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

\(a=\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x^3-3x-2\right)\left(x+\sqrt[]{x+2}\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+1\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x-2\right)\left(x+1\right)^2\left(x+\sqrt[]{x+2}\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}}{\left(x+1\right)\left(x+\sqrt[]{x+2}\right)}=...\)

\(b=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{1+2x}-x-1\right)+\left(x+1-\sqrt[3]{1+3x}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x^2}{\sqrt[]{1+2x}+x+1}+\dfrac{x^3+3x^2}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt[]{1+2x}+x+1}+\dfrac{x+3}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}\right)\)

\(=...\)

\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt[]{5+4x}-2x-3\right)+\left(2x+3-\sqrt[3]{7+6x}\right)}{x^3+x^2-x-1}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{5+4x-\left(2x+3\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(2x+3\right)^3-\left(7+6x\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4\left(x+1\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(x+1\right)^2\left(8x+20\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4}{2x+3+\sqrt[]{5+4x}}+\dfrac{8x+20}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{x-1}\)

\(=...\)

a.

ĐKXĐ: \(x\ge-1\)

\(7+12\sqrt{x+1}=x+4\sqrt{x^2+3x+2}\)

\(\Leftrightarrow4\sqrt{\left(x+1\right)\left(x+2\right)}-12\sqrt{x+1}+x-7=0\)

\(\Leftrightarrow4\sqrt{x+1}\left(\sqrt{x+2}-3\right)+x-7=0\)

\(\Leftrightarrow4\sqrt{x+1}\left(\dfrac{x-7}{\sqrt{x+2}+3}\right)+x-7=0\)

\(\Leftrightarrow\left(x-7\right)\left(\dfrac{4\sqrt{x+1}}{\sqrt{x+2}+3}+1\right)=0\)

\(\Leftrightarrow x-7=0\) (do \(\dfrac{4\sqrt{x+1}}{\sqrt{x+2}+3}+1>0;\forall x\ge-1\))

\(\Rightarrow x=7\)

b.

ĐKXĐ: \(x\ne-\dfrac{1}{3}\)

\(\Rightarrow3x^2+3x+2=\left(3x+1\right)\sqrt{x^2+x+2}\)

\(\Leftrightarrow x^2+x+2-\left(3x+1\right)\sqrt{x^2+x+2}+2x^2+2x=0\)

Đặt \(\sqrt{x^2+x+2}=t\)

\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)

\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)

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

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2x\left(x\ge0\right)\\\sqrt{x^2+x+2}=x+1\left(x\ge-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4x^2\left(x\ge0\right)\\x^2+x+2=x^2+2x+1\left(x\ge-1\right)\end{matrix}\right.\)

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

Bài 1:

Ta có: \(\left(3\sqrt{50}-5\sqrt{18}+3\sqrt{8}\right)\cdot\sqrt{2}\)

\(=\left(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\cdot\sqrt{2}\)

\(=6\sqrt{2}\cdot\sqrt{2}\)

=12

Bài 2: 

1) ĐKXĐ: \(x\le0\)

2) ĐKXĐ: \(x\le2\)

3) ĐKXĐ: \(x>\dfrac{-3}{2}\)

4) ĐKXĐ: x>0

5) ĐKXĐ: x<3

1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)

=>4x+8=3x-1

=>x=-9

2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)

=>8x-4=5x-7

=>3x=-3

=>x=-1

3: ĐKXD: x>=0

\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)

=>\(x+\sqrt{x}-6=x-1\)

=>căn x=-1+6=5

=>x=25

4: ĐKXĐ: x>=0

PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)

