giải BPT: √(x+3 ) + √(68-4x) ≥ x2 – 2x + 11 11/10/2021 Bởi Kinsley giải BPT: √(x+3 ) + √(68-4x) ≥ x2 – 2x + 11
theo quy tắc BPT √(x+3)+√(68-4x)≥ x2 – 2x +11 =√(x+3)+√(64x)≥x2-13x =√(x+3)+√(64x)≥x-11x =(x+1,5)+(32x)≥x-11x =x+30≥x-11x Bình luận
theo quy tắc BPT
√(x+3)+√(68-4x)≥ x2 – 2x +11
=√(x+3)+√(64x)≥x2-13x
=√(x+3)+√(64x)≥x-11x
=(x+1,5)+(32x)≥x-11x
=x+30≥x-11x