(9x^2-6x+1)+y^2+4

=(3x-1)^2+y^2+4

ta có (3x-1)^2>= 0

=>(3x-1)^2+y^2>=0

=>(3x-1)^2+y^2+4>=4

GTNN biểu thức là 4 và xảy ra khi 3x-1=0=>x=1/3, y=0

`2/[6x-5-9x^2]`

`=-2/[9x^2-6x+5]`

`=-2/[(3x-1)^2+4]`

Vì `(3x-1)^2 >= 0 AA x`

`<=>(3x-1)^2+4 >= 4 AA x`

`<=>1/[(3x-1)^2+4] <= 1/4`

`<=>-2/[(3x-1)^2+4] >= -1/2 AA x`

   `=>Mi n=-1/2`

Dấu "`=`" xảy ra `<=>3x-1=0<=>x=1/3`

Biểu thức này ko tồn tại cả min lẫn max

\(\dfrac{1}{M}=\dfrac{\sqrt{x}-1}{2\sqrt{x}+4}=\dfrac{-\dfrac{1}{4}\left(2\sqrt{x}+4\right)+\dfrac{\sqrt{x}}{2}}{2\sqrt{x}+4}=-\dfrac{1}{4}+\dfrac{\sqrt{x}}{4\left(\sqrt{x}+2\right)}\)

Do \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}+2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\sqrt{x}}{4\left(\sqrt{x}+2\right)}\ge0\)

\(\Rightarrow\dfrac{1}{M}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(x=0\)

Đề yêu cầu là gì bạn?

=x^2+7x+49/4-8133/4

=(x+7/2)^2-8133/4>=-8133/4

Dấu = xảy ra khi x=-7/2

\(x^2+7x-2021\\ =x^2+7x+12,25-2033,25\\ =\left(x+3,5\right)^2-2033,5\\ =-2033,5+\left(x+3,5\right)^2\)

\(Vì\) \(\left(x+3,5\right)^2\ge0\)

Nên GTNN của biểu thức là \(-2033,5\) khi \(x+3,5=0\Leftrightarrow x=-3,5\)

\(y=\dfrac{cos3x}{1-sinx}+tanx=\dfrac{cos3x}{1-sinx}+\dfrac{sinx}{cosx}\)

Hàm số xác định khi \(\left\{{}\begin{matrix}1-sinx\ne0\\cosx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sinx\ne1\\cosx\ne0\end{matrix}\right.\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\)

\(M=\left|3x-2\right|+\left|3x-6\right|=\left|3x-2\right|+\left|6-3x\right|\ge\left|3x-2+6-3x\right|=4\)

\(M_{min}=4\) khi \(\dfrac{3}{2}\le x\le2\)

Với \(x\ge\dfrac{1}{3}\Leftrightarrow B=9x^2-6x-4\left(3x-1\right)+6=9x^2-18x+10\)

\(B=9\left(x^2-2x+1\right)+1=9\left(x-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=1\left(1\right)\)

Với \(x< \dfrac{1}{3}\Leftrightarrow B=9x^2-6x+4\left(3x-1\right)+6=9x^2+6x+2\)

\(B=\left(9x^2+6x+1\right)+1=\left(3x+1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=-\dfrac{1}{3}\left(2\right)\)

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

\(C=2x^2+4x+2+3x^2+12x+12-4x^2-24x-36\\ C=x^2-8x-22=\left(x^2-8x+16\right)-38=\left(x-4\right)^2-38\ge-38\\ C_{min}=-38\Leftrightarrow x=4\)

Với \(x\ge\dfrac{1}{6}\Leftrightarrow A=5x^2-6x+1-1=5x^2-6x\)

\(A=5\left(x^2-2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{9}{5}=5\left(x-\dfrac{3}{5}\right)^2-\dfrac{9}{5}\ge-\dfrac{9}{5}\\ A_{min}=-\dfrac{9}{5}\Leftrightarrow x=\dfrac{3}{5}\left(1\right)\)

Với \(x< \dfrac{1}{6}\Leftrightarrow A=5x^2+6x-1-1=5x^2+6x-2\)

\(A=5\left(x^2+2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{19}{5}=5\left(x+\dfrac{3}{5}\right)^2-\dfrac{19}{5}\ge-\dfrac{19}{5}\\ A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\)

a: \(M=\dfrac{2\left(1-3x\right)\left(1+3x\right)}{3x\left(x+2\right)}\cdot\dfrac{3x}{2\left(1-3x\right)}=\dfrac{3x+1}{x+2}\)