a) x² - 4x - 5 = 0

=> x² - 5x + x - 5 = 0

=> x(x - 5) + (x - 5) = 0

=> (x + 1)(x - 5) = 0

=> \(\orbr{\begin{cases}x+1=0\\x-5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)

b) 4x² + 7x - 11 = 0

=> 4x² + 11x - 4x - 11 = 0

=> x(4x + 11) - (4x + 11) = 0

=> (x - 1)(4x + 11) = 0

=> \(\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}\)

c) -7x² + 6x + 1 = 0

=> -7x² + 7x - x + 1 = 0

=> -7x(x - 1) - (x - 1) = 0

=> (-7x - 1)(x - 1) = 0

=> \(\orbr{\begin{cases}-7x-1=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}-7x=1\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{1}{7}\\x=1\end{cases}}\)

d) -10x² + 7x + 3 = 0

=> -10x² + 10x - 3x + 3 = 0

=> -10x(x - 1) - 3(x - 1) = 0

=> (-10x - 3)(x - 1) = 0

=> \(\orbr{\begin{cases}-10x-3=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}-10x=3\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)

\(a,x^2-4x-5=0\)

\(\Rightarrow x^2-5x+x-5=0\)

\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

\(b,4x^2+7x-11=0\)

\(\Rightarrow4x^2-4x+11x-11=0\)

\(\Rightarrow4x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(4x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}}\)

\(c,-7x^2+6x+1=0\)

\(\Rightarrow-7x^2+7x-x+1=0\)

\(\Rightarrow-7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(-7x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\-7x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)

\(d,-10x^2+7x+3=0\)

\(\Rightarrow-10x^2+10x-3x+3=0\)

\(\Rightarrow-10x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(-10x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\-10x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{3}{10}\end{cases}}}\)

a: =>-2x=-8

hay x=4

b: =>7x=-21

hay x=-3

c: =>0,25x=-1,5

hay x=-6

d: =>5,3x=6,36

hay x=6/5

e: =>-4x=-12

hay x=3

f: =>-10x=-10

hay x=1

g: =>2x+2-3-2x=0

=>-1=0(vô lý)

h: =>3-3x+4x-3=0

=>x=0

a,

\(3-x=x-5\\ \Leftrightarrow3x-x+5=0\Leftrightarrow2x+5=0\)

\(\Rightarrow x=-\dfrac{5}{2}\)

b, \(\Rightarrow x=-\dfrac{21}{7}=-3\)

c, \(\Leftrightarrow x=\left(0-1,5\right):0,25=-6\)

a. <=> 2x=8 hay x=4

b.<=> x= -21/7 = -3

c. <=> x= -1,5/ 0,25=-6

d. <=> x= -6,36/-5,3=1,2

e.<=> 4x=12 hay x= 3

f. <=> 10x = 10 hay x = 1

g. <=> 2x +2 = 3 + 2x

<=> 2=3 ( vô lí )

h.<=> 3 - 3x + 4x -3 =0

<=> x=0

a) \(4x-3=11-3x\)

\(\Leftrightarrow4x+3x=11+3\)

\(\Leftrightarrow7x=14\)

\(\Leftrightarrow x=2\)

Vậy .............

b) \(x^3-4x^2+3x=0\)

\(\Leftrightarrow x\left(x^2-4x+3\right)=0\)

\(\Leftrightarrow x\left(x^2-x-3x+3\right)=0\)

\(\Leftrightarrow x\left[x\left(x-1\right)-3\left(x-1\right)\right]=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)

Vậy .................

P/s: câu c bn gõ lại dc ko

a: \(\left(3x-2\right)\cdot\left(\dfrac{2}{7}\left(x+3\right)-\dfrac{4x-3}{5}\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(\dfrac{2}{7}x+\dfrac{6}{7}-\dfrac{4}{5}x+\dfrac{3}{5}\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(-\dfrac{18}{35}x+\dfrac{51}{35}\right)=0\)

=>x=2/3 hoặc x=51/18=17/6

b \(\left(3.3-11x\right)\left(\dfrac{7x+2}{5}+\dfrac{2\left(1-3x\right)}{3}\right)=0\)

\(\Leftrightarrow\left(-10x+3\right)\left(21x+6+10-30x\right)=0\)

\(\Leftrightarrow\left(-10x+3\right)\left(-9x+16\right)=0\)

=>x=3/10 hoặc x=16/9

c: \(\dfrac{3}{7x-1}=\dfrac{1}{7x\left(3x-7\right)}\)

=>21x(3x-7)=7x-1

\(\Leftrightarrow63x^2-154x+1=0\)

\(\text{Δ}=\left(-154\right)^2-4\cdot63=23464\)

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

\(\left\{{}\begin{matrix}x_1=\dfrac{154-\sqrt{23464}}{126}\\x_2=\dfrac{154+\sqrt{23464}}{126}\end{matrix}\right.\)

\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)

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

\(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)

(5/7x-1/4)(-3/4x+1/2)=0

\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{7}x-\dfrac{1}{4}=0\\-\dfrac{3}{4}x+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{5}{7}x=\dfrac{1}{4}\\-\dfrac{3}{4}x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}:\dfrac{5}{7}\\x=\dfrac{1}{2}:\dfrac{3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}.\dfrac{7}{5}\\x=\dfrac{1}{2}.\dfrac{4}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{20}\\x=\dfrac{4}{6}=\dfrac{2}{3}\end{matrix}\right.\)

Vậy ....