![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta có: \(6.|3x-12|\ge0\forall x\)
\(\Rightarrow23+6.|3x-12|\ge23+0\forall x\)
Hay \(A\ge23\forall x\)
Dấu"=" xảy ra \(\Leftrightarrow3x-12=0\)
\(\Leftrightarrow x=4\)
Vậy Min A=23 \(\Leftrightarrow x=4\)
Bài 2:
Ta có: \(5.|14-7x|\ge0\forall x\)
\(\Rightarrow-5.|14-7x|\le0\forall x\)
\(\Rightarrow2019-5.|14-7x|\le2019-0\forall x\)
Hay \(B\le2019\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow14-7x=0\)
\(\Leftrightarrow x=2\)
Vậy Max B=2019 \(\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt |3x−1|=b(b≥0)|3x−1|=b(b≥0)
⇒B=b2−4b+5=(b−2)2+1≥1⇒B=b2−4b+5=(b−2)2+1≥1
Dấu "=" xảy ra khi b−2=0b−2=0
⇒|3x−1|=2⇒|3x−1|=2
⇒[3x−1=23x−1=−2⇒[3x−1=23x−1=−2
⇔⎡⎣x=1x=−13⇔[x=1x=−13
Vậy Min B = 1 khi x = 1 hoặc x = - 1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
Cái này là GTLN nha
Do \(-\left|3x-\dfrac{1}{4}\right|\le0\)
\(\Rightarrow D=-\left|3x-\dfrac{1}{3}\right|-\dfrac{1}{2}\le-\dfrac{1}{2}\)
\(maxD=-\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{12}\)
ta có:
\(-\left|3x-\dfrac{1}{4}\right|\) ≤ 0
=> \(-\left|3x-\dfrac{1}{4}\right|\)≤-1/2
GTNN:D= -1/2 khi 3x-1/4=0
3x=1/4
x=1/12
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Do \(\left|\dfrac{1}{3}x-2\right|\ge0\)
\(\Rightarrow B=-9+\left|\dfrac{1}{3}x-2\right|\ge-9\)
\(minB=-9\Leftrightarrow x-6\)
Cái dấu bằng xảy ra là x=6 nhé