Question

help mình mn ơi 

Answer 1

Bài 3:

a: Ta có: \(4x\left(x+1\right)=8\left(x+1\right)\)

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

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

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

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

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

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

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

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

Answer 2

2a) (3x-5y)(x+1)

b) (3x-2)(x-6)

c) (4y+1)(x-1)

d) (x-2)(x-3)(x-4)

e) (7x+1)(x-y)

Answer 3

d) pt<=> (x-2)(x-3)(x-4)=0

<=> x-2=0 hoặc x-3=0 hoặc x-4=0

<=> x=2 hoặc x=3 hoặc x=4

e) pt <=> (5x+1)(x-2)=0

<=> x=-1/5 hoặc x=2

g) pt <=> (5x-1)(x-2000)=0

<=> x=1/5 hoặc x=2000

h) pt <=> x(x-4)=0

<=> x =0 hoặc x=4

K) pt <=> x(x-1)=0

<=> x = 0 hoặc x=1

m) pt <=> x(6x+1)=0

<=> x = 0 hoặc x=-1/6

n) pt <=> (x+1)(1-x-1)=0

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

<=> x = 0 hoặc x=-1