a: Ta có: \(x^2+3x-10=0\)

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

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

b: Ta có: \(x^2-5x-6=0\)

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

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

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

x( 3x - 1 ) - 5( 1 - 3x ) = 0 <=> x( 3x - 1 ) + 5( 3x - 1 ) = 0 <=> ( 3x - 1 )( x + 5 ) = 0 <=> x = 1/3 hoặc x = -5

Trả lời:

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

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

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

Vậy x = 0; x = - 1; x = - 2 là nghiệm của pt.

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

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

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

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

Vậy x = 1/3; x = - 5 là nghiệm của pt.

a)  \(x^3+3x^2+3x+2=0\)

<=>  \(x^3+x^2+x+2x^2+2x+2=0\)

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

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

tự làm

b) \(x^4-2x^3+2x-1=0\)

<=>  \(\left(x^4-3x^3+3x^2-x\right)+\left(x^3-3x^2+3x-1\right)=0\)

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

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

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

tự làm

c)   \(x^4-3x^3-6x^2+8x=0\)

<=>   \(x\left(x^3-3x^2-6x+8\right)=0\)

<=>  \(x\left[\left(x^3+x^2-2x\right)-\left(4x^2+4x-8\right)\right]=0\)

<=>\(x\left[x\left(x^2+x-2\right)-4\left(x^2+x-2\right)\right]=0\)

<=>   \(x\left(x-4\right)\left(x^2+x-2\right)=0\)

<=> \(x\left(x-4\right)\left(x-1\right)\left(x+2\right)=0\)

tự làm

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

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

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

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

=> \(\hept{\begin{cases}3x=1\\x=2\end{cases}}\)

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

vậy x thuộc (1/3;2)

b. 4x(x-3)-2x+6=0

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

(4x-2)(x-3)

=>*4x-2=0

4x=2

x=1/2

*x-3=0

x=3

vậy x thuộc (1/2;3)

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

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

=>x-2=0 hoặc 3x-1=0

*x-2=0=>x=2

*3x-1=0=>x=\(\frac{1}{3}\)

Vậy x=2;x=\(\frac{1}{3}\)

b. 4x(x-3)-2x+6=0

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

=>(x-3)=0 hoặc 4x-2=0

*x-3=0=>x=3

*4x-2=0=>\(\frac{1}{2}\)

Vậy x=3; x=\(\frac{1}{2}\)

câu c, d,e thì chút nx mk gửi cho nha

b: \(3x^2-2x-1=0\)

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

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

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

a: Bạn ghi lại đề đi bạn

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

\(< =>\left(-x-1\right)\left(3x-1\right)=0< =>\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

b,\(\left(3x-5\right)^2-x\left(3x-5\right)=0< =>\left(3x-5\right)\left(3x-5-x\right)=0\)

\(< =>\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}\)

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

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow x=-1;x=\frac{1}{3}\)

b, \(\left(3x-5\right)^2-x\left(3x-5\right)=0\)

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

1,X=-1 hoặc 3

2,Tìm x sao cho (x+3) và (3x-2) ko bằng 0

a) \(2x\left(x-3\right)+6\left(3x-3\right)=0\)

\(\Leftrightarrow2x^2-6x+18x-18=0\)

\(\Leftrightarrow2x^2+12x-18=0\)

Mà \(2x^2\ge0\)

\(\Rightarrow x\in\varnothing\)

a)=>2x^2-6x+18x-18=0                           b)=>6x^2-15x-75-30x =????

=>2x^2+12x=0+18                                    

=>2x^2+12x=18

=>x.(2x+12)=18 (tự làm phần còn lai)

a, \(2x\left(x-3\right)+6\left(3x-3\right)=0\)

\(\Leftrightarrow2x^2-6x+18x-18=0\)

\(\Leftrightarrow2x^2+12x-18=0\)

=> Phương trình vô nghiệm 

b,Sử đề :  \(3x\left(2x-5\right)-15\left(5-2x\right)=0\)

\(\Leftrightarrow6x^2-15x-75+30x=0\)

\(\Leftrightarrow6x^2+15x-75=0\)

\(\Leftrightarrow3\left(x+5\right)\left(2x-5\right)=0\)

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

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

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

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

\(\Leftrightarrow2x^3-2x^2-x^2+2x+x-1=0\)

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

\(\Leftrightarrow2x\left(x^2-x+1\right)-\left(x^2-x+1\right)=0\)

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

\(TH1:2x-1=0\Leftrightarrow x=\frac{1}{2}\)

\(TH2:x^2-x+1=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\)

Mà \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)nên loại TH2

Vậy \(x=\frac{1}{2}\)

2x³ - 3x² + 3x - 1 = 0

(2x - 1)(x² - x + 1) = 0

Vì: x² - x + 1 > 0 nên:

2x - 1 = 0

2x = 0 + 1

2x = 1

x = 1/2

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

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

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

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

Vì \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\forall x\)

\(\Rightarrow2x-1=0\)

\(\Rightarrow x=\frac{1}{2}\)