vao olm thì biê liền OK

ăn clkt

HẾT RỒI NHÉ ĐÁP ÁN LÀ : 

+ Ta có: y '= 3x2 + 6x + m

      + Để hàm số đã cho đồng biến trên R thì y' ≥ 0,∀x ∈R

      + Yêu cầu bài toán trở thành tìm điều kiện của m để y' ≥ 0,∀x ∈R

Ta có y' = 3x2 + 6x + m, ta có: a = 3>0,Δ = 36 - 12m

Để y' ≥ 0,∀x ∈ R khi Δ ≤ 0 ⇔ 36 - 12m ≤ 0 ⇔ m ≥ 3

Vậy giá trị của tham số m cần tìm là m ≥ 3

me fan ó

0

(17+35-42+58-23-45):79=0

huhhuh

huhuhuhuhuhuhu

Số số hạng có trong dãy là:

\(\frac{90-11}{1}+1=80\)(số hạng)

\(11+12+13+14+...+90\)

\(=\frac{\left(90+11\right)\cdot80}{2}\)

\(=4040\)

Đáp số: \(4040\)

Trả lời

Số lượng số hạng của dãy số trên là:

    (90-11):1+1=80(số hạng)

Tổng của dãy số trên là:

   (90+11).80:2=4040

Vậy:

11+12+13+14+...+88+89+90=4040 nha !

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Thui khỏi bt lm òi * nhắn lun ko bt cho nhanh lại cn lm màu *

mik đồng ý vs ý kiến của bạn

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