a: \(\Leftrightarrow4\left(2x+1\right)-3\left(6x-1\right)=2x+1\)

=>8x+4-18x+3=2x+1

=>-10x+7=2x+1

=>-12x=-6

hay x=1/2

b: \(\Leftrightarrow4x^2-12x+7x-21-x^2=3x^2+6x\)

=>5x-21=6x

=>-x=21

hay x=-21

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

c: 2(x-1)+x=0

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x+3=4x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

c,

\(\Leftrightarrow2x-2+x=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

a) \(\dfrac{x+1}{4}-\dfrac{5+2x}{8}=\dfrac{3-4x}{2}\)

⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\) 

⇔ 2x + 2 - 5 - 2x = 12 -16x

⇔ 16x = 15 

⇔ x = 15/16

b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)

⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)

⇔ 8 - 6x - 4 + x = 5x + 10

⇔ 10x = -6

⇔ x = -6/10

Câu 1:

x + 1/4 - 5 + 2x/8 = 3 - 4x/2

<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8

<=> 2x + 2 - 5 - 2x = 12 - 16x

<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16

Câu 2:

4 - 3x/5 - 4 - x/10 = x + 2/2

<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10

<=> 8 - 6x - 4 + x = 5x + 10

<=> 4 - 5x = 5x + 10

<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

=>0,2x+0,4-0,5x=0,25-0,5x+0,25

=>0,2x+0,4=0,5

=>0,2x=0,1

=>x=1/2

\(\sqrt{x+3}-\sqrt{7-x}>\sqrt{2x-8}\)

⇔ \(\sqrt{x+3}>\sqrt{7-x}+\sqrt{2x-8}\)

⇔ \(\left\{{}\begin{matrix}4\le x\le8\\x+3>7-x+2x-8+2\sqrt{\left(7-x\right)\left(2x-8\right)}\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}4\le x\le8\\x+3>x-1+2\sqrt{\left(7-x\right)\left(2x+8\right)}\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}4\le x\le8\\4>2\sqrt{\left(7-x\right)\left(2x+8\right)}\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}4\le x\le8\\\sqrt{\left(7-x\right)\left(2x-8\right)}< 2\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}4\le x\le8\\-2x^2+22x-56< 2\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}4\le x\le8\\\left[{}\begin{matrix}x>\dfrac{11+\sqrt{5}}{2}\\x< \dfrac{11-\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}4\le x< \dfrac{11-\sqrt{5}}{2}\\\dfrac{11+\sqrt{5}}{2}< x\le8\end{matrix}\right.\)

Các giá trị nguyên của x thỏa mãn là S = {4 ; 7 ; 8}

Ấy chết sai điều kiện XĐ rồi, bạn sửa lại điều kiện thôi nhé

g: =>12x+1>=36x+12-24x-3

=>12x+1>=12x+9(loại)

h: =>6(x-1)+4(2-x)<=3(3x-3)

=>6x-6+8-4x<=9x-9

=>2x+2<=9x-9

=>-7x<=-11

=>x>=11/7

i: =>4x^2-12x+9>4x^2-3x

=>-12x+9>-3x

=>-9x>-9

=>x<1

\(\dfrac{x-3}{2+x}< 1\\ \Leftrightarrow\dfrac{x-3}{2+x}-1< 0\\ \Leftrightarrow\dfrac{x-3}{2+x}-\dfrac{2+x}{2+x}< 0\\ \Leftrightarrow\dfrac{x-3-2-x}{2+x}< 0\\ \Leftrightarrow\dfrac{-5}{2+x}< 0\)

Vì \(-5< 0\)

\(\Rightarrow2+x>0\\ \Rightarrow x>-2\)

Vậy \(x>-2\)

=>\(\dfrac{3x+5}{3}>\dfrac{-4}{6x+5}\)

=>\(\dfrac{18x^2+15x+30x+25+12}{3\left(6x+5\right)}>0\)

=>\(\dfrac{18x^2+45x+37}{3\left(6x+5\right)}>0\)

=>6x+5>0

=>x>-5/6