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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: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
(Pt trên là pt (1), pt dưới là pt (2))
Đk: \(x;y\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3=2x^3+x^2y\\3=2y^3+xy^2\end{matrix}\right.\)
\(\Rightarrow2\left(x^3-y^3\right)+\left(x^2y-xy^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+2xy+y^2\right)=0\)\(\Leftrightarrow\left(x-y\right)\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
TH1: \(x=y\) thay vào pt (1) \(\Rightarrow\dfrac{3}{y^2}=2y+y\)
\(\Leftrightarrow3=3y^3\) \(\Leftrightarrow y=1\) \(\Rightarrow x=y=1\) (TM)
TH2:\(x=-y\) thay vào pt (1) \(\Rightarrow\dfrac{3}{y^2}=-2y+y\)
\(\Leftrightarrow\dfrac{3}{y^2}=-1\left(L\right)\)
Vậy (x;y)=(1;1)
ĐKXĐ: ...
Cộng vế với vế: \(3\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)=3\left(x+y\right)\Rightarrow x+y=\dfrac{1}{x^2}+\dfrac{1}{y^2}\)
Trừ vế cho vế:
\(3\left(\dfrac{1}{x^2}-\dfrac{1}{y^2}\right)=x-y\)
\(\Leftrightarrow-3\left(\dfrac{x-y}{xy}\right)\left(\dfrac{x+y}{xy}\right)=x-y\)
\(\Leftrightarrow\left(x-y\right)\left(1+\dfrac{3\left(x+y\right)}{x^2y^2}\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(1+\dfrac{3\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)}{x^2y^2}\right)=0\)
\(\Leftrightarrow x-y=0\) (do \(1+\dfrac{3\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)}{x^2y^2}>0\))
Thế vào pt đầu:
\(\dfrac{3}{x^2}=3x\Leftrightarrow x^3=1\Leftrightarrow x=y=1\)
b) ĐKXĐ: \(x,y\neq 0\).
Ta có: \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{1}{x}-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-y=0\\xy=-1\end{matrix}\right.\\2y=x^3+1\end{matrix}\right.\).
Với x - y = 0 suy ra x = y. Do đó \(2x=x^3+1\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1=y\left(TMĐK\right)\\x=\pm\dfrac{\sqrt{5}-1}{2}=y\left(TMĐK\right)\end{matrix}\right.\).
Với xy = -1 suy ra \(y=-\dfrac{1}{x}\). Do đó \(x^3+\dfrac{2}{x}+1=0\Rightarrow x^4+x+2=0\). Phương trình vô nghiệm do \(x^4+x+2=\left(x^2-\dfrac{1}{2}\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}>0\).
Vậy...
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 :
\(\left\{{}\begin{matrix}0,3x-0,2y=0,5\\0,5x+0,4y=1,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
DK:\(y\ne0\)
PT (1) :\(3x^2+2y^2-4xy=11-\dfrac{1}{y}\left(2x+\dfrac{1}{y}\right)\)
\(\Leftrightarrow\left(x^2+\dfrac{2x}{y}+\dfrac{1}{y^2}\right)+2\left(x^2-2xy+y^2\right)=11\)
\(\Leftrightarrow\left(x+\dfrac{1}{y}\right)^2+2\left(x-y\right)^2=11\)
PT (2): \(2x+\dfrac{1}{y}-y=4\)
\(\Leftrightarrow\left(x+\dfrac{1}{y}\right)+\left(x-y\right)=4\)
Đặt \(a=x+\dfrac{1}{y};b=x-y\)
Hệ pt tt: \(\left\{{}\begin{matrix}a^2+2b^2=11\\a+b=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(4-b\right)^2+2b^2=11\\a=4-b\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}b=\dfrac{5}{3}\\b=1\end{matrix}\right.\\a=4-b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}b=\dfrac{5}{3}\\a=\dfrac{7}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}b=1\\a=3\end{matrix}\right.\end{matrix}\right.\)
TH1: \(a=\dfrac{7}{3};b=\dfrac{5}{3}\)\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{y}=\dfrac{7}{3}\\x-y=\dfrac{5}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+y=\dfrac{2}{3}\\x-y=\dfrac{5}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y^2-2y+3=0\left(vn\right)\\x-y=\dfrac{5}{3}\end{matrix}\right.\)
TH2:\(a=3;b=1\)\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{y}=3\\x-y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+y=2\\x-y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y^2-2y+1=0\\x-y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\) (thỏa mãn hệ)
Vậy hệ có nghiệm duy nhất (x;y)=(2;1).
Điều kiện \(\left\{{}\begin{matrix}\dfrac{4x-3x^2y-9xy^2}{x+3y}\ge0\\x+3y\ne0\end{matrix}\right.\)
Với \(3y\ge x\), hệ tương đương:
\(\left\{{}\begin{matrix}\left(x^4-2x^2+4\right)\left(x^2+2\right)=6x^5y\\\left(3y-x\right)^2=\dfrac{4x}{x+3y}-3xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^6+8=6x^5y\left(1\right)\\x^3+27y^3=4x\end{matrix}\right.\left(I\right)\)
Vì \(x=0\) thì hệ vô nghiệm nên \(x\ne0\), khi đó:
\(\left(I\right)\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{8}{x^6}=\dfrac{6y}{x}\\1+\dfrac{27y^3}{x^3}=\dfrac{4}{x^2}\end{matrix}\right.\)
Đặt \(\dfrac{3y}{x}=a,\dfrac{2}{x^2}=b\) ta được hệ:
\(\Leftrightarrow\left\{{}\begin{matrix}1+a^3=2b\\1+b^3=2a\end{matrix}\right.\)
Giải hệ này ta được \(a=b\Leftrightarrow\dfrac{3y}{x}=\dfrac{2}{x^2}\Leftrightarrow y=\dfrac{2}{3x}\)
\(\left(1\right)\Leftrightarrow x^6-4x^4+8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\\x=\sqrt{1+\sqrt{5}}\\x=-\sqrt{1+\sqrt{5}}\end{matrix}\right.\)
TH1: \(x=\sqrt{2}\Rightarrow y=\dfrac{\sqrt{2}}{3}\)
TH2: \(x=-\sqrt{2}\Rightarrow y=-\dfrac{\sqrt{2}}{3}\)
TH3: \(x=\sqrt{1+\sqrt{5}}\Rightarrow y=\dfrac{2}{3\sqrt{1+\sqrt{5}}}\)
TH4: \(x=-\sqrt{1+\sqrt{5}}\Rightarrow y=-\dfrac{2}{3\sqrt{1+\sqrt{5}}}\)
Đối chiếu với các điều kiện ta được \(\left(x;y\right)=\left(-\sqrt{1+\sqrt{5}};-\dfrac{2}{3\sqrt{1+\sqrt{5}}}\right)\)
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...
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
\(x^2y+2y+x=4xy< =>xy\left(x+3\right)=4xy< =>x+3=4< =>x=1\)
Thế x=1 vào 1 trong 2 phương trình => y=1