x= $\frac{8-m}{m+2}$ y= $\frac{5}{m+2}$ Tìm m sao cho S= $x^{2}$ – $y^{2}$ đạt GTNN 04/11/2021 Bởi Delilah x= $\frac{8-m}{m+2}$ y= $\frac{5}{m+2}$ Tìm m sao cho S= $x^{2}$ – $y^{2}$ đạt GTNN
$x=\dfrac{8-m}{m+2}\to {{x}^{2}}=\dfrac{{{m}^{2}}-16m+64}{{{\left( m+2 \right)}^{2}}}$ $y=\dfrac{5}{m+2}\to {{y}^{2}}=\dfrac{25}{{{\left( m+2 \right)}^{2}}}$ $S={{x}^{2}}-{{y}^{2}}$ $S=\dfrac{{{m}^{2}}-16m+64}{{{\left( m+2 \right)}^{2}}}-\dfrac{25}{{{\left( m+2 \right)}^{2}}}$ $S=\dfrac{{{m}^{2}}-16m+39}{{{\left( m+2 \right)}^{2}}}$ $S=\dfrac{{{m}^{2}}+4m+4}{{{\left( m+2 \right)}^{2}}}+\dfrac{-20m-40}{{{\left( m+2 \right)}^{2}}}+\dfrac{75}{{{\left( m+2 \right)}^{2}}}$ $S=\dfrac{{{\left( m+2 \right)}^{2}}}{{{\left( m+2 \right)}^{2}}}-\dfrac{20\left( m+2 \right)}{{{\left( m+2 \right)}^{2}}}+\dfrac{75}{{{\left( m+2 \right)}^{2}}}$ $S=1-\dfrac{20}{m+2}+\dfrac{75}{{{\left( m+2 \right)}^{2}}}$ $S=\dfrac{75}{{{\left( m+2 \right)}^{2}}}-\dfrac{20}{m+2}+1$ $S=\dfrac{75}{{{\left( m+2 \right)}^{2}}}-\dfrac{20}{m+2}+\dfrac{4}{3}-\dfrac{1}{3}$ $S={{\left( \dfrac{\sqrt{75}}{m+2}-\dfrac{2}{\sqrt{3}} \right)}^{2}}-\dfrac{1}{3}\,\,\,\ge \,\,\,-\dfrac{1}{3}$ Dấu “=” xảy ra khi và chỉ khi $\dfrac{\sqrt{75}}{m+2}=\dfrac{2}{\sqrt{3}}\Leftrightarrow m=\dfrac{11}{2}$ Vậy giá trị nhỏ nhất của $S$ bằng $-\dfrac{1}{3}$ khi $m=\dfrac{11}{2}$ Bình luận
Ta có : $x=\dfrac{8-m}{m+2} = \dfrac{10-(m+2)}{m+2} = \dfrac{10}{m+2} – 1$ Và $y=\dfrac{5}{m+2}$ với $ m \neq -2$ Đặt $\dfrac{5}{m+2} =a$. Khi đó ta có : $x^2-y^2 = (2a-1)^2-a^2$ $ = 4a^2-4a+1-a^2 = 3a^2-4a+1 =3.\bigg(a-\dfrac{2}{3}\bigg)^2-\dfrac{1}{3} ≥ -\dfrac{1}{3}$ Dấu “=” xảy ra $⇔a=\dfrac{2}{3} $ $⇔\dfrac{5}{m+2} = \dfrac{2}{3}$ $⇔m=\dfrac{11}{2}$ ( Thỏa mãn ) Vậy Min $S = \dfrac{-1}{3}$ khi $m=\dfrac{11}{2}$ Bình luận
$x=\dfrac{8-m}{m+2}\to {{x}^{2}}=\dfrac{{{m}^{2}}-16m+64}{{{\left( m+2 \right)}^{2}}}$
$y=\dfrac{5}{m+2}\to {{y}^{2}}=\dfrac{25}{{{\left( m+2 \right)}^{2}}}$
$S={{x}^{2}}-{{y}^{2}}$
$S=\dfrac{{{m}^{2}}-16m+64}{{{\left( m+2 \right)}^{2}}}-\dfrac{25}{{{\left( m+2 \right)}^{2}}}$
$S=\dfrac{{{m}^{2}}-16m+39}{{{\left( m+2 \right)}^{2}}}$
$S=\dfrac{{{m}^{2}}+4m+4}{{{\left( m+2 \right)}^{2}}}+\dfrac{-20m-40}{{{\left( m+2 \right)}^{2}}}+\dfrac{75}{{{\left( m+2 \right)}^{2}}}$
$S=\dfrac{{{\left( m+2 \right)}^{2}}}{{{\left( m+2 \right)}^{2}}}-\dfrac{20\left( m+2 \right)}{{{\left( m+2 \right)}^{2}}}+\dfrac{75}{{{\left( m+2 \right)}^{2}}}$
$S=1-\dfrac{20}{m+2}+\dfrac{75}{{{\left( m+2 \right)}^{2}}}$
$S=\dfrac{75}{{{\left( m+2 \right)}^{2}}}-\dfrac{20}{m+2}+1$
$S=\dfrac{75}{{{\left( m+2 \right)}^{2}}}-\dfrac{20}{m+2}+\dfrac{4}{3}-\dfrac{1}{3}$
$S={{\left( \dfrac{\sqrt{75}}{m+2}-\dfrac{2}{\sqrt{3}} \right)}^{2}}-\dfrac{1}{3}\,\,\,\ge \,\,\,-\dfrac{1}{3}$
Dấu “=” xảy ra khi và chỉ khi $\dfrac{\sqrt{75}}{m+2}=\dfrac{2}{\sqrt{3}}\Leftrightarrow m=\dfrac{11}{2}$
Vậy giá trị nhỏ nhất của $S$ bằng $-\dfrac{1}{3}$ khi $m=\dfrac{11}{2}$
Ta có : $x=\dfrac{8-m}{m+2} = \dfrac{10-(m+2)}{m+2} = \dfrac{10}{m+2} – 1$
Và $y=\dfrac{5}{m+2}$ với $ m \neq -2$
Đặt $\dfrac{5}{m+2} =a$. Khi đó ta có :
$x^2-y^2 = (2a-1)^2-a^2$
$ = 4a^2-4a+1-a^2 = 3a^2-4a+1 =3.\bigg(a-\dfrac{2}{3}\bigg)^2-\dfrac{1}{3} ≥ -\dfrac{1}{3}$
Dấu “=” xảy ra $⇔a=\dfrac{2}{3} $ $⇔\dfrac{5}{m+2} = \dfrac{2}{3}$
$⇔m=\dfrac{11}{2}$ ( Thỏa mãn )
Vậy Min $S = \dfrac{-1}{3}$ khi $m=\dfrac{11}{2}$