tìm giá trị nhỏ nhất của e=2x^2+3x-1 f=(2x^2+1)^2+3x^2+10 03/07/2021 Bởi Gabriella tìm giá trị nhỏ nhất của e=2x^2+3x-1 f=(2x^2+1)^2+3x^2+10
Đáp án: `E=2x^2+3x-1` `=2(x^2+3/2x)-1` `=2(x^2+3/4x+3/4x+9/16-9/16)-1` `=2[x(x+3/4)+3/4(x+3/4)]-9/8-1` `=2(x+3/4)(x+3/4)-17/8` `=2(x+3/4)^2-17/8>=-17/8` Dấu “=” xảy ra khi `x=-3/4` `F=(2x^2+1)^2+3x^2+10` Vì `2x^2>=0=>2x^2+1>=1` `=>F>=1+0+10=11` Dấu “=” xảy ra khi `x=0` Bình luận
$E=2x^2+3x-1\\=2\left(x^2+\dfrac{3}{2}x-\dfrac{1}{2}\right)\\=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}\right)\\=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{17}{8}$ Nhận thấy $\left(x+\dfrac{3}{4}\right)^2≥0$ $→2\left(x+\dfrac{3}{4}\right)^2≥0\\→E≥-\dfrac{17}{8}\\→\min E=-\dfrac{17}{8}$ $→$ Dấu “=” xảy ra khi $x+\dfrac{3}{4}=0$ $↔x=-\dfrac{3}{4}$ Vậy $\min E=-\dfrac{17}{8}$ khi $x=-\dfrac{3}{4}$ $\\\\$ $F=(2x^2+1)^2+3x^2+10\\=4x^4+4x^2+1+3x^2+10\\=4x^4+7x^2+10\\=(2x^2)^2+2.2x^2.\dfrac{7}{4}+\dfrac{49}{16}+\dfrac{111}{16}\\=\bigg(2x^2+\dfrac{7}{4}\bigg)^2+\dfrac{111}{16}$ Vì $x^2≥0→2x^2≥0$ $→2x^2+\dfrac{7}{4}≥\dfrac{7}{4}\\→\bigg(2x^2+\dfrac{7}{4}\bigg)^2≥\dfrac{49}{16}\\→F≥\dfrac{49}{16}+\dfrac{111}{16}\\→F≥10\\→\min F=10$ $→$ Dấu “=” xảy ra khi $2x^2=0$ $↔x^2=0\\↔x=0$ Vậy $\min F=10$ khi $x=0$ Bình luận
Đáp án:
`E=2x^2+3x-1`
`=2(x^2+3/2x)-1`
`=2(x^2+3/4x+3/4x+9/16-9/16)-1`
`=2[x(x+3/4)+3/4(x+3/4)]-9/8-1`
`=2(x+3/4)(x+3/4)-17/8`
`=2(x+3/4)^2-17/8>=-17/8`
Dấu “=” xảy ra khi `x=-3/4`
`F=(2x^2+1)^2+3x^2+10`
Vì `2x^2>=0=>2x^2+1>=1`
`=>F>=1+0+10=11`
Dấu “=” xảy ra khi `x=0`
$E=2x^2+3x-1\\=2\left(x^2+\dfrac{3}{2}x-\dfrac{1}{2}\right)\\=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}\right)\\=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{17}{8}$
Nhận thấy $\left(x+\dfrac{3}{4}\right)^2≥0$
$→2\left(x+\dfrac{3}{4}\right)^2≥0\\→E≥-\dfrac{17}{8}\\→\min E=-\dfrac{17}{8}$
$→$ Dấu “=” xảy ra khi $x+\dfrac{3}{4}=0$
$↔x=-\dfrac{3}{4}$
Vậy $\min E=-\dfrac{17}{8}$ khi $x=-\dfrac{3}{4}$
$\\\\$
$F=(2x^2+1)^2+3x^2+10\\=4x^4+4x^2+1+3x^2+10\\=4x^4+7x^2+10\\=(2x^2)^2+2.2x^2.\dfrac{7}{4}+\dfrac{49}{16}+\dfrac{111}{16}\\=\bigg(2x^2+\dfrac{7}{4}\bigg)^2+\dfrac{111}{16}$
Vì $x^2≥0→2x^2≥0$
$→2x^2+\dfrac{7}{4}≥\dfrac{7}{4}\\→\bigg(2x^2+\dfrac{7}{4}\bigg)^2≥\dfrac{49}{16}\\→F≥\dfrac{49}{16}+\dfrac{111}{16}\\→F≥10\\→\min F=10$
$→$ Dấu “=” xảy ra khi $2x^2=0$
$↔x^2=0\\↔x=0$
Vậy $\min F=10$ khi $x=0$