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ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-2\\y\le\frac{16}{3}\end{matrix}\right.\)
\(2x^2-\left(3y-6\right)x+y^2-8y-20=0\)
\(\Delta=\left(3y-6\right)^2-8\left(y^2-8y-20\right)=y^2+28y+196=\left(y+14\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3y-6+y+14}{4}=y+2\\x=\frac{3y-6-y-14}{4}=\frac{y-10}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x-2\\y=2x+10\end{matrix}\right.\)
- Với \(y=2x+10\ge-2.2+10=6>\frac{16}{3}\) ko phù hợp ĐKXĐ (loại)
- Với \(y=x-2\)
\(4\sqrt{x+2}+\sqrt{22-3x}=x^2+8\)
\(\Leftrightarrow x^2+8-4\sqrt{x+2}-\sqrt{22-3x}=0\)
\(\Leftrightarrow x^2-x-2+\frac{4}{3}\left(x+4-3\sqrt{x+2}\right)+\frac{1}{3}\left(14-x-3\sqrt{22-3x}\right)=0\)
\(\Leftrightarrow x^2-x-2+\frac{4}{3}\left(\frac{x^2-x-2}{x+4+3\sqrt{x+2}}\right)+\frac{1}{3}\left(\frac{x^2-x-2}{14-x+3\sqrt{22-3x}}\right)=0\)
\(\Leftrightarrow\left(x^2-x-2\right)\left(....\right)=0\) (ngoặc phía sau luôn dương)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\Rightarrow y=-3\\x=2\Rightarrow y=0\end{matrix}\right.\)
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{2x+y}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\) thì pt đầu trở thành:
\(\dfrac{a^2-b^2}{2}-4b^2+3b=a\Leftrightarrow a^2-9b^2+6b=2a\)
\(\Leftrightarrow\left(a-3b\right)\left(a+3b\right)-2\left(a-3b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a+3b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=3b\\a=2-3b\end{matrix}\right.\) \(\Rightarrow...\)
a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)
TH1 : \(x\le-3\) ( LĐ )
TH2 : \(x\ge0\)
BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)
\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)
\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow x\ge0\)
Vậy \(S=R/\left(-3;0\right)\)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
\(\sqrt{x^2-y+3}+\sqrt{y-x+1}=2\)
Xét \(pt\left(1\right)\Leftrightarrow2x^2+y^2-3xy-4x+3y+2=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(2x-y-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x-1\\y=2x-2\end{matrix}\right.\)
*)\(y=x-1\) thay vao \(pt(2)\) :
\(pt\Leftrightarrow\sqrt{x^2-x+4}=2\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=0\end{matrix}\right.\)
*)\(y=2x-2\) thay vao \(pt(2)\):
\(pt\Leftrightarrow\sqrt{x^2-2x+5}+\sqrt{x-1}=2\)
\(\Leftrightarrow\dfrac{x^2-2x+1}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x-1}{\sqrt{x^2-2x+5}+2}+\dfrac{1}{\sqrt{x-1}}\right)=0\)
\(\Leftrightarrow x=1\)\(\Leftrightarrow y=0\)
Câu hỏi của tiên lê - Toán lớp 10 | Học trực tuyến