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a, ĐK: \(x,y\ge0\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\sqrt{y}}{\sqrt{x+3}-\sqrt{x}}=3\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=\sqrt{x+3}\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+3}=x+1\)
\(\Leftrightarrow x+3=x^2+2x+1\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\left(l\right)\end{matrix}\right.\)
Thay \(x=1\) vào hệ phương trình đã cho ta được \(y=1\)
Vậy pt đã cho có nghiệm \(x=y=1\)
b, \(hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(y+\dfrac{1}{2}\right)^2\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y=-1\end{matrix}\right.\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2-3x=0\end{matrix}\right.\left(1\right)\\\left\{{}\begin{matrix}x+y=-1\\x^2+y^2=-3\end{matrix}\right.\left(vn\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=y=3\\x=y=0\end{matrix}\right.\)
Vậy ...
a, \(\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^3+y^3\right)=280\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^2+y^2-xy\right)=70\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(16-2xy\right)\left(16-3xy\right)=70\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\3x^2y^2-40xy+93=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left[{}\begin{matrix}xy=\dfrac{31}{3}\\xy=3\end{matrix}\right.\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=4\\xy=\dfrac{31}{3}\end{matrix}\right.\)
Phương trình này vô nghiệm
Vậy hệ đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(1;3\right);\left(3;1\right)\right\}\)
b, ĐK: \(xy>0\)
\(\left\{{}\begin{matrix}\sqrt{\dfrac{2x}{y}}+\sqrt{\dfrac{2y}{x}}=3\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x}{y}+\dfrac{2y}{x}+4=9\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2+y^2\right)=5xy\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(x-2y\right)=0\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}2x=y\\x=2y\end{matrix}\right.\\x-y+xy=3\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}y=2x\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2x\\2x^2-x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2x\\\left(x+1\right)\left(2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=-2\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=3\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x=2y\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2+y-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
a.
\(\left\{{}\begin{matrix}x^4+y^4=34\\y=2-x\end{matrix}\right.\)
\(\Rightarrow x^4+\left(x-2\right)^4=34\)
Đặt \(x-1=t\)
\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=34\)
\(\Leftrightarrow t^4+6t^2-16=0\Rightarrow\left[{}\begin{matrix}t^2=2\\t^2=-8\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}t=\sqrt{2}\Rightarrow x=\sqrt{2}+1\Rightarrow y=1-\sqrt{2}\\t=-\sqrt{2}\Rightarrow x=1-\sqrt{2}\Rightarrow y=1+\sqrt{2}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}xy^2-x^2y+6x-y^2-y-6=0\\x^2y-xy^2+6y-x^2-x-6=0\end{matrix}\right.\) (1)
Lần lượt cộng 2 vế và trừ 2 vế ta được:
\(\left\{{}\begin{matrix}-x^2-y^2+5x+5y-12=0\\2xy\left(y-x\right)+7\left(x-y\right)+\left(x-y\right)\left(x+y\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-5\left(x+y\right)+12=0\\\left(y-x\right)\left(2xy-x-y-7\right)=0\end{matrix}\right.\)
Th1: \(\left\{{}\begin{matrix}x=y\\x^2+y^2-5\left(x+y\right)+12=0\end{matrix}\right.\)
\(\Rightarrow2x^2-10x+12=0\Rightarrow...\)
TH2: \(\left\{{}\begin{matrix}2xy-\left(x+y\right)-7=0\\x^2+y^2-5\left(x+y\right)+12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2xy-\left(x+y\right)-7=0\\\left(x+y\right)^2-2xy-5\left(x+y\right)+12=0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2v-u-7=0\\u^2-2v-5u+12=0\end{matrix}\right.\)
\(\Rightarrow u^2-6u+5=0\)
\(\Leftrightarrow...\)
a/ \(\lim\limits_{x\rightarrow\sqrt{2}}f\left(x\right)=\lim\limits_{x\rightarrow\sqrt{2}}\frac{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}{x-\sqrt{2}}=\lim\limits_{x\rightarrow\sqrt{2}}\left(x+\sqrt{2}\right)=2\sqrt{2}\)
\(\Rightarrow\lim\limits_{x\rightarrow\sqrt{2}}f\left(x\right)=f\left(\sqrt{2}\right)\Rightarrow\) hàm số liên tục tại \(x=\sqrt{2}\)
b/ \(\lim\limits_{x\rightarrow5^+}f\left(x\right)=\lim\limits_{x\rightarrow5^+}\frac{x-5}{\sqrt{2x-1}-3}=\frac{\left(x-5\right)\left(\sqrt{2x-1}+3\right)}{2\left(x-5\right)}=\lim\limits_{x\rightarrow5^+}\frac{\sqrt{2x-1}+3}{2}=3\)
\(f\left(5\right)=\lim\limits_{x\rightarrow5^-}f\left(x\right)=\lim\limits_{x\rightarrow5^-}\left[\left(x-5\right)^2+3\right]=5\)
\(\Rightarrow\lim\limits_{x\rightarrow5^+}f\left(x\right)=\lim\limits_{x\rightarrow5^-}f\left(x\right)=f\left(5\right)\Rightarrow\) hàm số liên tục tại \(x=5\)
a.
