1/ x² - 5x + 6 = 0 
⇔ x² - 2x - 3x + 6 = 0 
⇔ x(x - 2) - 3(x - 2) = 0 
⇔ (x - 2)(x - 3) = 0 
⇒S = {2 ; 3}.

1) \(x^2+5x+6=0\)

\(\Leftrightarrow x^2+2x+3x+6=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\end{array}\right.\)

2) \(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\2-x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=2\end{array}\right.\)

3) \(x^2+4x+3=0\)

\(\Leftrightarrow x^2+x+3x+3=0\)

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

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

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

4) \(2x^2-3x-5=0\)

\(\Leftrightarrow2x^2+2x-5x-5=0\)

\(\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\\2x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=\frac{5}{2}\end{array}\right.\)

a, \(x\left(x-2\right)+x-2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

b, \(x^3+x^2+x+1=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x^2=-1\left(voly\right)\end{cases}\Leftrightarrow}x=-1\)

c, \(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\2-x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

d, \(2x\left(3x-5\right)=10-6x\)

\(\Leftrightarrow6x^2-10x-10+6x=0\)

\(\Leftrightarrow\left(6x^2+6x\right)-\left(10x+10\right)=0\)

\(\Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(6x-10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\6x-10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\6x=10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)

a. \(x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)

\(x^3-25x-\left(x^3+8\right)=17\)

\(x^3-25x-x^3-8=17\)

\(-25x=25\)

\(x=-1\)

c. \(6x^2-\left(6x^2-4x+15x-10\right)=7\)

\(6x^2-6x^2-11x+10=7\)

\(-11x=-3\)

\(x=\frac{3}{11}\)

1) -3x²+5x=0

-x(3x-5)=0

suy ra hoặc x=0 hoặc 3x-5=0. giải ra ta có nghiệm phương trình là 0 và 3/5

2) x²+3x-2x-6=0

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

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

suy ra hoặc x-2=0 hoặc x+3=0. giải ra ta có nghiệm là 2 và -3

3) x²+6x-x-6=0

x(x+6)-(x+6)=0

(x-1)(x+6)=0. vậy nghiệm là 1 và -6

4) x²+2x-3x-6=0

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

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

vậy nghiệm là -2 và 3

5) x(x-6)-4(x-6)=0

(x-4)(x-6)=0. vậy nghiệm là 4 và 6

6)x(x-8)-3(x-8)=0

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

suy ra nghiệm là 3 và 8

7) x²-5x-24=0

x²-8x+3x-24=0

x(x-8)+3(x-8)=0

(x+3)(x-8)=0

vậy nghiệm là -3 và 8

câu 1:  -3x² + 5x = 0

suy ra -x(3x-5)=0

sung ra x = 0 hoặc 3x-5=0 suy ra 3x = 5 suy ra x = 5/3

a)

\(x\left(x-2\right)+x-2=0\)

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

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

Vậy x = 2 ; x = - 1

b)

\(x^3+x^2+x+1=0\)

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

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

Vì x²+1 > 0

=> x + 1 = 0

=> x = - 1

Vậy x = - 1

c)

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

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

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

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

Vậy x = 1 ; x = - 3

d)

\(2x\left(3x-5\right)=10-6x\)

\(\Leftrightarrow2x\left(3x-5\right)+2\left(3x-5\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{3}\\x=-\frac{1}{2}\end{array}\right.\)

Vậy x = 5 / 3 ; x = - 1 / 2

a) \(5x\left(x+4\right)-x\left(5x+1\right)=0\)

\(\Leftrightarrow x\left[5\left(x+4\right)-5x-1\right]=0\)

\(\Leftrightarrow x\left(5x+20-5x-1\right)=0\Leftrightarrow x=0\)

b) \(3x\left(5-x\right)+4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(4-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)

c) \(x\left(x-3\right)+4x-12=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

d) \(x^2-36=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

e) \(x^2+3x+1=2\)

\(\Leftrightarrow x^2+3x+1-2=0\)

\(\Leftrightarrow x^2+3x-1=0\)

\(\Leftrightarrow x^2+3x+\frac{3}{2}-\frac{5}{2}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{2}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}+\frac{\sqrt{5}}{\sqrt{2}}\right)\left(x+\frac{3}{2}-\frac{\sqrt{5}}{\sqrt{2}}\right)=0\)

Còn lại ........... Tự lm nất nha 

mk đưa về pt tích, phần tiếp theo bạn làm tiếp

a)  \(2x.\left(x-3\right)-\left(3-x\right)=0\)

<=>  \(\left(x-3\right)\left(2x+1\right)=0\)

..................

b)  \(3x\left(x+5\right)-6\left(x+5\right)=0\)

<=>  \(3\left(x+5\right)\left(x-2\right)=0\)

.....................

c)  \(x^4-x^2=0\)

<=> \(x^2\left(x^2-1\right)=0\)

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

..................

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

<=> (x-3)(2x-1)=0

=>\(\hept{\begin{cases}x-3=0\\2x-1=0\end{cases}< =>\hept{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}}\)

đối với mấy bài này bạn nên chú ý đổi dấu mấy bài này cũng ko khó

b) 3x(x+5)-6(x+5)=0

<=> (x+5)(3x-6)=0

=>\(\hept{\begin{cases}x+5=0\\3x-6=0\end{cases}=>\hept{\begin{cases}x=-5\\x=2\end{cases}}}\)

c)  x⁴-x²

⁼=x²(x²-1)=0

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

mình giải xong rùi đó

hok tốt nha

a, pt <=> 3.(x^2-2x+1)-3x^2+15x=2

<=>3x^2-6x+3-3x^2+15x=2

<=>9x+3=2

<=>9x=2-3=-1

<=>x=-1/9

b,pt <=> (x^2-x)+(5x-5)=0

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

<=>x-1=0 hoặc x+5=0

<=>x=1 hoặc x=-5

k mk nha

1) \(x^2-6x+9=\left(5-3x\right)^2\)

\(\left(x-3\right)^2=\left(5-3x\right)^2\)

\(\Rightarrow x-3=5-3x\)

\(\Rightarrow x+3x=5+3\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=2\)

\(3x\left(2x-3\right)=5\left(3-2x\right)\)

\(3x\left(2x-3\right)+5\left(2x-3\right)=0\)

\(\left(3x+5\right)\left(2x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+5=0\\2x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{3}\\x=\frac{3}{2}\end{cases}}\)

3) \(x^2-2x-15=0\)

\(x^2-2x+1-16=0\)

\(\left(x-1\right)^2-4^2=0\)

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

\(\left(x-5\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)