Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

1/

a)5x – 20y=5(x-4y)

b) 5x.(x –  1) –  3x(x – 1)=2x(x-1)

c) x.(x+y) – 5x – 5y=c) x.(x+y) – 5(x+y)=(x-5)(x+y)

2/

a)x² + xy + x = x(x+y+1)=77.(77+22+1)=77.100=7700

b)  x . ( x – y ) + y . ( y – x )=(x-y)(x-y)=(x-y)²=(53-3)²=2500

3/

a) X + 5x² = 0

⇒x(x+5)=0

⇒hoặc x=0

x+5=0⇒x=-5

b)x + 1 = ( x + 1 )² 

⇒(x + 1)-( x + 1 )² =0

⇒x(x+1)=0

⇒ hoặc x=0

hoặc x+1=0⇒x=-1

4/

a) 97 . 13 + 130 . 0,3 = 97.13+13.10.0,3=97.13+13.3=100.13=1300

b)86 . 153 – 530 . 8,6=86.153–53.10.8,6=86.153-53.86=86.100=8600

C) 85 .12,7 + 5,3 . 12,7= 12,7(85+5,3)=12,7.90,3=1146,81

D)52.143 – 52 . 39 – 8.26=52(143-39)-8,26=52.104-8,26=5399,74

a) x= + - 5
b) x\(\in\)\(\left\{-1;-7\right\}\)

a/ \(x^2-25=0\)

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

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

b/ \(x\left(x+7\right)+x+7=0\)

\(x\left(x+7\right)+\left(x+7\right)=0\)

\(\left(x+7\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+7=0\Rightarrow x=-7\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)

a) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

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

a: (x+1)(y-2)=0

=>\(\begin{cases}x+1=0\\ y-2=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=2\end{cases}\)

b: (x-5)(y-7)=1

=>(x-5;y-7)∈{(1;1);(-1;-1)}

=>(x;y)∈{(6;8);(4;6)}

c: (x+4)(y-2)=2

=>(x+4;y-2)∈{(1;2);(2;1);(-1;-2);(-2;-1)}

=>(x;y)∈{(-3;4);(-2;3);(-5;0);(-6;1)}

d: (x+3)(y-6)=-4

=>(x+3;y-6)∈{(1;-4);(-4;1);(-1;4);(4;-1);(2;-2);(-2;2)}

=>(x;y)∈{(-2;2);(-7;7);(-4;10);(1;5);(-1;4);(-5;8)}

e: (x+7)(5-y)=-6

=>(x+7)(y-5)=6

=>(x+7;y-5)∈{(1;6);(6;1);(-1;-6);(-6;-1);(2;3);(3;2);(-2;-3);(-3;-2)}

=>(x;y)∈{(-6;11);(-1;6);(-8;-1);(-13;4);(-5;8);(-4;7);(-9;2);(-10;3)}

f: (12-y)(6-y)=-2

=>(x-12)(y-6)=2

=>(x-12;y-6)∈{(1;2);(2;1);(-1;-2);(-2;-1)}

=>(x;y)∈{(13;8);(14;7);(11;4);(10;5)}

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

vì (x-2016)^2016 >= 0 vs mọi x

    (y-2017)^2018>= 0 vs mọi y

    /x+y-z/ >= 0 vs mọi x,y,z

mà (x-2016)^2016+(y-2017)^2018+/x-y+z/=\(\hept{\begin{cases}\left(x-2016\right)^{2016}=0\\^{\left(-2017\right)^{2018}}=0\\x+y-z=0\end{cases}}\)0 nên ​^{_{\(\hept{\begin{cases}x-2016=0\\y-2017=0\\x+y-z\end{cases}}\)\(\hept{\begin{cases}x=2016\\y=2017\\x+y-z=0\end{cases}}\)}}

mà x+y=2016+2017=4033

\(\Rightarrow\)4033-z=0

z=4033

vậy x=2016 y=2017 z=4033

`1)`

`a)3x^2-6xy+3y^2=3(x^2-2xy+y^2)=3(x-y)^2`

`b)(x-y)^2-4x^2=(x-y-2x)(x-y+2x)=(-x-y)(3x-y)`

`2)`

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

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

`<=>(x-3)(2x-1)=0`

`<=>[(x=3),(x=1/2):}`

`b)x^2+5x+6=0`

`<=>x^2+2x+3x+6=0`

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

`<=>[(x=-2),(x=-3):}`