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

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

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

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

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

\(y_{min}=-3\) 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 \(x=-2\)

a, \(y=3-4sin^2x.cos^2x=3-sin^22x\)

Đặt \(sin2x=t\left(t\in\left[-1;1\right]\right)\).

\(\Rightarrow y=f\left(t\right)=3-t^2\)

\(\Rightarrow y_{min}=minf\left(t\right)=2\)

\(y_{max}=maxf\left(t\right)=3\)

b, \(y=f\left(t\right)=\dfrac{-2}{3t-5}\left(t\in\left[0;1\right]\right)\)

\(\Rightarrow y_{min}=minf\left(t\right)=\dfrac{2}{5}\)

\(y_{max}=maxf\left(t\right)=1\)

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

Biểu thức nào em?

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

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

\(\left|2x-1\right|+3\ge3\Leftrightarrow\dfrac{3+\left|2x-1\right|}{14}\ge\dfrac{3}{14}\)

Dấu \("="\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

\(\dfrac{-4x^2+4x}{15}=\dfrac{-4x^2+4x-1+1}{15}=\dfrac{-\left(2x-1\right)^2+1}{15}\)

Ta có \(-\left(2x-1\right)^2+1\le1\Leftrightarrow\dfrac{-\left(2x-1\right)^2+1}{15}\le\dfrac{1}{15}\)

Dấu \("="\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

=>3x-9-10x+2=-4

=>-7x-7=-4

=>-7x=3

=>x=-3/7

b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)

=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)

=>10-2x+7x-14=4x-4+x

=>5x-4=5x-4

=>0x=0(luôn đúng)

Vậy: S=R\{0;2}