\left\{\begin{matrix}x\in \begin{bmatrix}-\sqrt{-\left(y-5\right)\left(y+1\right)},\sqrt{-\left(y-5\right)\left(y+1\right)}\end{bmatrix}\text{, }&y\geq -1\text{ and }y\leq \frac{3-\sqrt{17}}{2}\\x=\sqrt{\left(5-y\right)\left(y+1\right)}\text{, }&y=\frac{\sqrt{17}+3}{2}\\x\in \begin{bmatrix}y-1,\sqrt{-\left(y-5\right)\left(y+1\right)}\end{bmatrix}\text{, }&y>\frac{3-\sqrt{17}}{2}\text{ and }
y<\frac{\sqrt{17}+3}{2}\end{matrix}\right.
\left\{\begin{matrix}y=2\text{, }&x\geq 1\text{ and }x\leq 3\\y\in \begin{bmatrix}-\sqrt{9-x^{2}}+2,\sqrt{9-x^{2}}+2\end{bmatrix}\text{, }&x\geq \frac{\sqrt{17}+1}{2}\text{ and }x<3\\y=x+1\text{, }&x=\frac{1-\sqrt{17}}{2}\\y\in \begin{bmatrix}-\sqrt{9-x^{2}}+2,x+1\end{bmatrix}\text{, }&\left(x>\frac{1-\sqrt{17}}{2}\text{ and }x<\frac{\sqrt{17}+1}{2}\text{ and }|x|<3\right)\text{ or }\left(x\geq 1\text{ and }x<\frac{\sqrt{17}+1}{2}\right)\end{matrix}\right.
Объяснение:
\left\{\begin{matrix}x\in \begin{bmatrix}-\sqrt{-\left(y-5\right)\left(y+1\right)},\sqrt{-\left(y-5\right)\left(y+1\right)}\end{bmatrix}\text{, }&y\geq -1\text{ and }y\leq \frac{3-\sqrt{17}}{2}\\x=\sqrt{\left(5-y\right)\left(y+1\right)}\text{, }&y=\frac{\sqrt{17}+3}{2}\\x\in \begin{bmatrix}y-1,\sqrt{-\left(y-5\right)\left(y+1\right)}\end{bmatrix}\text{, }&y>\frac{3-\sqrt{17}}{2}\text{ and }
y<\frac{\sqrt{17}+3}{2}\end{matrix}\right.
\left\{\begin{matrix}y=2\text{, }&x\geq 1\text{ and }x\leq 3\\y\in \begin{bmatrix}-\sqrt{9-x^{2}}+2,\sqrt{9-x^{2}}+2\end{bmatrix}\text{, }&x\geq \frac{\sqrt{17}+1}{2}\text{ and }x<3\\y=x+1\text{, }&x=\frac{1-\sqrt{17}}{2}\\y\in \begin{bmatrix}-\sqrt{9-x^{2}}+2,x+1\end{bmatrix}\text{, }&\left(x>\frac{1-\sqrt{17}}{2}\text{ and }x<\frac{\sqrt{17}+1}{2}\text{ and }|x|<3\right)\text{ or }\left(x\geq 1\text{ and }x<\frac{\sqrt{17}+1}{2}\right)\end{matrix}\right.
1а)7,9 б)-3,5 в)6 2.а)3 б)12 в)3 г)20 3. а)х=±0,8 б)х=±√17 4.а)2у в 4 степени; б)-28 5. 6,1∠√38∠6,2 6. х=3
Объяснение:√196=14, √0,36=0,6
а)1/2 *14+1,5* 0,6=7+0,9=7,9
б)1,5-7 * 5/7=1,5 -5=-3,5
в)(2√1,5)²=2²*(√1,5)²=4* 1,5= 6
2.а) √0,36*25=√0,36 *√25=0,6*5=3
б)√8*√18=√(4*2*2*9)=4*3=12
в)√27/√3=√(27/3)=√9=3
г)√〖2^4〗*〖5^2〗=2²*5-4*5=20
3.а) х²=0,64
х=±0,8
б)х²=17
х=±√17
4.а) у³√4у²=у³*2у=2 у∧4
б)7а √(16/а²)=-7а* (4/а)=-28
5. 6²=36
(6,1)²=37,21
(6,2)²=38,44
6,1∠√38∠6,2
6.√(х-2)=1 поднесем до квадрата обе части уравнения
х-2=1
х=1+2
х=3