\(\dfrac{7}{P}\) chỉ có GTLN chứ ko có GTNN

a)ĐK:`-3x+5>=0`

`<=>5>=3x`

`<=>x<=5/3`

b)ĐK:`5/(2x+7)>=0(x ne -7/2)`

Mà `5>0`

`=>2x+7>0`

`<=>2x> -7`

`<=>x> -7/2`

c)ĐK:`(-4x+12)/(-8)>=0`

`<=>(-4(x-3))/(-4.2)>=0`

`<=>(x-3)/2>=0`

`<=>x-3>=0`

`<=>x>=3`

a, ĐKXĐ : \(\dfrac{-3x+5}{5}\ge0\)

\(\Leftrightarrow-3x+5\ge0\)

\(\Leftrightarrow x\le\dfrac{5}{3}\)

Vậy ..

b, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{5}{2x+7}\ge0\\2x+7\ne0\end{matrix}\right.\)

\(\Leftrightarrow2x+7>0\)

\(\Leftrightarrow x>-\dfrac{7}{2}\)

Vậy ...

c, ĐKXĐ : \(\dfrac{-4x+12}{-8}\ge0\)

\(\Leftrightarrow-4x+12\le0\)

\(\Leftrightarrow x\ge3\)

Vậy ...

Bài 2: 

Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)

ĐKXĐ: \(\left\{{}\begin{matrix}5x-2>0\\3-2x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{2}{5}\\x\le\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow\dfrac{2}{5}< x\le\dfrac{3}{2}\)

\(a,\dfrac{-5}{x+6}\ge0\\ mà\left(-5< 0\right)\\ \Rightarrow x+6< 0\\ \Rightarrow x< -6\\ b,\dfrac{2}{6-x}\ge0\\ mà\left(2>0\right)\\ \Rightarrow6-x>0\\ \Rightarrow x< 6\\ c,\dfrac{-x+3}{-6}\ge0\\ mà-6< 0\\ \Rightarrow-x+3< 0\\ \Rightarrow x>3\\\)

\(d,\dfrac{7x-1}{-9}\ge0\\mà-9< 0\\ \Rightarrow 7x-1\le0\\ \Rightarrow x\le\dfrac{1}{7}\\ e,\dfrac{x+2}{x^2+2x+1}\ge0\\ mà\left(x^2+2x+1\right)>0\forall x\\ \Rightarrow x+2\ge0\\ \Rightarrow x\ge-2\\ f,\dfrac{x-2}{x^2-2x+4}\ge0\\ mà\left(x^2-2x+4\right)>0\forall x\\ \Rightarrow x-2\ge0\\ \Rightarrow x\ge2\)

Chứng minh : \(x^2-2x+4>0\\ x^2-2x+1+3=\left(x-1\right)^2+3\ge3>0\)

a: ĐKXĐ: \(\dfrac{-5}{x+6}>=0\)

=>x+6<0

=>x<-6

b: ĐKXĐ: (-2)/(6-x)>=0

=>6-x<0

=>x>6

c: ĐKXĐ: (-x+3)/(-6)>=0

=>-x+3<=0

=>-x<=-3

=>x>=3

d: ĐKXĐ: (7x-1)/-9>=0

=>7x-1<=0

=>x<=1/7

e: ĐKXĐ: (x+2)/(x^2+2x+1)>=0

=>x+2>=0

=>x>=-1

f: ĐKXĐ: (x-2)/(x^2-2x+4)>=0

=>x-2>=0

=>x>=2

ĐKXĐ: \(\left\{{}\begin{matrix}2x-1\ge0\\x+2>0\\3-x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x>-2\\x\le3\end{matrix}\right.\) 

\(\Rightarrow\dfrac{1}{2}\le x\le3\)

a) Biểu thức xác định `<=> x^2-2x-1>0`

`<=>(x^2-2x+1)-2>0`

`<=>(x-1)^2-(\sqrt2)^2>0`

`<=>(x-1+\sqrt2)(x-1-\sqrt2)>0`

`<=>` \(\left[{}\begin{matrix}x< 1-\sqrt{2}\\x>1+\sqrt{2}\end{matrix}\right.\)

`D=(-∞; 1-\sqrt2) \cup (1+\sqrt2 ; +∞)`

b) Biểu thức xác định `<=> x-\sqrt(2x+1)>0`

`<=> x>\sqrt(2x+1)`

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\2x+1\ge0\\x^2>2x+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\\\left[{}\begin{matrix}x< 1-\sqrt{2}\\x>1+\sqrt{2}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow x>1+\sqrt{2}\)

`D=(1+\sqrt2 ; +∞)`

a: ĐKXĐ: \(x\ge1\)

b: ĐKXĐ: \(x< 0\)

c: ĐKXĐ: \(\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

1) ĐKXĐ: \(\left\{{}\begin{matrix}2x+11\ge0\\x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)

2) ĐKXĐ: \(\left\{{}\begin{matrix}-5x\ge0\\x\ne0\end{matrix}\right.\)\(\Leftrightarrow x< 0\)

3) ĐKXĐ: \(7x^2+1\ge0\left(đúng\forall x\right)\Leftrightarrow x\in R\)

4) ĐKXĐ: \(x^2-14x+33\ge0\Leftrightarrow\left(x-11\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-11\ge0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-11\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

5) ĐKXĐ: 

+) \(-x^2+6x+16\ge0\)

\(\Leftrightarrow-\left(x^2-6x+9\right)+25\ge0\)

\(\Leftrightarrow\left(x-3\right)^2\le25\Leftrightarrow-5\le x-3\le5\)

\(\Leftrightarrow-2\le x\le8\)

+) \(3x^2\ne0\Leftrightarrow x\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}-2\le x\le8\\x\ne0\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\8-2x>0\\\end{matrix}\right.\) \(\Rightarrow0\le x< 4\)

a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x< -2\end{matrix}\right.\)

b) ĐKXĐ: \(-\sqrt{2}\le x\le\sqrt{2}\)

c) ĐKXĐ: \(x\ge1\)