Đáp án:`-1/2<=M<=3/2`
Giải thích các bước giải:
`M=y^2+\sqrt{3}xy`
`<=>2M=2y^2+2\sqrt{3}xy`
Xét `2M+1`
`2M+1=2y^2+2\sqrt{3}xy+1`
`2M+1=2y^2+2\sqrt{3}xy+x^2+y^2`
`2M+1=3y^2+2\sqrt{3}y.x+x^2`
`2M+1=(\sqrt{3}y+x)^2>=0`
`<=>2M+1>=0`
`<=>2M>=-1`
`<=>M>=-1/2`
Dấu "=" xảy ra khi `{(x^2+y^2=1),(-\sqrt{3}y=x):}`
`<=>{(3y^2+y^2=1),(-\sqrt{3}y=x):}`
`<=>{(4y^2=1),(-\sqrt{3}y=x):}`
`<=>` \(\left[ \begin{array}{l}\begin{cases}y=\dfrac12\\x=-\sqrt{3}y=\dfrac{-\sqrt{3}}{2}\\\end{cases}\\\begin{cases}y=-\dfrac12\\x=-\sqrt{3}y=\dfrac{\sqrt{3}}{2}\\\end{cases}\end{array} \right.\)
Xét `2M-3`
`2M-3=2y^2+2\sqrt{3}xy-3`
`2M-3=2y^2+2\sqrt{3}xy-3(x^2+y^2)`
`2M-3=-y^2+2\sqrt{3}xy-3x^2`
`2M-3=-(y^2-2\sqrt{3}xy+3x^2)`
`2M-3=-(y-\sqrt{3}x)^2<=0`
`<=>2M-3<=0`
`<=>2M<=3`
`<=>M<=3/2.`
Dấu "=" xảy ra khi
`{(x^2+y^2=1),(y=\sqrt{3}x):}`
`<=>{(3x^2+x^2=1),(\sqrt{3}x=y):}`
`<=>{(4x^2=1),(\sqrt{3}x=y):}`
`<=>` \(\left[ \begin{array}{l}\begin{cases}x=\dfrac12\\y=\sqrt{3}x=\dfrac{\sqrt{3}}{2}\\\end{cases}\\\begin{cases}x=-\dfrac12\\y=-\sqrt{3}x=\dfrac{-\sqrt{3}}{2}\\\end{cases}\end{array} \right.\)
