GHPT
\(\left\{{}\begin{matrix}7\sqrt{16-y^2}+6=x^2+5x\\\left(x+2\right)^2+2\left(y-4\right)^2=9\end{matrix}\right.\)
GHPT :
\(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\3x\left(y-7\right)+10=\sqrt{10x-2}+2\sqrt{8y-3}\end{matrix}\right.\)
\(ĐK:x\ge\dfrac{1}{5};y\ge\dfrac{3}{8}\)
\(PT\left(1\right)\Leftrightarrow\dfrac{3x^2-3y^2}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}=3\left(x+y\right)\\ \Leftrightarrow3\left(x+y\right)\left(\dfrac{x-y}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+y=0\\\dfrac{x-y}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x-y=\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}\\ \Leftrightarrow\left(x-y\right)=\dfrac{3\left(x^2-y^2\right)}{\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}}\\ \Leftrightarrow\left(x-y\right)\left[\dfrac{3\left(x+y\right)}{\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}}-1\right]=0\)
\(\Leftrightarrow x=y\)
Với \(x+y=0\Leftrightarrow x=-y\), thay vào PT 2
\(\Leftrightarrow3\left(-y\right)\left(y-7\right)+10=\sqrt{10\left(-y\right)-2}+2\sqrt{8y-3}\\ \Leftrightarrow3y\left(7-y\right)+10=\sqrt{-10y-2}+2\sqrt{8y-3}\)
ĐK: \(\left\{{}\begin{matrix}-10y-2\ge0\\8y-3\ge0\end{matrix}\right.\Leftrightarrow y\in\varnothing\)
Với \(x-y=0\Leftrightarrow x=y\), thay vào PT 2
\(\Leftrightarrow3x^2-21x+10=\sqrt{10x-2}+2\sqrt{8x-3}\left(x\ge\dfrac{3}{8}\right)\\ \Leftrightarrow3x^2-24x+9=\sqrt{10x-2}-\left(x+1\right)+2\sqrt{8x-3}-2x\)
\(\Leftrightarrow3\left(x^2-8x+3\right)=\dfrac{-x^2+8x-3}{\sqrt{10x-2}+\left(x+1\right)}+\dfrac{2\left(-x^2+8x-3\right)}{\sqrt{8x-3}+x}\\ \Leftrightarrow\left(x^2-8x+3\right)\left(3+\dfrac{1}{\sqrt{10x-2}+x+1}+\dfrac{2}{\sqrt{8x-3}+x}\right)=0\)
Dễ thấy ngoặc lớn vô nghiệm với \(x\ge\dfrac{3}{8}>0\)
\(\Leftrightarrow x^2-8x+3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{13}\left(n\right)\\x=4-\sqrt{13}\left(n\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=4+\sqrt{13}\\y=4-\sqrt{13}\end{matrix}\right.\)
Vậy HPT có nghiệm \(\left(x;y\right)\in\left\{\left(4+\sqrt{13};4+\sqrt{13}\right);\left(4-\sqrt{13};4-\sqrt{13}\right)\right\}\)
1/Ghpt\(\left\{{}\begin{matrix}x^2+y^2+x^2y^2=1+2xy\\\left(x-y\right)\left(1+xy\right)=1-xy\end{matrix}\right.\)
2/Ghpt\(\left\{{}\begin{matrix}x^2y+y+xy^2+x=18xy\\x^4y^2+y^2+x^2y^4+x^2=208x^2y^2\end{matrix}\right.\)
3/Ghpt\(\left\{{}\begin{matrix}\sqrt{x+3}+\sqrt{y+3}=4\\\dfrac{1}{x}+\dfrac{1}{y}=2\end{matrix}\right.\)
4/ Cho x,y là nghiệm của hệ phương trình
