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 ...

\(a,\dfrac{3}{\sqrt{12x-1}}\) xác định \(\Leftrightarrow12x-1>0\Leftrightarrow12x>1\Leftrightarrow x>\dfrac{1}{12}\)

\(b,\sqrt{\left(3x+2\right)\left(x-1\right)}\) xác định \(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}3x+2\ge0\\x-1\ge0\end{matrix}\right.\\\left[{}\begin{matrix}3x+2\le0\\x-1\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-\dfrac{2}{3}\\x\ge1\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{2}{3}\\x\le1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le-\dfrac{2}{3}\\x\ge1\end{matrix}\right.\)

\(c,\sqrt{3x-2}.\sqrt{x-1}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}3x-2\ge0\\x-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{2}{3}\\x\ge1\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)

\(d,\sqrt{\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}}\) xác định \(\Leftrightarrow-x+5>0\Leftrightarrow x< 5\)

Bài 2: 

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

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

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

\(TC:\left(\sqrt{5}+\sqrt{3}\right)^2=8+2\sqrt{15}\)

\(3^2=9=8+1=8+\sqrt{1}\)

vi \(15>1\Rightarrow\sqrt{15}>\sqrt{1}\Leftrightarrow\sqrt{15}>1\Rightarrow2\sqrt{15}>1\)

\(\Rightarrow8+2\sqrt{15}>8+1\Leftrightarrow8+2\sqrt{15}>9\)

\(\Rightarrow\sqrt{8+2\sqrt{15}}>\sqrt{9}\)

\(\Rightarrow\sqrt{5}+\sqrt{3}>3\)

DKXD: x khac 7

Ta có : 

\(\sqrt{5}+\sqrt{3}>\sqrt{4}+\sqrt{1}=2+1=3\)

Vậy \(\sqrt{5}+\sqrt{3}>3\)

Ngô Ngọc Hải bài 2 là tìm x để biểu thức có nghĩa ạ

x∈[0, ∞)

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: \(x\ge1\)

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

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

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

1.

6x + 1 ≥0

<=>6x≥-1

<=>x≥-1/6

2.

3x - 5 > 0 

<=> 3x > 5

<=> x > 5/3

3.

x - 7 > 0

<=> x > 7

4. 

-3x ≥0

<=>x≤0

5.

√5 - √3 . x ≥0

<=> √3 . x ≤ √5

<=> x ≤ √5/3 = (√15)/3

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

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

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