Question

[huyền trang bùi thị]

1. phân tích các đa thức sau thành nhân tử:

a) 5x - 15y

b) 3/5x² + 5x⁴ - x²y

c) 14x²y² - 21xy² + 28x²y

d) 2/7x (3y - 1) - 2/7y (3y - 1)

e) x³ - 3x² + 3x - 1

f) ( x + y )² - 4x²

g) 27x³ + 1/8

h) ( x + y )³ - ( x - y )³

2. Tìm x, biết:

a) x²( x + 1 ) + 2x ( x + 1 ) = 0

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

c) 4/9 - 25x² = 0

d) x² - x = 1/4 = 0

Answer 1

[Đời về cơ bản là buồn... cười!!!]

Bài 1:

a) \(5x-15y=5\left(x-3y\right)\)

b) \(\dfrac{3}{5}x^2+5x^4-x^2y=x^2\left(\dfrac{3}{5}+5x^2-y\right)\)

c) \(14x^2y^2-21xy^2+28x^2y=7xy\left(2xy-3y+4x\right)\)

d) \(\dfrac{2}{7}x\left(3y-1\right)-\dfrac{2}{7}y\left(3y-1\right)=\dfrac{2}{7}\left(3y-1\right)\left(x-y\right)\)

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

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

g) \(27x^3+\dfrac{1}{8}=\left(3x+\dfrac{1}{2}\right)\left(6x^2+1,5x+\dfrac{1}{4}\right)\)

h) \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)

\(=6x^2y+2y^3=2y\left(3x^2+y\right)\)

Bài 2:

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\Rightarrow x=-1\\x+2=0\Rightarrow x=-2\end{matrix}\right.\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}3x-2=0\Rightarrow x=\dfrac{2}{3}\\x+5=0\Rightarrow x=-5\end{matrix}\right.\)

c) \(\dfrac{4}{9}-25x^2=0\)

\(\Rightarrow\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-5x=0\Rightarrow x=\dfrac{2}{15}\\\dfrac{2}{3}+5x=0\Rightarrow x=\dfrac{-2}{15}\end{matrix}\right.\)

d) Có tới 2 dấu "=".

Answer 2

[L-girl Minh Châu]

bài 1 dễ mk ko lm nữa nhé

bafi2:

a,x(x+1)(x+2)=0

x=0 ; x=-1 ; x=-2

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

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

x=-5 ; x=2/3

c,

(2/3)²- (5x)²=0

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

x=+-2/15

d, X²-2*1/2x+(1/2)²=0

(X-1/2)²2=0

X=1/2