(x- 2)(3x2- 10x+ 3) =0
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a) 3x² - 6x = 0
3x(x - 2) = 0
3x = 0 hoặc x - 2 = 0
*) 3x = 0
x = 0
*) x - 2 = 0
x = 2
Vậy x = 0; x = 2
b) x² + 10x + 25 = 64
x² + 2.x.5 + 5² = 64
(x + 5)² = 64
x + 5 = -8 hoặc x + 5 = 8
*) x + 5 = -8
x = -8 - 5
x = -13
*) x + 5 = 8
x = 8 - 5
x = 3
Vậy x = -13; x = 3
c) (2 - 3x)(x + 4) + 3x - 2 = 0
(2 - 3x)(x + 4) - (2 - 3x) = 0
(2 - 3x)(x + 4 - 1) = 0
(2 - 3x)(x + 3) = 0
2 - 3x = 0 hoặc x + 3 = 0
*) 2 - 3x = 0
3x = 2

*) x + 3 = 0
x = 0 - 3
x = -3
Vậy:

a: \(x^2+3y^2-4x+6y+7=0\)
\(\Leftrightarrow x^2-4x+4+3y^2+6y+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\)
\(\Leftrightarrow\left(x,y\right)=\left(-2;1\right)\)
b) Ta có: 5x3 – 3x2 + 10x – 6 = (5x3 + 10x )+ ( -3x2– 6)
= 5x(x2 + 2) – 3(x2 + 2) = (x2 + 2)(5x – 3)
Vậy (x2 + 2)(5x – 3) = 0 ⇒ 5x – 3 = 0 (vì x2 + 2 ≥ 0, với mọi x)
⇒x = 3/5
`a,3x^2+7x+2=0`
`<=>3x^2+6x+x+2=0`
`<=>3x(x+2)+x+2=0`
`<=>(x+2)(3x+1)=0`
`<=>x=-2\or\x=-1/3`
d) Ta có: (x-1)(x+2)=70
\(\Leftrightarrow x^2+2x-x-2-70=0\)
\(\Leftrightarrow x^2+x-72=0\)
\(\Leftrightarrow x^2+9x-8x-72=0\)
\(\Leftrightarrow x\left(x+9\right)-8\left(x+9\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=8\end{matrix}\right.\)
Vậy: S={8;-9}
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2




\(\left(x-2\right)\left(3x^2-10x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x^2-9x-x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)