\(\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)
Nhân vế:
\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)
\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)
\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4y}{7}\\y=-3x\end{matrix}\right.\)
Thế vào \(y^2=5x^2+4...\)
b. Đề bài không hợp lý ở \(4x^2\)
c.
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)
Trừ vế:
\(x^3-y^3-3x^2-6y^2=9-3x+12y\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+2\right)^3\)
\(\Leftrightarrow x-1=y+2\)
\(\Leftrightarrow y=x-3\)
Thế vào \(x^2=2y^2=x-4y\) ...
a, Cộng vế theo vế hai phương trình ta được:
\(x^2+y^2+2xy+x+y=2\)
\(\Leftrightarrow\left(x+y\right)^2+x+y-2=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x+y+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=1\\x+y=-2\end{matrix}\right.\)
TH1: \(x+y=1\)
\(pt\left(2\right)\Leftrightarrow xy+1=-1\Leftrightarrow xy=-2\)
Ta có hệ: \(\left\{{}\begin{matrix}x+y=1\\xy=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\xy=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\end{matrix}\right.\)
TH2: \(x+y=-2\)
\(pt\left(2\right)\Leftrightarrow xy-2=-1\Leftrightarrow xy=1\)
Ta có hệ: \(\left\{{}\begin{matrix}x+y=-2\\xy=1\end{matrix}\right.\Leftrightarrow x=y=-1\)
b, \(\left\{{}\begin{matrix}x^3-y^3=7\left(x-y\right)\\x^2+y^2=x+y+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2+xy-7\right)=0\\x^2+y^2=x+y+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x^2+y^2+xy=7\end{matrix}\right.\\x^2+y^2=x+y+2\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x=y\\x^2+y^2=x+y+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^2-x-1=0\end{matrix}\right.\)
\(\Leftrightarrow x=y=\dfrac{1\pm\sqrt{5}}{2}\)
TH2: \(\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^2+y^2=x+y+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=7\\\left(x+y\right)^2-2xy-x-y=2\end{matrix}\right.\)
Đặt \(x+y=u;xy=v\)
Hệ trở thành: \(\left\{{}\begin{matrix}u^2-v=7\\u^2-2v-u=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-7\\u^2-2\left(u^2-7\right)-u=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-7\\u^2+u-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-7\\\left[{}\begin{matrix}u=3\\u=-4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}v=2\\u=3\end{matrix}\right.\\\left\{{}\begin{matrix}v=9\\u=-4\end{matrix}\right.\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}v=2\\u=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=2\\x+y=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}v=9\\u=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=9\\x+y=-4\end{matrix}\right.\left(vn\right)\)
a.
Với \(y=0\) không phải nghiệm
Với \(y\ne0\Rightarrow\left\{{}\begin{matrix}3x+2=\dfrac{5}{y}\\2x\left(x+y\right)+y=\dfrac{5}{y}\end{matrix}\right.\)
\(\Rightarrow3x+2=2x\left(x+y\right)+y\)
\(\Leftrightarrow2x^2+\left(2y-3\right)x+y-2=0\)
\(\Delta=\left(2y-3\right)^2-8\left(y-2\right)=\left(2y-5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2y+3+2y-5}{4}=-\dfrac{1}{2}\\x=\dfrac{-2y+3-2y+5}{4}=-y+2\end{matrix}\right.\)
Thế vào pt đầu ...
Câu b chắc chắn đề sai