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\(\left\{{}\begin{matrix}x^2+y^2=25\\x.y=10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=25\\x=\dfrac{10}{y}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{10}{y}\right)^2+y^2=25\\x=\dfrac{10}{y}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{100}{y^2}+y^2=25\\x=\dfrac{10}{y}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}100+y^4-25y^2=0\\x=\dfrac{10}{y}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y^2=20\\y^2=5\end{matrix}\right.\\x=\dfrac{10}{y}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=\pm\sqrt{20}\\y=\pm\sqrt{5}\end{matrix}\right.\\x=\dfrac{10}{y}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\sqrt{20};x=\sqrt{5}\\y=-\sqrt{20};x=-\sqrt{5}\\y=-\sqrt{5};x=-\sqrt{20}\\y=\sqrt{5};x=\sqrt{20}\end{matrix}\right.\)
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix}
\frac{1}{x}+\frac{1}{y}=\frac{3}{8}\\
\frac{1}{y}+\frac{1}{z}=\frac{3}{4}\\
\frac{1}{z}+\frac{1}{x}=\frac{5}{6}\end{matrix}\right.\Rightarrow 2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{8}+\frac{3}{4}+\frac{5}{6}\)
\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{47}{48}\)
\(\Rightarrow \left\{\begin{matrix} \frac{1}{z}=\frac{47}{48}-\frac{3}{8}\\ \frac{1}{x}=\frac{47}{48}-\frac{3}{4}\\ \frac{1}{y}=\frac{47}{48}-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{48}{29}\\ y=\frac{48}{11}\\ z=\frac{48}{7}\end{matrix}\right.\)
ĐKXĐ : \(x;y\ne0\)
Khi đó \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2x^2-xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-\dfrac{x-y}{xy}\\2x^2-xy=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(\dfrac{xy+1}{xy}\right)=0\\2x^2-xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\xy=-1\end{matrix}\right.\\2x^2-xy=1\end{matrix}\right.\)
Với x = y thì 2x2 - xy = 1
<=> 2x2 - x2 = 1
<=> x2 = 1
<=> x = \(\pm1\) (tm)
Khi x = -1 => y = -1
x = 1 => y = 1
Với xy = - 1 thì 2x2 - xy = 1
<=> 2x2 - (-1) = 1
<=> x2 = 0
<=> x = 0 (ktm)
Vậy hệ có 2 nghiệm (x;y) = (1; 1) ; (-1 ; -1)
\(\left\{{}\begin{matrix}\sqrt{x}+\dfrac{3}{\sqrt{x}}=\sqrt{y}+\dfrac{3}{\sqrt{y}}\left(1\right)\\2x-\sqrt{xy}-1=0\left(2\right)\end{matrix}\right.\) đk : x>=; y>=0
Ta có (1) <=> \(\left(\sqrt{x}-\sqrt{y}\right)-\left(\dfrac{3}{\sqrt{y}}-\dfrac{3}{\sqrt{x}}\right)=0\)
<=> \(\left(\sqrt{x}-\sqrt{y}\right)-3\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}=0\)
<=> \(\left(\sqrt{x}-\sqrt{y}\right)\left(1-\dfrac{3}{\sqrt{xy}}\right)=0\)
<=> \(\left[{}\begin{matrix}x=y\\\sqrt{xy}=3\end{matrix}\right.\)
+) với x=y, thay vào (2) ta có:
\(2x-\sqrt{x^2}-1=0\)
<=> 2x- x-1=0(do x>0)
<=> x=1 => y =1(t/m)
+) với \(\sqrt{xy}=3\) thay vào (2) ta có :
2x - 3-1 =0
<=> x= 2 (tm) => y = 9/2
Vậy hệ có nghiệm (x;y) là (1;1), (2;\(\dfrac{9}{2}\) )
