Bài 2:

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

  Dấu bằng xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

  Vậy \(A_{Min}=-2\) khi \(x=-1\)

Bài 1:

a) Ta có: \(2x^2-6=0\)

\(\Leftrightarrow2x^2=6\)

\(\Leftrightarrow x^2=3\)

hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

Vậy: \(S=\left\{\sqrt{3};-\sqrt{3}\right\}\)

1.

\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)

\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)

\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max

2.

\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)

\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)

\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)

\(E_{min}=-1\) khi \(x=0\)

\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)

\(G_{min}=-2\) khi \(x=2\)

Đặt \(x+1=t\Rightarrow x=t-1\)

\(P=\dfrac{3\left(t-1\right)^2-2\left(t-1\right)-1}{t^2}=\dfrac{3t^2-8t+4}{t^2}=\dfrac{4}{t^2}-\dfrac{8}{t}+3=4\left(\dfrac{1}{t}-1\right)^2-1\ge-1\)

\(P_{min}=-1\) khi \(t=1\Rightarrow x=0\)

Giusp e ạ !

Áp dụng bất đẳng thức AM - GM ta có:

\(f\left(x\right)=x+\left(2x+\dfrac{1}{2x}\right)\ge1+2\sqrt{2x.\dfrac{1}{2x}}=3\).

Dấu "=" xảy ra khi và chỉ khi x = 1.

Vậy...

a.

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

\(y\left(1\right)=0\) ; \(y\left(2\right)=\dfrac{1}{2}\) ; \(y\left(5\right)=\dfrac{2}{5}\)

\(\Rightarrow y_{min}=y\left(1\right)=0\)

\(y_{max}=y\left(2\right)=\dfrac{1}{2}\)

b.

\(y'=\dfrac{1-3x}{\sqrt{\left(x^2+1\right)^3}}< 0\) ; \(\forall x\in\left[1;3\right]\Rightarrow\) hàm nghịch biến trên [1;3]

\(\Rightarrow y_{max}=y\left(1\right)=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)

\(y_{min}=y\left(3\right)=\dfrac{6}{\sqrt{10}}=\dfrac{3\sqrt{10}}{5}\)

c.

\(y=1-cos^2x-cosx+1=-cos^2x-cosx+2\)

Đặt \(cosx=t\Rightarrow t\in\left[-1;1\right]\)

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

\(f'\left(t\right)=-2t-1=0\Rightarrow t=-\dfrac{1}{2}\)

\(f\left(-1\right)=2\) ; \(f\left(1\right)=0\) ; \(f\left(-\dfrac{1}{2}\right)=\dfrac{9}{4}\)

\(\Rightarrow y_{min}=0\) ; \(y_{max}=\dfrac{9}{4}\)

d.

Đặt \(sinx=t\Rightarrow t\in\left[-1;1\right]\)

\(y=f\left(t\right)=t^3-3t^2+2\Rightarrow f'\left(t\right)=3t^2-6t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\notin\left[-1;1\right]\end{matrix}\right.\)

\(f\left(-1\right)=-2\) ; \(f\left(1\right)=0\) ; \(f\left(0\right)=2\)

\(\Rightarrow y_{min}=-2\) ; \(y_{max}=2\)

1: Khi x=2 thì \(A=\dfrac{4\cdot2+1}{2-1}=9\)

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

a: \(A=\left|3x-9\right|+1.5\ge1.5\forall x\)

Dấu '=' xảy ra khi x=3

b: \(B=\left|x-7\right|-14\ge-14\forall x\)

Dấu '=' xảy ra khi x=7

a, Ta có \(A=\left|3x-9\right|+1,5\)

Ta thấy: \(\left|3x-9\right|\ge0\Rightarrow\left|3x-9\right|+1,5\ge1,5\Rightarrow A\ge1,5\)

Dấy"=" xảy ra \(\Leftrightarrow3x-9=0\Leftrightarrow3x=9\Leftrightarrow x=3\)

Vậy \(A_{min}=1,5\Leftrightarrow x=0\)

b, Ta có \(B=\left|x-7\right|-14\)

Ta thấy: \(\left|x-7\right|\ge0\Rightarrow\left|x-7\right|-14\ge-14\Rightarrow B\ge-14\)

Dấu "=" xảy ra \(\Leftrightarrow x-7=0\Leftrightarrow x=7\)

Vậy \(B_{min}=-14\Leftrightarrow x=7\)

c, Ta có: \(C=-\left|\dfrac{1}{2}x-4\right|+13\Rightarrow C=13-\left|\dfrac{1}{2}x-4\right|\)

Ta thấy: \(\left|\dfrac{1}{2}x-4\right|\ge0\Rightarrow13-\left|\dfrac{1}{2}x-4\right|\le13\Rightarrow C\le13\)

Dấu "=" xảy ra \(\Leftrightarrow\dfrac{1}{2}x-4=0\Leftrightarrow\dfrac{1}{2}x=4\Leftrightarrow x=8\)

Vậy \(C_{max}=13\Leftrightarrow x=8\)

d, Ta có: \(D=-\left|1,5-x\right|-14\Rightarrow D=-14-\left|1,5-x\right|\)

Ta thấy: \(\left|1,5-x\right|\ge0\Rightarrow-14-\left|1,5-x\right|\le-14\Rightarrow D\le-14\)

Dấu "=" xảy ra\(\Leftrightarrow1,5-x=0\Rightarrow x=1,5\)

Vậy \(D_{max}=-14\Leftrightarrow x=1,5\)

Hoctot