Toán Cho 4x+y=1 Chứng minh rằng: $4x^{2}$+ $y^{2}$ ≥ $\frac{1}{5}$ 28/10/2021 By Kennedy Cho 4x+y=1 Chứng minh rằng: $4x^{2}$+ $y^{2}$ ≥ $\frac{1}{5}$
Ta có $4x+y=1$ $\to y = 1-4x$ Ta có : $4x^2+y^2$ $ = 4x^2+(1-4x)^2$ $ = 20x^2-8y+1$ $ 20.\bigg(x^2-\dfrac{2}{5}x+\dfrac{1}{20}\bigg)$ $ = 20.\bigg(x^2-2.x.\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{100}\bigg)$ $ = 20.\bigg(x-\dfrac{1}{5}\bigg)^2 +\dfrac{1}{5} ≥ \dfrac{1}{5}$ Dấu “=” xảy ra $⇔x=y=\dfrac{1}{5}$ Trả lời
$4x+y=1⇒y=1-4x$ $⇒4x^2+y^2$ $=4x^2+(1-4x)^2$ $=4x^2+1-8x+16x^2$ $=20x^2-8x+1$ $=20(x^2-\frac{2}{5}x+\frac{1}{20})$ $=20(x-\frac{1}{5})^2+\frac{1}{5}≥\frac{1}{5}(đpcm)$. Trả lời
Ta có $4x+y=1$
$\to y = 1-4x$
Ta có : $4x^2+y^2$
$ = 4x^2+(1-4x)^2$
$ = 20x^2-8y+1$
$ 20.\bigg(x^2-\dfrac{2}{5}x+\dfrac{1}{20}\bigg)$
$ = 20.\bigg(x^2-2.x.\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{100}\bigg)$
$ = 20.\bigg(x-\dfrac{1}{5}\bigg)^2 +\dfrac{1}{5} ≥ \dfrac{1}{5}$
Dấu “=” xảy ra $⇔x=y=\dfrac{1}{5}$
$4x+y=1⇒y=1-4x$
$⇒4x^2+y^2$
$=4x^2+(1-4x)^2$
$=4x^2+1-8x+16x^2$
$=20x^2-8x+1$
$=20(x^2-\frac{2}{5}x+\frac{1}{20})$
$=20(x-\frac{1}{5})^2+\frac{1}{5}≥\frac{1}{5}(đpcm)$.