=>x-2*căn x-3=x-4

=>-2căn x-3=-4

=>2căn x+3=4

=>2căn x=1

=>căn x=1/2

=>x=1/4

a) Để \(\sqrt{\dfrac{x}{3}}\) có nghĩa thì \(\dfrac{x}{3}\ge0\Leftrightarrow x\ge0\)

b) Để \(\sqrt{-5x}\) có nghĩa thì \(-5x\ge0\Leftrightarrow x\le0\)

c) Để \(\sqrt{4-x}\) có nghĩa thì \(4-x\ge0\Leftrightarrow x\le4\)

d) Để \(\sqrt{3x+7}\) có nghĩa thì \(3x+7\ge0\Leftrightarrow x\ge-\dfrac{7}{3}\)

e) Để \(\sqrt{-3x+4}\) có nghĩa thì \(-3x+4\ge0\Leftrightarrow x\le\dfrac{4}{3}\)

f) Để \(\sqrt{\dfrac{1}{-1+x}}\) có nghĩa thì \(\left\{{}\begin{matrix}\dfrac{1}{-1+x}\ge0\\-1+x\ne0\end{matrix}\right.\)

\(\Leftrightarrow-1+x>0\Leftrightarrow x>1\)

g) Để \(\sqrt{1+x^2}\) có nghĩa thì \(1+x^2\ge0\left(đúng\forall x\right)\)

h) \(\sqrt{\dfrac{5}{x-2}}\) có nghĩ thì \(\left\{{}\begin{matrix}\dfrac{5}{x-2}\ge0\\x-2\ne0\end{matrix}\right.\)

\(\Leftrightarrow x-2>0\Leftrightarrow x>2\)

a. \(x\ge0\)

b. \(x< 0\)

c. \(x\le4\)

d. \(x\ge\dfrac{-7}{3}\)

e. \(x\le\dfrac{4}{3}\)

f. \(x>1\)

g. Mọi x

h. \(x>2\)

a.ĐKXĐ: \(x\ge0\)
 \(\sqrt{2x}< \dfrac{1}{3}\) \(\Leftrightarrow2x< \dfrac{1}{3}\Leftrightarrow6x< 1\Leftrightarrow x< \dfrac{1}{6}\)

b. ĐKXĐ: \(x\ge\dfrac{1}{6}\)

\(\sqrt{-3x+\dfrac{1}{2}}\ge5\Leftrightarrow-3x+\dfrac{1}{2}\ge25\Leftrightarrow x=-\dfrac{49}{6}\) 

c. ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{-2x+1}>7\) \(\Leftrightarrow-2x+1>49\Leftrightarrow x=-24\)

d. ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}\le\dfrac{3}{2}\Leftrightarrow2x-1\le\dfrac{9}{4}\Leftrightarrow x=\dfrac{13}{8}\)

a: Ta có: \(\sqrt{2x}< \dfrac{1}{3}\)

\(\Leftrightarrow2x< \dfrac{1}{9}\)

\(\Leftrightarrow x< \dfrac{1}{18}\)

Kết hợp ĐKXĐ, ta được: \(0\le x< \dfrac{1}{18}\)

b: Ta có: \(\sqrt{-3x+\dfrac{1}{2}}\ge5\)

\(\Leftrightarrow-3x+\dfrac{1}{2}\ge25\)

\(\Leftrightarrow-3x\ge\dfrac{49}{2}\)

hay \(x\le-\dfrac{49}{6}\)

c: Ta có: \(\sqrt{-2x+1}>7\)

\(\Leftrightarrow-2x+1>49\)

\(\Leftrightarrow-2x>48\)

hay x<-24

a) ĐKXĐ: \(10-5x< 0\Leftrightarrow5x>10\Leftrightarrow x>2\)

b) ĐKXĐ: \(7-3x>0\Leftrightarrow3x< 7\Leftrightarrow x< \dfrac{7}{3}\)

c) ĐKXĐ: \(-5-2x\ge0\Leftrightarrow2x\le-5\Leftrightarrow x\le-\dfrac{5}{2}\)

a) \(x>2\)

b) \(x< \dfrac{7}{3}\)

c) \(x\le-\dfrac{5}{2}\)

a: ĐKXĐ: x>2

b: ĐKXĐ: \(x< \dfrac{7}{3}\)