rồi thế vào 1 trong các pt đầu
Nhưng em có nhầm đề ko mà con số xấu kinh khủng vậy nhỉ? Số \(\sqrt{7}\) kia cho xấu 1 cách ko cần thiết, nó ko ảnh hưởng đến cách giải mà chỉ khiến cho việc tính toán khó khăn 1 cách cơ học khá vớ vẩn
1.Gpt: \(\dfrac{6}{x-3\sqrt{x-2}+7}=\dfrac{1}{\sqrt{x-2}}+\dfrac{\sqrt{3}}{3\sqrt{2\sqrt{x-2}}-3}\)
2.Ghpt: \(\left\{{}\begin{matrix}x^2-y-z=0\\x^3-y^2-z^2+2=0\end{matrix}\right.\)
Giải hệ pt
a) \(\left\{{}\begin{matrix}x+\dfrac{y}{\sqrt{1+x^2}+x}+y^2=0\\\dfrac{x^2}{y^2}+2\sqrt{x^2+1}+y^2=3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{x^2+4}+\sqrt{y^2+2y-4}=4\\\sqrt{x^2+9}+y=5\end{matrix}\right.\)