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

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

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

nha

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

<=> 2(x+3) -x.(x+3) = 0 

<=> (2-x).(x+3) = 0 

<=> 2-x = 0 hoặc x+3 = 0

<=> x= 2 hoặc x=  -3 

Vậy _____________

2(x + 3) - x² - 3x = 2(x + 3) - x(x + 3) = (2 - x)(x + 3) = 0

=> 2 - x = 0 hoặc x + 3 = 0 => x = 2 ; -3

a) x.(x-1) - 2(1-x) =0

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

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

b) (x-3)^3 + (3-x) =0

=> (x-3)^3 - (x-3) =0

=> (x-3)​​[(x-3)^2 - 1 ] =0

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

=> (x-3)(x-4)(x-2) =0

(x-3)³ + 3 -x =0

=>x³-9x²+26x-24=0

=>x³-7x²+12x-2x²+14x-24=0

=>x(x²-7x+12)-2(x²-7x+12)=0

=>(x-2)(x²-7x+12)=0

=>(x-2)[x²-4x-3x+12]=0

=>(x-2)[x(x-4)-3(x-4)]=0

=>(x-2)(x-3)(x-4)=0

=>x-2=0 hoặc x-3=0 hoặc x-4=0

=>x=2 hoặc 3 hoặc 4

Vậy tập nghiệm của pt là S={2;3;4}

= (x-3)³ - (x-3) =0

(x-3)((x-3)² -1)=0

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

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

x = 3;2;4

đơn giản,dễ hiểu, vận dụng hđt đáng nhớ, có ai giỏi =em k

8: \(=\left(x-2y\right)\cdot x\cdot\left(x+3\right)\)

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

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

=2(2x+1)(x-3)

3: \(=2\left(x+2\right)\left(25x-15-x\right)\)

\(=2\left(x+2\right)\left(24x-15\right)\)

=6(x+2)(8x-5)

= (-2a ^2 +4a)(1-x)

= -2 (a^2 -2a)(1-x)

= -2a(a-2)(1-x)

= 2a(a-2)(x-1)

\(-2a^2\left(x-1\right)+4a\left(1-x\right)\)

\(=-a\cdot2a\left(x-1\right)-2\cdot2a\left(x-1\right)\)

\(=2a\left(x-1\right)\left(-a-2\right)\)

\(7xy^5\left(x-1\right)-3x^2y^4\left(1-x\right)+5xy^3\left(x-1\right)\)

\(=7xy^5\left(x-1\right)+3x^2y^4\left(x-1\right)+6xy^3\left(x-1\right)\)

\(=\left(x-1\right)\left(7xy^5+3x^2y^4-6xy^3\right)=xy\left(x-1\right)\left(7y^4+3xy^3-6y^2\right)\)

 Trả lời:

7xy⁵(x - 1) - 3x²y⁴(1 - x) + 5xy³(x - 1)

= 7xy⁵(x - 1) + 3x²y⁴(x - 1) + 5xy³(x - 1)

= (7xy⁵ + 3x²y⁴ + 5xy³)(x - 1)

= xy(7y⁴ + 3xy³ + 5y²)(x - 1)

\(3x.\left(x-10\right)=\left(x-10\right)\)

\(\Rightarrow3x-\left(x-10\right)-\left(x-10\right)=0\)

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

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