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27 tháng 10 2021

vao olm thì biê liền OK

3 tháng 4 2019

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

29 tháng 4 2021

me fan ó

29 tháng 4 2021
Why i don"t like blackpink
9 tháng 3 2023

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

3 tháng 7 2019

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: \(4040\)

3 tháng 7 2019

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 !

4 tháng 2 2020

mình 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4 tháng 2 2020

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

8 tháng 3 2022

mik đồng ý vs ý kiến của bạn

22 tháng 3 2022

đúng r