\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)

\(y_{min}=-3\) khi \(x=1\)

\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)

\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)

\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)

\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)

\(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\)

vì \(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6\le0,\forall x\inℝ\)

\(\Rightarrow B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\le3\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{4}{9}x-\dfrac{2}{15}=0\Rightarrow\dfrac{4}{9}x=\dfrac{2}{15}\Rightarrow x=\dfrac{9}{15}\)

Vậy \(GTLN\left(B\right)=3\left(tạix=\dfrac{9}{15}\right)\)

\(A=\left(2x+\dfrac{1}{3}\right)^4-1\)

vì \(\left(2x+\dfrac{1}{3}\right)^4\ge0,\forall x\inℝ\)

\(\Rightarrow A=\left(2x+\dfrac{1}{3}\right)^4-1\ge-1\)

Dấu "=" xảy ra khi và chỉ khi

\(2x+\dfrac{1}{3}=0\Rightarrow2x=-\dfrac{1}{3}\Rightarrow x=-\dfrac{1}{6}\)

\(\Rightarrow GTNN\left(A\right)=-1\left(tạix=-\dfrac{1}{6}\right)\)

\(A=\left|x+1\right|-3\\ min_A=-3.khi.x+1=0\Leftrightarrow x=-1\\ B=-\left|x-\dfrac{3}{7}\right|-\dfrac{1}{4}\\ max_B=-\dfrac{1}{4}.khi.\left(x-\dfrac{3}{7}\right)=0\Leftrightarrow x=\dfrac{3}{7}\)

a)

A = |x + 1| - 3 ≥ 0 - 3 = -3

Dấu "=" xảy ra khi x + 1 = 0 hay x = -1

Do đó A đạt GTNN là -3 khi x = -1

b)

\(B=-\left|x-\dfrac{3}{7}\right|-\dfrac{1}{4}\le-0-\dfrac{1}{4}=-\dfrac{1}{4}\)

Dấu "=" xảy ra khi khi \(x-\dfrac{3}{7}=0\) hay \(x=\dfrac{3}{7}\)

Do đó B đạt GTLN là \(-\dfrac{1}{4}\) khi \(x=\dfrac{3}{7}\)

C=|2x-3/5|+4/3>=4/3

Dấu = xảy ra khi x=3/10

D=|x-3|+|-x-2|>=|x-3-x-2|=5

Dấu = xảy ra khi -2<=x<=3

\(D=\dfrac{\left|x\right|+2023}{\left|x\right|+2022}=\dfrac{\left|x\right|+2022}{\left|x\right|+2022}+\dfrac{1}{\left|x\right|+2022}\\ =1+\dfrac{1}{\left|x\right|+2022}\)

Nhận thấy : \(\left|x\right|\ge0\forall x\inℝ\)

\(\Rightarrow\left|x\right|+2022\ge2022\)

\(\Rightarrow\dfrac{1}{\left|x\right|+2022}\le\dfrac{1}{2022}\)

\(\Rightarrow D=1+\dfrac{1}{\left|x\right|+2022}\le1+\dfrac{1}{2022}=\dfrac{2023}{2022}\)

Dấu = xảy ra khi : \(\left|x\right|=0\Rightarrow x=0\)

Vậy GTLN của D là : \(\dfrac{2023}{2022}\) tại x=0

Biểu thức nào em?

\(y=4cos^2\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)-7=2\left[cos\left(x-\dfrac{\pi}{6}\right)+1\right]-7=2cos\left(x-\dfrac{\pi}{6}\right)-5\)

Đặt \(x-\dfrac{\pi}{6}=t\Rightarrow t\in\left[-\dfrac{\pi}{6};\dfrac{5\pi}{6}\right]\)

\(\Rightarrow y=2cost-5\)

Do \(t\in\left[-\dfrac{\pi}{6};\dfrac{5\pi}{6}\right]\Rightarrow cost\in\left[-\dfrac{\sqrt{3}}{2};1\right]\)

\(\Rightarrow y\in\left[-5-\sqrt{3};-3\right]\)

\(y_{max}=-3\) khi \(t=0\) hay \(x=\dfrac{\pi}{6}\)

\(y_{min}=-5-\sqrt{3}\) khi \(y=\dfrac{5\pi}{6}\) hay \(x=\pi\)

\(a,B=4,2+\left|x+1,5\right|\ge4,2\\ B_{min}=4,2\Leftrightarrow x+1,5=0\Leftrightarrow x=-1,5\\ b,C=\dfrac{4}{5}-\left|2x+1\right|\le\dfrac{4}{5}\\ C_{max}=\dfrac{4}{5}\Leftrightarrow2x+1=0\Leftrightarrow x=-\dfrac{1}{2}\)

a, Do |x +1,5| ≥ 0 ⇒ 4,2 + |x + 1,5| ≥ 4,2

Dấu "=" xảy ra ⇔ x + 1,5 = 0 ⇔  x = - 1,5

Vậy B_min=  4,2 ⇔ x= -1,5

b, Do |2x + 1| ≥ 0 ⇒ \(\dfrac{4}{5}-\left|2x+1\right|\le\dfrac{4}{5}\)

Dấu "=" xảy ra ⇔ 2x + 1 = 0 ⇔ 2x = -1 ⇔ \(x=-\dfrac{1}{2}\)

Vậy C_max = \(\dfrac{4}{5}\Leftrightarrow x=-\dfrac{1}{2}\)

1:

a: \(A=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

căn x+1>=1

=>2/căn x+1<=2

=>-2/căn x+1>=-2

=>A>=-2+1=-1

Dấu = xảy ra khi x=0

b: