Tìm giá trị nhỏ nhất: b) B = 3x2 + y2 + 4x – y .
Giải thích
b) Ta có: B = 3x2 + y2 + 4x – y
=3x2+43x+49+y2−y+14−1912
=3x+232+y−122−1912
⇒B=3x+232+y−122−1912≥−1912.
Dấu = xảy ra khi x+23=0 ; y−12=0 .
Vậy Bmin=−1912⇔x=−23y=12 .
b) Ta có: B = 3x2 + y2 + 4x – y
=3x2+43x+49+y2−y+14−1912
=3x+232+y−122−1912
⇒B=3x+232+y−122−1912≥−1912.
Dấu = xảy ra khi x+23=0 ; y−12=0 .
Vậy Bmin=−1912⇔x=−23y=12 .