a) Ta có: \(\left|x\right|\ge0\left(\forall x\in Z\right)\)

\(\Rightarrow A=\left|x\right|+\frac{6}{13}\ge\frac{6}{13}\)

Dấu "=" xảy ra "=" |x| = 0 <=> x = 0

Vậy A_min = 6/13 khi và chỉ khi x = 0

b) Ta có: \(\left|x+2,8\right|\ge0\left(\forall x\in Z\right)\)

\(\Rightarrow B=\left|x+2,8\right|-7,9=\left|x+2,8\right|+\left(-7,9\right)\ge-7,9\)

Dấu "=" xảy ra <=> |x+2,8| = 0 <=> x + 2,8 = 0 <=> x = -2,8

Vậy B_min = -7,9 khi và chỉ khi x = -2,8

c) Ta có: \(\left|x+1,5\right|\ge0\left(\forall x\in Z\right)\)

\(\Rightarrow C=\left|x+1,5\right|-5,7=\left|x+1,5\right|+\left(-5,7\right)\ge-5,7\)

Dấu "=" xảy ra <=> |x+1,5| = 0 <=> x + 1,5 = 0 <=> x = -1,5

Vậy C_min = -5,7 khi và chỉ khi x = -1,5

đợi tý

Đã trả lời rồi còn độ tí đồ ngull

Ai lm đc câu nào thì giúp mk với , cảm ơn !!

\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)

\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)

\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)

a) Ta có: \(\left(2x-1\right)^2\ge0\forall x\)

\(\Rightarrow-3\left(2x-1\right)^2\le0\forall x\)

\(\Rightarrow-3\left(2x-1\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi 2x-1=0

\(\Leftrightarrow2x=1\)

hay \(x=\dfrac{1}{2}\)

Vậy: Giá trị lớn nhất của biểu thức \(A=5-3\left(2x-1\right)^2\) là 5 khi \(x=\dfrac{1}{2}\)

\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)

a: \(B=\left|2-x\right|+1.5>=1.5\)

Dấu '=' xảy ra khi x=2

b: \(B=-5\left|1-4x\right|-1\le-1\)

Dấu '=' xảy ra khi x=1/4

g: \(C=x^2+\left|y-2\right|-5>=-5\)

Dấu '=' xảy ra khi x=0 và y=2

a, \(A=\left|x-2017\right|+\left|2018-x\right|\ge\left|x-2017+2018-x\right|=1\)

Vậy \(Min=1\Leftrightarrow2017\le x\le2018\)

b, \(B=\dfrac{x^2+4+8}{x^2+4}=1+\dfrac{8}{x^2+4}\)

Thấy : \(x^2+4\ge4\)

\(\Rightarrow B=1+\dfrac{8}{x^2+4}\le3\)

Vậy \(Max=3\Leftrightarrow x=0\)

là GTNN á

\(A=2x^2-2\ge-2\)

Dấu "=" xảy ra khi: \(x=0\)

\(B=\left|x+\dfrac{1}{3}\right|-\dfrac{1}{6}\ge-\dfrac{1}{6}\)

Dấu "=" xảy ra khi: \(x=-\dfrac{1}{3}\)

\(C=\dfrac{\left|x\right|+2017}{2018}\ge\dfrac{2017}{2018}\)

Dấu "=" xảy ra khi: \(x=0\)

\(D=3-\left(x+1\right)^2\le3\)

Dấu "=" xảy ra khi: \(x=-1\)

\(E-\left|0,1+x\right|-1,9\le-1,9\)

Dấu "=" xảy ra khi: \(x=-0,1\)

\(F=\dfrac{1}{\left|x\right|+2017}\le\dfrac{1}{2017}\)

Dấu "=" xảy ra khi: \(x=0\)