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

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

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

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

Biểu thức nào em?

cả hai ạ

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

\(y_{min}=-3\) 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 \(x=-2\)

\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)

Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)

Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{21}{15}\)

=>-10<=x<=-13/7

hay \(x\in\left\{-10;-9;...;-2\right\}\)

b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{-11}{12}\)

=>-13/9<=x<=11/18

hay \(x\in\left\{-1;0\right\}\)

a: \(\Leftrightarrow15\left(x-1\right)-2\left(7x+3\right)\le10\left(2x+1\right)+6\left(3-2x\right)\)

\(\Leftrightarrow15x-15-14x-6\le20x+10+18-12x\)

=>x-21<=8x+28

=>-7x<=49

hay x>=-7

b: \(\Leftrightarrow20\left(2x+1\right)-15\left(2x^2+3\right)< 10x\left(5-3x\right)-12\left(4x+1\right)\)

\(\Leftrightarrow40x+20-30x^2-45< 50x-30x^2-48x-12\)

=>40x-25<2x-12

=>38x<13

hay x<13/38

\(a,\dfrac{x-1}{2}-\dfrac{7x+3}{15}\le\dfrac{2x+1}{3}+\dfrac{3-2x}{5}\\ \Leftrightarrow\dfrac{15\left(x-1\right)}{30}-\dfrac{2\left(7x+3\right)}{30}\le\dfrac{10\left(2x+1\right)}{30}+\dfrac{6\left(3-2x\right)}{30}\\ \Leftrightarrow15x-15-14x-6\le20x+10+18-12x\\ \Leftrightarrow x-21\le8x+28\\ \Leftrightarrow7x+49\ge0\\ \Leftrightarrow x\ge-7\)

\(b,\dfrac{2x+1}{-3}-\dfrac{2x^2+3}{-4}>\dfrac{x\left(5-3x\right)}{-6}-\dfrac{4x+1}{-5}\\ \Leftrightarrow\dfrac{20\left(2x+1\right)}{-60}-\dfrac{15\left(2x^2+3\right)}{-60}>\dfrac{10x\left(5-3x\right)}{-60}-\dfrac{12\left(4x+1\right)}{-60}\\ \Leftrightarrow40x+20-30x^2-45>50x-30x^2-48x-12\\ \Leftrightarrow38x-13>0\\ \Leftrightarrow x>\dfrac{13}{38}\)

2. \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{4}\)

\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-3}{2}\right)=\dfrac{-21}{4}.\dfrac{2}{7}\)

\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-15}{10}\right)=\dfrac{-3}{2}\)

\(\Leftrightarrow x.\dfrac{6}{5}=\dfrac{-3}{2}\)

\(\Leftrightarrow x=\dfrac{-3}{2}:\dfrac{6}{5}\)

\(\Leftrightarrow x=\dfrac{-3}{2}.\dfrac{5}{6}\)

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

3.\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=1\\2x-\dfrac{3}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1+\dfrac{3}{4}\\2x=\left(-1\right)+\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=\dfrac{-7}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}.\dfrac{1}{2}\\x=\dfrac{-7}{3}.\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{-7}{6}\end{matrix}\right.\)

vậy \(x\in\left\{\dfrac{7}{6};\dfrac{-7}{6}\right\}\)