Lời giải:
a. 

PT $\Leftrightarrow -5x^2+15x-5+x+5x^2=x-2$
$\Leftrightarrow 16x-5=x-2$

$\Leftrightarrow 15x=3$

$\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}$

b.

PT $\Leftrightarrow -4x^2+20x+7x^2-28x-3x^2=12$

$\Leftrightarrow -8x=12$

$\Leftrightarrow x=\frac{-3}{2}$

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

2: x=3

3: x=18

1) \(x=51\)

2) \(x=3\)

3) \(x=18\)

4) \(x=2\)

5) \(x=7\)

\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a.

4x(x-5) - (x-1)(4x-3)-5=0

 4x^2-20x-4x^2+3x+4x+3=0

(4x^2-4x^2)+(-20x+3x+4x)+3=0

 13x+3 = 0

13x=-3

x=-3/13

b,

(3x-4)(x-2)-3x(x-9)+3=0

3x^2-6x-4x+8 - 3x^2+27x+3=0

(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0

17x+11=0

17x=-11

x=-11/17

c, 2(x+3)-x^2-3x=0

2(x+3) - x(x+3)=0

(x+3)(2-x)=0

TH1: x+3 = 0; x=-3

TH2: 2-x=0;x=2

\(a,\dfrac{3}{7}-x=\dfrac{1}{2}x-3\)

\(\Rightarrow-x-\dfrac{1}{2}x=-3-\dfrac{3}{7}\)

\(\Rightarrow-\dfrac{3}{2}x=-\dfrac{24}{7}\)

\(\Rightarrow x=-\dfrac{24}{7}:\left(-\dfrac{3}{2}\right)\)

\(\Rightarrow x=\dfrac{16}{7}\)

\(b,5x-\dfrac{2}{3}=\dfrac{5}{3}-2x\)

\(\Rightarrow5x+2x=\dfrac{5}{3}+\dfrac{2}{3}\)

\(\Rightarrow7x=\dfrac{7}{3}\)

\(\Rightarrow x=\dfrac{7}{3}:7\)

\(\Rightarrow x=\dfrac{1}{3}\)

#Toru

a: 3/7-x=1/2x-3

=>-3/2x=-3+3/7

=>-1/2x=-1+1/7=-6/7

=>1/2x=6/7

=>x=6/7*2=12/7

b: =>5x+2x=5/3+2/3

=>7x=7/3

=>x=1/3

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

b: =>4x^2+8x-8x^2+5x-10=0

=>-4x^2+13x-10=0

=>x=2 hoặc x=5/4

c: =>2x^2-5x+6x-15=2x^2+8x

=>x-15=8x

=>-7x=15

=>x=-15/7

d: =>3x^2+15x-2x-10-3x^2-12x=5

=>x-10=5

=>x=15

e: =>x^2-3x+2x^2+2x=3x^2-12

=>-x=-12

=>x=12