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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: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
1. \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=5\\x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=9\end{matrix}\right.\) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y+xy^2+x+y=5xy\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^4y^2+x^2y^4+x^2+y^2=25x^2y^2\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\)\(\Leftrightarrow0=16x^2y^2\)
\(\Rightarrow\) phương trình vô nghiệm
a)\(\left\{{}\begin{matrix}2x-3y=1\\x+2y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\cdot\left(3-2y\right)-3y=1\\x=3-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6-7y=1\\x=3-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{7}\\x=3-2\cdot\dfrac{5}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{7}\\x=\dfrac{11}{7}\end{matrix}\right.\)b) Biểu diễn lại một biến theo một biến như pt trên rồi giải, ta có :
\(\left\{{}\begin{matrix}2x+4y=5\\4x-2y=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{10}\\y=\dfrac{4}{5}\end{matrix}\right.\)
c) Cách làm tương tự như pt a ta có :
\(\left\{{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}y=\dfrac{2}{3}\\\dfrac{1}{3}x-\dfrac{3}{4}y=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{8}\\y=-\dfrac{1}{6}\end{matrix}\right.\)
d) Tương tự ta có :
\(\left\{{}\begin{matrix}0,3x-0,2y=0,5\\0,5x+0,4y=1,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\left(x+y\right)^2+\left(x-y\right)^2+\dfrac{3}{\left(x+y\right)^2}=\dfrac{85}{3}\\\left(x+y\right)+\left(x-y\right)+\dfrac{1}{x+y}=\dfrac{13}{3}\end{matrix}\right.\)
\(a=x+y\); \(b=x-y\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a^2+b^2+\dfrac{3}{a^2}=\dfrac{85}{3}\\a+b+\dfrac{1}{a}=\dfrac{13}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\left(a+\dfrac{1}{a}\right)^2-6+b^2=\dfrac{85}{3}\\a+\dfrac{1}{a}=\dfrac{13}{3}-b\end{matrix}\right.\)
\(\Rightarrow3\left(\dfrac{13}{3}-b\right)^2-6+b^2=\dfrac{85}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}b=1\\b=\dfrac{11}{2}\end{matrix}\right.\)đến đây tự làm nha
HPT \(\Leftrightarrow\left\{{}\begin{matrix}3\left(x^2+y^2\right)+2xy+\dfrac{1}{\left(x-y\right)^2}=20\\\left(x-y\right)+\left(x+y\right)+\dfrac{1}{x-y}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)^2+\left(x-y\right)^2+\dfrac{1}{\left(x-y\right)^2}=20\\\left(x-y\right)+\left(x+y\right)+\dfrac{1}{x-y}=5\end{matrix}\right.\)
Đặt \(a=x+y;b=x-y\)
\(\Rightarrow\left\{{}\begin{matrix}2a^2+b^2+\dfrac{1}{b^2}=20\\a+b+\dfrac{1}{b}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2a^2+\left(b+\dfrac{1}{b}\right)^2=22\\b+\dfrac{1}{b}=5-a\end{matrix}\right.\)
\(\Rightarrow2a^2+\left(a-5\right)^2=22\)
\(\)Đến đây thì dễ rồi tự làm nhé
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 \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}=u\\x-y+\dfrac{1}{x-y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=\dfrac{16}{3}\\u^2+v^2=\dfrac{136}{9}\end{matrix}\right.\)
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