\(1,\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

\(2,\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

\(x^2-2x=0\)

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

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

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

\(5,\)

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

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)

b: =>-6<x<5

1) (x+6)(3x-1)+x+6=0

⇔(x+6)(3x-1)+(x+6)=0

⇔(x+6)(3x-1+1)=0

⇔3x(x+6)=0

2) (x+4)(5x+9)-x-4=0

⇔(x+4)(5x+9)-(x+4)=0

⇔(x+4)(5x+9-1)=0

⇔(x+4)(5x+8)=0

3)(1-x)(5x+3)÷(3x-7)(x-1)

=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)

2)  \(x^3-6x^2+11x-6=0\)

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

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

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

bn giải tiếp nha

3)   \(x^3-4x^2+x+6=0\)

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

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

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

lm tiếp nha

4)  \(x^3-3x^2+4=0\)

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

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

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

lm tiếp nha

Mk làm mẫu 1 bài cho nha !

1. <=> (x^3-x^2)+(5x^2-5x)+(6x-6) = 0

<=> (x-1).(x^2+5x+6) = 0

<=> (x-1).[(x^2+2x)+(3x+6)] = 0

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

<=> x-1=0 hoặc x+2=0 hoặc x+3=0

<=> x=1 hoặc x=-2 hoặc x=-3

Vậy ..............

Tk mk nha

2.  x3−6x2+11x−6=0

⇔x3−3x2−3x2+9x+2x−6=0

⇔(x−3)(x2−3x+2)=0

⇔(x−3)(x−2)(x−1)=0

bn giải tiếp nha

3)   x3−4x2+x+6=0

⇔x3−3x2−x2+3x−2x+6=0

⇔(x−3)(x2−x−2)=0

⇔(x−3)(x−2)(x+1)=0

lm tiếp nha

4)  x3−3x2+4=0

⇔x3+x2−4x2−4x+4x+4=0

⇔(x+1)(x2−4x+4)=0

⇔(x+1)(x−2)2=0

lm tiếp nha

1: =>(x+3)(x-5)=0

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

9: =>(x-2)(x-4)=0

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2

mình cần gấp

a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)

\(\Leftrightarrow2x^2-2x-2x^2=-6\)

\(\Leftrightarrow x=3\)

b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

(2x - 7) + 17 = 6

=> 2x - 7 = 6 - 17

=> 2x - 7 = -11

=> 2x = -11 + 7

=> 2x = -4

=> x = -4 : 2

=> x = -2

+) 12 -2(3 - 3x)= -2

=> 2(3 - 3x) = 12 + 2

=> 2(3 - 3x) = 14

=> 3 - 3x = 14 : 2

=> 3 - 3x = 7

=> 3x = 3 - 7

=> 3x = -4

=> x = -4/3

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

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

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

Vậy...

\(\left(2x-7\right)+17=6\)

\(\left(2x-7\right)=6-17\)

\(2x-7=-11\)

\(2x=-11+7\)

\(2x=-4\)

\(x=-4:2\)

\(\Rightarrow x=-2\)

\(V\text{ậy x = -2}\)

      c.   x^2-5x +6 = 0

<=> x^2 - 5x = -6

<=> - 4x = -6

<=> x= -6/-4

 Mình chỉ phân tích đa thức thành nhân tử thôi , phần còn lại bạn tự tính nha keo dài lắm

A)  2x²(x+3) - x(x+3) = 0  <=> x(x - 3)(2x-1)=0

B)  (2x+5)² - (x+2)²=0  <=>  (x+3)(3x+7)=0

C)  (x²-2x) - (3x-6)=0  <=> (x-2)(x-3)=0

D)  (2x-7)(2x-7-6x+18)=0   <=> (2x-7)(-4x+11)=0

E)  (x-2)(x+1) - (x-2)(x+2)=0   <=>  (x-2)*(-1)=0   <=> x-2=0

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

H)  (1-x)(5x+3+3x-7)=0     <=>  (1-x)(8x-4)=0

F)   (x+6)*3x=0

I)  (x-3)(4x-1-5x-2)=0  <=>  (x-3)(-x-3)=0

K)   (x+4)(5x+8)=0

H)  (x+3)(4x-9)=0

B> <2X+5>²-<X+2>²=0

<2X+5-X-2><2X+X+2>=0

<X+3><3X+7>=0

X+3=0 HOẶC 3X+7=0

X=-3 HOẶC X=-7/3

C>X²-5X+6=0

X²-4X+4-X+2=0

<X-2>²-<X-2>=0

<X-2.><X-3>=0

X-2=0 HOẶC X-3=0

X=2 HOẶC X=3

D> <2X-7><2X-7-6<X-3>>=0

<2X-7><-4X+11>=0

2X-7=0 HOẶC -4X+11=0

X=7/2 HOẶC X=11/4

E><X-2><X+1>=X²-4

<X-2><X+1>-<X²-4>=0

<X-2><X+1>-<X-2><X+2>=0

-X+2=0

X=2

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