Question

[Nguyễn Xuân Đăng]

Tìm giá trị nhỏ nhất

M=3x²-7x

N=x⁴-7x² +6

Answer 1

[tthnew]

\(M=3x^2-7x=3\left(x^2-2.x.\frac{7}{6}+\left(\frac{7}{6}\right)^2-\left(\frac{7}{6}\right)^2\right)\)

\(=3\left(x-\frac{7}{6}\right)^2-\frac{49}{12}\ge-\frac{49}{12}\)

Đẳng thức xảy ra khi x = 7/6

N = \(\left(x^4-7x^2+6\right)=x^4-2.x^2.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\left(\frac{7}{2}\right)^2+6\)

\(=\left(x^2-\frac{7}{2}\right)^2-\frac{25}{4}\ge-\frac{25}{4}\)

đẲng thức xảy ra khi \(x^2=\frac{7}{2}\Leftrightarrow x=\pm\sqrt{\frac{7}{2}}\)

Answer 2

[Zore]

Lời giải:

a, Ta có:

M = 3x² - 7x = 3 . ( x² - \(\frac{7}{3}\)x ) = 3 . [ x² - 2.\(\frac{7}{6}\)x + \(\left(\frac{7}{6}\right)^2\) - \(\left(\frac{7}{6}\right)^2\)]

= 3 . [ \(\left(x-\frac{7}{6}\right)^2\) - \(\frac{49}{36}\)]

= 3.\(\left(x-\frac{7}{6}\right)^2\) - \(\frac{49}{12}\) ≥ - \(\frac{49}{12}\) . Vì: 3.\(\left(x-\frac{7}{6}\right)^2\) ≥ 0 ∀x => M_min = - \(\frac{49}{12}\)

<=> 3.\(\left(x-\frac{7}{6}\right)^2\) = 0

<=> \(\left(x-\frac{7}{6}\right)^2\) = 0

<=> \(x-\frac{7}{6}\) = 0

<=> x = \(\frac{7}{6}\)

Vây: M_min = -\(\frac{49}{12}\) tại x = \(\frac{7}{6}\).

b, Ta có:

N = x⁴ - 7x² + 6 = [(x²)² - 2 . \(\frac{7}{2}\) . x² + \(\left(\frac{7}{2}\right)\)² ] - [\(\left(\frac{7}{2}\right)\)² - 6]

= ( x² - \(\frac{7}{2}\))² - \(\frac{25}{4}\) ≥ -\(\frac{25}{4}\) . Vì: ( x² - \(\frac{7}{2}\))² ≥ 0 ∀x => N_min= -\(\frac{25}{4}\)

<=> ( x² - \(\frac{7}{2}\))² = 0

<=> x² - \(\frac{7}{2}\) = 0

<=> x² = \(\frac{7}{2}\)

<=> x = \(\sqrt{\frac{7}{2}}\)

Vây: N_min = - \(\frac{25}{4}\) tại x = \(\sqrt{\frac{7}{2}}\)

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