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

=>x=3 hoặc x=-1

b: =>x(x-3)=0

=>x=0 hoặc x=3

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

=>x=5 hoặc x=-1

d: =>5x^2+7x-5x-7=0

=>(5x+7)(x-1)=0

=>x=1 hoặc x=-7/5

e: =>x^2-4=0

=>x=2 hoặc x=-4

h: =>x^2-4x+4-3=0

=>(x-2)^2=3

=>\(x=2\pm\sqrt{3}\)

1) Ta có: \(x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}

\(2x^3-3x^2+3x+8=0\)\(\Leftrightarrow\left(2x^3+2x^2\right)-\left(5x^2+5x\right)+\left(8x+8\right)=0\)

\(\Leftrightarrow2x^2\left(x+1\right)-5x\left(x+1\right)+8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-5x+8\right)=0\)

\(\Leftrightarrow x+1=0\)( vì \(2x^2-5x+8>0\)) \(\Leftrightarrow x=-1\)

Vậy tập nghiệm của phương trình là \(S=\left\{-1\right\}\)

\(2x^3-3x^2+3x+8=0\)

\(\Leftrightarrow2x^3+2x^2-5x^2-5x+8x+8=0\)

\(\Leftrightarrow2x^2\left(x+1\right)-5x\left(x+1\right)+8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-5x+8\right)=0\)

\(\Leftrightarrow x+1=0\left(2x^2-5x+8>0\forall x\right)\)

\(\Leftrightarrow x=-1\)

`3x+7=0`

`<=>3x=-7`

`<=>x=-7/3`

Vậy `S={-7/3}`

______________________

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

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

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

`@TH1:2x=0<=>x=0`

`@TH2: 3-2x=0<=>2x=3<=>x=3/2`

Vậy `S={0;3/2}`

3x+7=0

\(\Leftrightarrow3x=-7\Leftrightarrow x=-\dfrac{7}{3}\)

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

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

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

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

3x + 7 = 0

\(\Leftrightarrow\) 3x = -7

\(\Leftrightarrow\) x = -7/3

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

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

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

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

1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)

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

\(\Leftrightarrow2x^2+18=0\left(loại\right)\)

2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

4: Ta có: \(2x\left(x-5\right)-3x+15=0\)

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

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

5: Ta có: \(3x\left(x+4\right)-2x-8=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)

6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3

Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) đẻ được hỗ trợ tốt hơn. Viết như thế kia rất khó đọc => khả năng bị bỏ qua bài cao.

a: =>3x=3

=>x=1

b: =>12x-2(5x-1)=3(8-3x)

=>12x-10x+2=24-9x

=>2x+2=24-9x

=>11x=22

=>x=2

c: =>2x-3(2x+1)=x-6x

=>-5x=2x-6x-3=-4x-3

=>-x=-3

=>x=3

d: =>2x-5=0 hoặc x+3=0

=>x=5/2 hoặc x=-3

e: =>x+2=0

=>x=-2

1: \(\Leftrightarrow6\left(3x-1\right)+3\left(6x-2\right)=4\left(1-3x\right)\)

=>18x-6+18x-6=4-12x

=>36x-12=4-12x

=>48x=16

hay x=1/3

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

=>(2x-1)(3x-4)=0

=>x=1/2 hoặc x=4/3