Question

tìm gtnn các bước sau :

A = x²+4x+3

B = 4x²+2x-5

C = 3x²-7x+1

giúp mình với ạ

Answer 1

\(A=x^2+4x+3=x^2+2.x.2+2^2-1=\left(x+2\right)^2-1\ge-1\)

Dấu "=" xảy ra khi : \(x+2=0\)

\(\Leftrightarrow x=-2\)

Answer 2

A=x²+4x+3

=x²+2.x.2+2²-2²+3

=(x²+2.x.2+2²)-4+3

=(x+2)²-1

Ta có:(x+2)²≥0

⇒ (x+2)²-1≥-1

A _min=-1 tại (x+2)²=0 ⇒x=-2.

B=4x²+2x-5

=(2x)²+2.2x.\(\frac{1}{2}\)+(\(\frac{1}{2}\))²-(\(\frac{1}{2}\))²-5

=[(2x)²+2.2x.\(\frac{1}{2}\)+(\(\frac{1}{2}\))² ]-\(\frac{1}{4}\)-5

=(2x+\(\frac{1}{2}\))²-\(\frac{21}{4}\)

Ta có:(2x+\(\frac{1}{2}\))²≥0

⇒(2x+\(\frac{1}{2}\))²-\(\frac{21}{4}\)≥-\(\frac{21}{4}\)

Vậy B_min=-\(\frac{21}{4}\)tại (2x+\(\frac{1}{2}\))²=0 ⇒ x=-\(\frac{1}{4}\)

Answer 3

C=3x²-7x+1

=3.(x²-\(\frac{7}{3}\)x+\(\frac{1}{3}\))

=3.[x²-2.x.\(\frac{7}{6}\)+(\(\frac{7}{6}\))²-(\(\frac{7}{6}\))²+\(\frac{1}{3}\))

=3.[(x-\(\frac{7}{6}\))²-\(\frac{37}{36}\)]

=3(x-\(\frac{7}{6}\))²-\(\frac{37}{12}\)

Ta có:3(x-\(\frac{7}{6}\))²≥0

⇒3(x-\(\frac{7}{6}\))²-\(\frac{37}{12}\)≥-\(\frac{37}{12}\)

Vậy C _min=-\(\frac{37}{12}\)tại x=\(\frac{7}{6}\)