1) \(\Leftrightarrow\left\{{}\begin{matrix}a=2+b\\b\left(2+b\right)-\left(2+b\right)-34=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2+b\\b^2+b-36=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2+b\\b=\frac{-1\pm\sqrt{145}}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2+\frac{-1\pm\sqrt{145}}{2}\\b=\frac{-1\pm\sqrt{145}}{2}\end{matrix}\right.\)
Thay \(x=\dfrac{3}{4}y\) vào phương trình dưới, ta có:
\(\dfrac{1}{2}\left(\dfrac{3}{4}y+3\right)\left(y-2\right)-\dfrac{1}{2}.\dfrac{3}{4}y^2=9\)
\(\Leftrightarrow\dfrac{3}{8}y^2-\dfrac{3}{4}y+\dfrac{3}{2}y-3-\dfrac{3}{8}y^2=9\\ \Leftrightarrow\dfrac{3}{4}y=12\\ \Leftrightarrow y=18\Rightarrow x=12\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(12;18\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2xy+y^2=xy+3y-1\\\left(x+y\right)\left(x^2+1\right)=x^2+y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y^2+\left(x-3\right)y+x^2+1=0\\x^3+x+x^2y-x^2-1=0\end{matrix}\right.\)
Trừ vế cho vế:
\(\Rightarrow y^2-\left(x^2-x+3\right)y-x^3+2x^2-x+2=0\)
\(\Delta=\left(x^2-x+3\right)^2-4\left(-x^3+2x^2-x+2\right)=\left(x^2+x-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{x^2-x+3+x^2+x-1}{2}=x^2+1\\y=\dfrac{x^2-x+3-x^2-x+1}{2}=-x+2\end{matrix}\right.\)
Thế vào pt dưới:
\(\left[{}\begin{matrix}x+x^2+1=2\\x-x+2=\dfrac{x^2+1-x+2}{x^2+1}\end{matrix}\right.\)
\(\Leftrightarrow...\)
a) \(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=-1\end{matrix}\right.\)(Đk: \(x\ne-1;y\ne-1\))
Đặt \(\dfrac{x}{x+1}\) là A
\(\dfrac{y}{y+1}\) là B
Ta có HPT mới : \(\left\{{}\begin{matrix}2A+B=2\\A+3B=-1\end{matrix}\right.\)(1)
Giải HPT (1) ta được A= \(\dfrac{7}{5}\) ; B=\(-\dfrac{4}{5}\)
+Với A=\(\dfrac{7}{5}\) ta có:
\(\dfrac{x}{x+1}=\dfrac{7}{5}\)
<=>\(5x=7x+7\)
<=>-2x=7
<=> x=\(-\dfrac{7}{2}\)
+Với B = \(-\dfrac{4}{5}\) ta có:
\(\dfrac{y}{y+1}=-\dfrac{4}{5}\)
<=>5y=-4y-4
<=>9y=-4
<=>y=\(-\dfrac{4}{9}\)
Vậy HPT có nghiệm (x;y) = \(\left\{-\dfrac{7}{2};-\dfrac{4}{9}\right\}\)
`{([10x+y]/[x+y]=6),(xy+25=10y+x):}` `ĐK: x \ne -y`
`<=>{(10x+y=6x+6y),(xy+25=10y+x):}`
`<=>{(y=4/5x),(x. 4/5x+25=10. 4/5x+x):}`
`<=>{(y=4/5x),(4/5x^2-9x+25=0):}`
`<=>{(y=4/5x),([(x=25/4),(x=5):}):}`
`<=>[({(x=25/4),(y=4/5 . 25/4=5):}),({(x=5),(y=4/5 .5=4):}):}` (t/m)
\(\left\{{}\begin{matrix}\dfrac{10x+y}{x+y}=6\\xy+25=10y+x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x+y=6\left(x+y\right)\\xy-10y-x=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x-5y=0\\xy-10y-x=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}\\xy-10y-x=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}\\\dfrac{5y}{4}y-10y-\dfrac{5y}{4}=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}\\\dfrac{5y^2-45y}{4}=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}\\5y^2-45y+100=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}\\y_1=5\\y_2=4\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left\{{}\begin{matrix}x=\dfrac{25}{4}\\y=5\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=5\\y=4\end{matrix}\right.\)