Question

bài 4 thôi

Answer 1

4:

a: A=|x-7/4|+8/5

|x-7/4|>=0

=>|x-7/4|+8/5>=8/5

Dấu = xảy ra khi x=7/4

b: B=|x-5|+|x+3/4|

=>B=|x+3/4|+|5-x|>=|x+3/4+5-x|=23/4

Dấu = xảy ra khi (x-5)(x+3/4)<=0

=>-3/4<=x<=5

Answer 2

a)\(A=\left|x-\dfrac{7}{4}\right|+1\dfrac{3}{5}\)

\(A=\left|x-\dfrac{7}{4}\right|+\dfrac{8}{5}\) 

Mà: \(\left|x-\dfrac{7}{4}\right|\ge0\forall x\) nên
\(\Rightarrow A=\left|x-\dfrac{7}{4}\right|+\dfrac{8}{5}\ge\dfrac{8}{5}\)

Dấu "=" xảy ra:

\(\left|x-\dfrac{7}{4}\right|+\dfrac{8}{5}=\dfrac{8}{5}\)

\(\Rightarrow x=\dfrac{7}{4}\)

Vậy: \(A_{min}=\dfrac{8}{5}\) khi \(x=\dfrac{7}{4}\)

b) \(B=\left|x-5\right|+\left|x+\dfrac{3}{4}\right|\)

Mà: \(B=\left|5-x\right|+\left|x+\dfrac{3}{4}\right|\ge\left|5-x+x+\dfrac{3}{4}\right|=\dfrac{23}{4}\)

Dấu "=" xảy ra:

\(\left(5-x\right)\left(x+\dfrac{3}{4}\right)\le0\)

\(\Rightarrow\left\{{}\begin{matrix}x\le5\\x\ge-\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow-\dfrac{3}{4}\le x\le5\)

Vậy: \(B_{min}=\dfrac{23}{4}\) khi \(-\dfrac{3}{4}\le x\le5\)