a)  (x - 3)² - 5.(x - 2) + 5 = 0.

<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0

<=> x^2 - 11x + 24 = 0

<=> (x-3)(x-8)=0

<=> x = 3 hoặc x = 8

b) (2x - 1)² - 3.(x - 2).(x + 2) - 25 = 0.

<=> 4x^2 - 4x + 1 - 3x^2 + 12 - 25 = 0

<=> x² - 4x - 12 = 0

<=> (x+2)(x-6) = 0

<=> x = -2 hoặc x = 6

d)  x² - 4x + 5 = 0.

<=> (x - 2)² = -1 (vô lý)

Vậy phương trình vô nghiệm

a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

\(\left(x^2-5\right)\left(x^2+1\right)=0\)

<=> \(\hept{\begin{cases}x^2-5=0\\x^2+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x^2=5\\x^2=-1\end{cases}}\)

<=> \(\hept{\begin{cases}x=\sqrt{5};x=-\sqrt{5}\\x\in\varnothing\end{cases}}\)

câu còn lại tương tự nha

a) (x - 3)(5 - 2x) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\5-2x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=\frac{5}{2}\end{matrix}\right.\)

b) (x + 5)(x - 1) - 2x(x - 1) = 0

<=> (x - 1)(x + 5 - 2x) = 0

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

<=> \(\left[{}\begin{matrix}x-1=0\\5-x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

c) 5(x + 3)(x - 2) - 3(x + 5)(x - 2) = 0

<=> (x - 2)[5(x + 3) - 3(x + 5)] = 0

<=> (x - 2)(5x + 3 - 3x - 15) = 0

<=> (x - 2)(2x - 12) = 0

<=> \(\left[{}\begin{matrix}x-2=0\\2x-12=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

d) (x - 6)(x + 1) - 2(x + 1) = 0

<=> (x + 1)(x - 6 - 2) = 0

<=> (x + 1)(x - 8) = 0

<=> \(\left[{}\begin{matrix}x+1=0\\x-8=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)

Câu e thì để mình nghĩ đã :)

#Học tốt!

Giúp luôn Đức Hải Nguyễn câu e:

e, (x - 1)² + 2(x - 1)(x + 2) + (x + 2)² = 0

\(\Leftrightarrow\) (x - 1 + x + 2)² = 0

\(\Leftrightarrow\) (2x + 1)² = 0

\(\Leftrightarrow\) 2x + 1 = 0

\(\Leftrightarrow\) x = \(\frac{-1}{2}\)

Vậy S = {\(\frac{-1}{2}\)}

Chúc bn học tốt!!

câu e nó là hàng đẳng thức đó (a+b)^2 với a là (x-1) B là x+2 ta có  (a+b)^2 = a^2+2.a.b+b^2

1. \(\Leftrightarrow\left(x-6\right)\left(x+7\right)+5\left(x-6\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left[\left(x+7\right)+5\left(3x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-6\right)\left(16x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\16x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-\frac{1}{8}\end{matrix}\right.\)

4. \(\Leftrightarrow\left(x+5\right)^2\left(3x+2\right)^2-x^2\left(x+5\right)^2=0\)

\(\Leftrightarrow\left(x+5\right)^2\left[\left(3x+2\right)^2-x^2\right]=0\)

\(\Leftrightarrow\left(x+5\right)^2\left(2x+2\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=0\\2x+2=0\\4x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x=-2\\4x=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-1\\x=-\frac{1}{2}\end{matrix}\right.\)

Lời giải:
a. $x^2-4x-5=0$
$\Leftrightarrow (x+1)(x-5)=0$

$\Leftrightarrow x+1=0$ hoặc $x-5=0$

$\Leftrightarrow x=-1$ hoặc $x=5$

b. 

$5x^2-9x-2=0$
$\Leftrightarrow (x-2)(5x+1)=0$

$\Leftrightarrow x-2=0$ hoặc $5x+1=0$

$\Leftrightarrow x=2$ hoặc $x=\frac{-1}{5}$

c.

$(x^2+1)-5(x^2+1)+6=0$

$\Leftrightarrow a^2-5a+6=0$ (đặt $x^2+1=a$)

$\Leftrightarrow (a-2)(a-3)=0$

$\Leftrightarrow a-2=0$ hoặc $a-3=0$

$\Leftrightarrow x^2-1=0$ hoặc $x^2-2=0$

$\Leftrightarrow (x-1)(x+1)=0$ hoặc $(x-\sqrt{2})(x+\sqrt{2})=0$

$\Leftrightarrow x\in\left\{\pm 1; \pm \sqrt{2}\right\}$

d.

$(x^2+6x)-2(x+3)^2-17=0$

$\Leftrightarrow (x^2+6x+9)-2(x+3)^2-26=0$

$\Leftrightarrow (x+3)^2-2(x+3)^2-26=0$
$\Leftrightarrow -(x+3)^2-26=0$

$\Leftrightarrow (x+3)^2=-26<0$ (vô lý)

Do đó không tồn tại $x$ thỏa mãn.

b) x(x-4) - 2x+8 = 0
    x(x-4) - 2(x-4) = 0
    (x-2) (x-4) = 0
TH1: x-2=0              TH2: x-4=0
            x=2                          x=4
Vậy x\(\in\){2;4}

\(b,\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ c,\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ d,\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ e,\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ f,\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow x=2\)

b) x(x-4)-2x+8=0

x(x-4)-2(x-4)=0

(x-4)(x-2)=0

th1: x-4=0

x=4

th2: x-2=0

x=2

Vậy x thuộc tập hợp 4;-2

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

=>(x+5)(x-3)+8=x^2-1

=>x^2+2x-15+8=x^2-1

=>2x-7=-1

=>x=3(loại)

b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)

=>(x-4)(x+1)+x^2+3+5(x-1)=0

=>x^2-3x-4+x^2+3+5x-5=0

=>2x^2+2x-6=0

=>x^2+x-3=0

=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)

e: =>x^2-2x+1+2x+2=5x+5

=>x^2+3=5x+5

=>x^2-5x-2=0

=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)

g: (x-3)(x+4)*x=0

=>x=0 hoặc x-3=0 hoặc x+4=0

=>x=0;x=3;x=-4

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

b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)

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

g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)