a^2+b^2leq9text{and}b^2-4ab+a^2leq0 Denklemini Çözme

$$a ^ { 2 } + b ^ { 2 } \leq 9 \text { and } b ^ { 2 } - 4 a b + a ^ { 2 } \leq 0$$

b için çöz

$$\left\{\begin{matrix}b=\sqrt{9-a^{2}}\text{, }&a=\frac{3\sqrt{2}+3\sqrt{6}}{4}\\b=-\sqrt{3}a+2a\text{, }&a=0\text{ or }a=\frac{-3\sqrt{2}-3\sqrt{6}}{4}\\b\in \begin{bmatrix}-\sqrt{9-a^{2}},\sqrt{3}|a|+2a\end{bmatrix}\text{, }&a>\frac{-3\sqrt{2}-3\sqrt{6}}{4}\text{ and }a\leq \frac{3\sqrt{2}-3\sqrt{6}}{4}\\b\in \begin{bmatrix}-\sqrt{3}|a|+2a,\sqrt{3}|a|+2a\end{bmatrix}\text{, }&\left(a>0\text{ and }a<\frac{3\sqrt{6}-3\sqrt{2}}{4}\right)\text{ or }\left(a>\frac{3\sqrt{2}-3\sqrt{6}}{4}\text{ and }a<0\right)\\b\in \begin{bmatrix}-\sqrt{3}|a|+2a,\sqrt{9-a^{2}}\end{bmatrix}\text{, }&a\geq \frac{3\sqrt{6}-3\sqrt{2}}{4}\text{ and }a<\frac{3\sqrt{2}+3\sqrt{6}}{4}\end{matrix}\right,$$

a için çöz

$$\left\{\begin{matrix}a=\sqrt{9-b^{2}}\text{, }&b=\frac{3\sqrt{2}+3\sqrt{6}}{4}\\a=-\sqrt{3}b+2b\text{, }&b=0\text{ or }b=\frac{-3\sqrt{2}-3\sqrt{6}}{4}\\a\in \begin{bmatrix}-\sqrt{9-b^{2}},\sqrt{3}|b|+2b\end{bmatrix}\text{, }&b>\frac{-3\sqrt{2}-3\sqrt{6}}{4}\text{ and }b\leq \frac{3\sqrt{2}-3\sqrt{6}}{4}\\a\in \begin{bmatrix}-\sqrt{3}|b|+2b,\sqrt{3}|b|+2b\end{bmatrix}\text{, }&\left(b>0\text{ and }b<\frac{3\sqrt{6}-3\sqrt{2}}{4}\right)\text{ or }\left(b>\frac{3\sqrt{2}-3\sqrt{6}}{4}\text{ and }b<0\right)\\a\in \begin{bmatrix}-\sqrt{3}|b|+2b,\sqrt{9-b^{2}}\end{bmatrix}\text{, }&b\geq \frac{3\sqrt{6}-3\sqrt{2}}{4}\text{ and }b<\frac{3\sqrt{2}+3\sqrt{6}}{4}\end{matrix}\right,$$