\(\left\{{}\begin{matrix}x+y=m\\x^2+y^2=2m\end{matrix}\right.\)
Tìm min và max của A=xy
5/cho x,y,z thỏa mãn đk
\(\left\{{}\begin{matrix}xy+yz+xz=1\\x^2+y^2+z^2=2\end{matrix}\right.\)
Chứng minh rằng: \(\dfrac{-4}{3}\le x,y,z\le\dfrac{4}{3}\)
6/Ghpt bằng 3 cách\(\left\{{}\begin{matrix}x+y+z=1\\\\x^2+y^2+z^2=1\\x^3+y^3+z^3=1\end{matrix}\right.\)
7/Ghpt\(\left\{{}\begin{matrix}x^3+1=2y\\y^3+1=2x\end{matrix}\right.\)
8/Ghpt\(\left\{{}\begin{matrix}x^2-3y=-2\\y^2-3x=-2\end{matrix}\right.\)
9/Ghpt bằng 2 cách\(\left\{{}\begin{matrix}x+\sqrt{y+3}=3\\y+\sqrt{x+3}=3\end{matrix}\right.\)
10/Ghpt\(\left\{{}\begin{matrix}x+\dfrac{2}{y}=\dfrac{3}{x}\\y+\dfrac{2}{x}=\dfrac{3}{y}\end{matrix}\right.\)
11/Ghpt\(\left\{{}\begin{matrix}\sqrt[3]{3x+5}=y+1\\\sqrt[3]{3y+5}=x+1\end{matrix}\right.\)
12/Ghpt\(\left\{{}\begin{matrix}3x^2y-y^2-2=0\\3y^2x-x^2-2=0\end{matrix}\right.\)
13/Giải các phương trình sau bằng cách đứa về hệ pt đối xứng loại II:
a)\(\left(x^2-3\right)^2-x-3=0\)
b)\(x^2-2=\sqrt{x+2}\)
14/Ghpt:\(\left\{{}\begin{matrix}x^2+y^2+xy=3\\x^2-y^2+xy=1\end{matrix}\right.\)
Ghpt \(\left\{{}\begin{matrix}x^2+2y=xy+4\\x^2-x-3-x\sqrt{6-x}=\left(y-3\right)\sqrt{y-3}\end{matrix}\right.\)
\(ĐK:x\le6;y\ge3\\ \left\{{}\begin{matrix}x^2+2y=xy+4\left(1\right)\\x^2-x-3-x\sqrt{6-x}=\left(y-3\right)\sqrt{y-3}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2-4+2y-xy=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)-y\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-y+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=y-2\end{matrix}\right.\)
Từ đó thế vào PT(2)
Với \(x=y-2\Leftrightarrow x+2=y\)
\(\left(2\right)\Leftrightarrow x^2-x+3-x\sqrt{6-x}=\left(x-1\right)\sqrt{x-1}\left(1\le x\le6\right)\\ \Leftrightarrow2x^2-2x+6-2x\sqrt{6-x}=2\left(x-1\right)\sqrt{x-1}\\ \Leftrightarrow\left(x-\sqrt{6-x}\right)^2+x\left(x-1\right)=2\left(x-1\right)\sqrt{x-1}\\ \Leftrightarrow\left(x-\sqrt{6-x}\right)^2+\left(x-1\right)\left(x-2\sqrt{x-1}\right)=0\\ \Leftrightarrow\left(\dfrac{x^2-6+x}{x+\sqrt{6-x}}\right)^2+\dfrac{\left(x-1\right)\left(x^2-4x+4\right)}{x^2+2\sqrt{x-1}}=0\\ \Leftrightarrow\left[\dfrac{\left(x-2\right)\left(x+3\right)}{x+\sqrt{6-x}}\right]^2+\dfrac{\left(x-1\right)\left(x-2\right)^2}{x^2+2\sqrt{x-1}}=0\\ \Leftrightarrow\left(x-2\right)^2\left[\left(\dfrac{x+3}{x+\sqrt{6-x}}\right)^2+\dfrac{x-1}{x^2+2\sqrt{x-1}}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\left(\dfrac{x+3}{x+\sqrt{6-x}}\right)^2+\dfrac{x-1}{x^2+2\sqrt{x-1}}=0\left(1\right)\end{matrix}\right.\)
