\(\dfrac{2x-6}{x+1}\ge0\)

`<=> 2x-6 >= 0`

`<=> 2x >=6`

`<=> x>=3`

Vật bpt đã cho có tập nghiệm \(S=\left\{x|x\ge3\right\}\)

=>2x-6>=0 hoặc x+1<0

=>x>=3 hoặc x<-1

c: \(\Leftrightarrow2x+1=3\)

hay x=1

a) \(6x^2-2x-6x^2+13=0\\ -2x=-13\\ x=\dfrac{13}{2}\)

b: =>2x-2x-1=x-6x

=>-5x=-1

hay x=1/5

Lời giải:
a. 

$2x(3x-1)=6x^2-13$

$\Leftrightarrow 6x^2-2x=6x^2-13$

$\Leftrightarrow 2x=13$

$\Leftrightarrow x=\frac{13}{2}$

b. 

$\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x$

$\Leftrightarrow \frac{2x-(2x+1)}{6}=\frac{-5}{6}x$

$\Leftrightarrow \frac{-1}{6}=\frac{-5}{6}x$
$\Leftrightarrow x=\frac{-1}{6}: \frac{-5}{6}=\frac{1}{5}$

`|2x+1|-3=x+4`

`<=>|2x+1|=x+4+3=x+7(x>=-7)`

`**2x+1=x+7`

`<=>x=7-1=6(tm)`

`**2x+1=-x-7`

`<=>3x=-6`

`<=>x=-2(tm)`

`|3x-5|=1-3x(x<=1/3)`

`**3x-5=1-3x`

`<=>6x=6`

`<=>x=1(l)`

`**3x-5=3x-1`

`<=>-5=-1` vô lý

`|2x+2|+|x-1|=10`

Nếu `x>=1`

`pt<=>2x+2+x-1=10`

`<=>3x+1=10`

`<=>3x=9`

`<=>x=3(tm)`

Nếu `x<=-1`

`pt<=>-2x-2+1-x=10`

`<=>-1-3x=10`

`<=>-11=3x`

`<=>x=-11/3(tm)`

Nếu `-1<=x<=1`

`pt<=>2x+2+1-x=10`

`<=>x+3=10`

`<=>x=7(l)`

Vậy `S={3,-11/3}`

d) 

+) Với \(x< -4\), PT \(\Rightarrow3-x-x-4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{17}{4}\)  (Nhận)

+) Với \(-4\le x\le-3\), PT \(\Rightarrow3-x+x+4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{9}{2}\) (Loại)

+) Với \(-3< x\le3\), PT \(\Rightarrow3-x+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{2}\) (Nhận)

+) Với \(x>3\), PT \(\Rightarrow x+3+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{4}\) (Loại)

  Vậy \(x\in\left\{-\dfrac{3}{2};-\dfrac{17}{4}\right\}\) 

Ta có: \(\sqrt{2x+7}-6=x\)

\(\Leftrightarrow\sqrt{2x+7}=x+6\)

\(\Leftrightarrow x^2+12x+36-2x-7=0\)

\(\Leftrightarrow x^2+10x+29=0\)(Vô lý)

Vậy: \(S=\varnothing\)

Điều kiện : x ≥ 0

\(\sqrt{2x+97}-6=x\text{⇔}\sqrt{2x+97}=x+6\\ \text{⇔}2x+97=x^2+12x+36\text{⇔}x^2+10x-61=0\\ \text{⇔}\left[{}\begin{matrix}x=-5+\sqrt{86}\\x=-5-\sqrt{86}\end{matrix}\right.\)

\(\sqrt{2x+97}-6=x\)

\(\Leftrightarrow\sqrt{2x+97}=x+6\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+6\ge0\\2x+97=\left(x+6\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\x^2+10x-61=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\\left[{}\begin{matrix}x=-5+\sqrt{86}\\x=-5-\sqrt{86}\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=-5+\sqrt{86}\)

Vậy..,

b) \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)

<=> \(\frac{13\left(x+1\right)}{12}-\frac{5x+3}{6}=\frac{x+7}{12}\)

<=> 13(x + 1) - 2(5x + 3) = x + 7

<=> 13x + 13 - 10x - 6 = x + 7

<=> 3x + 7 = x + 7

<=> 3x + 7 - x = 7

<=> 2x + 7 = 7

<=> 2x = 7 - 7

<=> 2x = 0

<=> x = 0

c) 2x + 4(x - 2) = 5

<=> 2x + 4x - 8 = 5

<=> 6x - 8 = 5

<=> 6x = 5 + 8

<=> 6x = 13

<=> x = 13/6

\(\left(2x-6\right)\times\left(5-x\right)=0\)

\(2x-6=0;5-x=0\)

\(x=3;x=5\)

Vậy: \(x=3;x=5\)

$\left(2x-6\right)\times \left(5-x\right)=0$

$2x-6=0;5-x=0$

$x\; =3,\; x\; =\; 5$

Vậy x = 3 và 5.

a,<=> 3x+1/4-2x-3/5=1

<=> x-7/20=1

<=> x= 27/20

a, \(\left(3x+\frac{1}{4}\right)-\frac{1}{3}\left(6x+\frac{9}{5}\right)=1\)

\(3x+\frac{1}{4}-\frac{6}{3}x-\frac{3}{5}=1\)

\(x-\frac{7}{20}=1\Leftrightarrow x=\frac{27}{20}\)

b,ĐKXĐ : x \(\ne\)-1/2 ; 1/2 

 \(\left(\frac{5}{2x+1}\right)-\left(\frac{2x}{1-2x}\right)=1-\left(\frac{6-4x}{4x^2-1}\right)\)

\(\frac{5}{2x+1}-\frac{2x}{1-2x}=1-\frac{6-4x}{4x^2-1}\)

\(\frac{5}{2x+1}-\frac{2x}{1-2x}=1-\frac{2\left(3-2x\right)}{\left(2x+1\right)\left(2x-1\right)}\)

\(\frac{5\left(1-2x\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}-\frac{2x\left(2x+1\right)^2\left(2x-1\right)}{\left(1-2x\right)\left(2x+1\right)^2\left(2x-1\right)}=\frac{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}-\frac{2\left(3-2x\right)\left(2x+1\right)\left(1-2x\right)}{\left(2x+1\right)\left(2x-1\right)^2\left(2x-1\right)\left(1-2x\right)}\)

\(22x-5-20x^2-8x^3=18x-7-8x^3-4x^2\)

lm nốt nha,bị troll rồi ko vt đc nữa.