a) x.(x-1)=0

\(\Rightarrow\)x=0 hoặc x-1=0

\(\Rightarrow\)x=0+1

\(\Rightarrow\)x=1

vậy x=1 hoặc x=0

b) -x.(x+3)=0

\(\Rightarrow\)-x = 0 hoặc x+3 = 0

\(\Rightarrow\)x= 0-3

\(\Rightarrow\)x=-3

vậy x=0 hoặc x=-3

c) (2x-4).(x+2)=0

(2x-4)= 0

2x=0+4

2x=4

x=4:2

x=2

hoặc (x+2)=0

x= 0-2

x=-2

vậy x=2 hoặc x=-2

d) (3-x).|x+5|=0

3-x = 0

x= 3-0

x=3

hoặc |x+5|=0

x+ 5=0

x=0-5

x=-5

vậy x=3 hoặc x=-5

e) (|x|+1).( 4-2x) = 0

(|x|+1) =0

|x|= 0-1

|x|=-1

hoặc( 4-2x) = 0

2x=4-0

2x=4

x=4:2

x=2

g) x²+5x=0

x²=0

x=0

hoặc 5x=0

x= 0: 5

x=0

vậy x=0

2)

a) (x+3).(y-5)= 7

(x+3)và (y-5)\(\in\)Ư(7)=\(\left\{1;-1;7;-7\right\}\)

b) xy + 3x - 2y= 11

x( y+3) -2y=11

x(y-3)- 2( y+3) +6 = 11

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

vì x,y thuộc Z \(\Leftrightarrow\)y+3 và x-2 \(\in\)Z

do đó y+3 và x-2 \(\in\)Ư ( 5)= \(\left\{1;5;-1;-5\right\}\)

\(\in\)\(\in\)

c) xy + 3x - 7y= 21

x( y+3) -7y= 21

x( y+3) - 7( y+3)+21= 21

(y+3)( x-7) =0

1.

a) \(=3x\left(x-3y\right)\left(12x+1\right)\)

b) \(=\left(x-1\right)^2-4y^2=\left(x-2y-1\right)\left(x+2y-1\right)\)

c) \(=y\left(x-z\right)+3\left(x-z\right)=\left(x-z\right)\left(y+3\right)\)

d) \(=\left(x-1\right)^2-49=\left(x-8\right)\left(x+6\right)\)

e) Chịu!
f) Chịu!
3.

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

⇔ \(\left(x-5\right)\left(x-10\right)=0\)

⇔ \(\left[{}\begin{matrix}x=5\\x=10\end{matrix}\right.\)

b) \(A=x\left(5x-3y\right)-z\left(3y-5x\right)=\left(x+z\right)\left(5x-3y\right)\)

Thay x = 2015,2016; y = 1,4; = -2015,2016 vào A ta có

A = 0

1,

a, =36x^2.(x-3y)+3x.(x-3y)

=(36x^2+3x)(x-3y)=3x(12x+1)(x-3y)

b, =(x^2-2x+1)-4y^2

=(x-1)^2-4y^2

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

C, =xy+3x-yz-3z

=x(y+3)-z(y+3)=(x-z)(y+3)

d, x^2-2x+1-49

=(x-1)^2-7^2

e, chịu

f, chịu

2, a, chịu

b, =x(5x-3y)+z(5x-3y)

=(x+z)(5x-3y), thay vào ta có

=(2015,2016+-2015,2016)(5.2015,2016-3.-2015,2016)

=0

Mấy bài chịu để suy nghĩ đã sorry

f, =x^2-2.5.x+5^2-5(x-5)=0

=(x-5)(x-5)-5(x-5)=0

=(x-5-5)(x-5)=0

=(x-10)(x-5)=0

=> {X-10=0 =>{x=10

{x-5=0 {x=5

Bài2:

a: \(Q=\left(\dfrac{2x+1}{x\left(x-5\right)}-\dfrac{2x}{x\left(x+5\right)}\right)\cdot\dfrac{x\left(x-5\right)\left(x+5\right)}{21x-2}\)

\(=\dfrac{2x^2+11x+5-2x^2+10x}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x-5\right)\left(x+5\right)}{21x-2}\)

\(=\dfrac{21x+5}{21x-2}\)

b: Để Q là số nguyên thì \(21x-2+7⋮21x-2\)

\(\Leftrightarrow21x-2\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{\dfrac{1}{7};\dfrac{1}{21};\dfrac{3}{7};-\dfrac{5}{21}\right\}\)