Dễ thấy \(\left(1\right)>0\) với \(x\ge1\)
Do đó \(x=2\Leftrightarrow y=4\)
Vậy HPT có nghiệm \(\left(x;y\right)=\left(2;4\right)\)
GHPT\(\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x^2-y^2\right)=144\\\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=y\end{matrix}\right.\)
\(PT\left(2\right)\Leftrightarrow\dfrac{x^2+y^2-x^2+y^2}{\sqrt{x^2+y^2}+\sqrt{x^2-y^2}}-y=0\\ \Leftrightarrow\dfrac{2y^2}{\sqrt{x^2+y^2}+\sqrt{x^2-y^2}}-y=0\\ \Leftrightarrow y\left(\dfrac{2y}{\sqrt{x^2+y^2}+\sqrt{x^2-y^2}}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y=0\\\dfrac{2y}{\sqrt{x^2+y^2}+\sqrt{x^2-y^2}}=1\left(3\right)\end{matrix}\right.\\ \left(3\right)\Leftrightarrow\sqrt{x^2+y^2}+\sqrt{x^2-y^2}=2y\\ \Leftrightarrow x^2+\sqrt{x^4-y^4}=2y^2\\ \Leftrightarrow\sqrt{x^4-y^4}=\left(2y^2-x^2\right)^2\\ \Leftrightarrow x^4-y^4=4y^4-4x^2y^2+x^4\\ \Leftrightarrow5y^4-4x^2y^2=0\\ \Leftrightarrow y^2\left(5y^2-4x^2\right)=0\Leftrightarrow\left[{}\begin{matrix}y=0\left(trùng.n_o\right)\\5y^2=4x^2\end{matrix}\right.\)
\(\Leftrightarrow x^2=\dfrac{5}{4}y^2\)
Từ đó thế 2 trường hợp vào PT(1)
Ghpt:\(\left\{{}\begin{matrix}\left(4x^2+1\right).x+\left(y-3\right)\sqrt{5-2y}=0\\4x^2+y^2+2\sqrt{3-4x}=7\end{matrix}\right.\)
GHPT: \(\left\{{}\begin{matrix}\left(4x^2+1\right).x+\left(y-3\right)\sqrt{5-2y}=0\\4x^2+y^2+2\sqrt{3-4x}=7\end{matrix}\right.\)
\(Ghpt:\left\{{}\begin{matrix}\sqrt{3x}\left(1+\dfrac{1}{x+y}\right)=2\\\sqrt{7y}\left(1-\dfrac{1}{x+y}\right)=4\sqrt{2}\end{matrix}\right.\)
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
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. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
GHPT: \(\left\{{}\begin{matrix}2x+\dfrac{1}{x+y}+\dfrac{1}{x-y}=\dfrac{16}{3}\\2\left(x^2+y^2\right)+\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{\left(x-y\right)^2}=\dfrac{100}{9}\end{matrix}\right.\)
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}+x-y+\dfrac{1}{x-y}=\dfrac{16}{3}\\\left(x+y\right)^2+\dfrac{1}{\left(x+y\right)^2}+\left(x-y\right)^2+\dfrac{1}{\left(x-y\right)^2}=\dfrac{100}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}+x-y+\dfrac{1}{x-y}=\dfrac{16}{3}\\\left(x+y+\dfrac{1}{x+y}\right)^2+\left(x-y+\dfrac{1}{x-y}\right)^2=\dfrac{136}{9}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}=u\\x-y+\dfrac{1}{x-y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=\dfrac{16}{3}\\u^2+v^2=\dfrac{136}{9}\end{matrix}\right.\)
Hệ cơ bản, chắc bạn tự giải quyết phần còn lại được