1) Theo đề bài ta có `{(CD bot a),(CD bot b):}`

`=> a //b ` ( từ vuông góc đến song song)

2) Ta có  `a//b`

`=> hat{DBA} = hat{A_2}` ( hai góc so le trong)

`=>  hat{DBA} = hat{A_2} =45^o`

Lại có : ` hat{A_1} + hat{A_2} =180^0` ( kề bù)`

`=> hat{A_1} = 180^0 -  hat{A_2} = 180^0 -45^0 = 135^0`

mà `hat{A_1}  =hat{A_3} ` ( đối đỉnh)

`=> hat{A_1}  =hat{A_3} = 135^0`

Vậy `hat{A_1} =135^0`

       `hat{A_3} =135^0`

        `hat{A_2} = 45^0`

`3x^2(2x+1)-6x(1+2x)=0 `

`<=> (3x^2 -6x ) (2x+1) =0`

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

`@TH1 : 3x =0 => x=0`

`@TH2 : x-2 =0 => x =2`

`@TH3 : 2x+1 =0 => 2x=-1 => x =-1/2`

Vậy `S={0;2;-1/2}`

sai thì thôi nhó :q

`a)Đ k : x ne +- 3`

\(A=\sqrt{\dfrac{1}{x^2-6x+9}}-\dfrac{6}{x^2-9}\)

\(A=\dfrac{\sqrt{1}}{\sqrt{\left(x-3\right)^2}}-\dfrac{6}{\left(x+3\right)\left(x-3\right)}=\dfrac{1}{\left|x-3\right|}-\dfrac{6}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{\left(x+3\right)\left(x-3\right)-\left(x-3\right)6}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{\left(x+3-6\right)\left(x-3\right)}{\left(x-3\right)^2\left(x+3\right)}\)

\(A=\dfrac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{1}{x+3}\)

`a) x(x+1)+4(x+1)=0`

`<=> (x+4)(x+1) =0`

`<=> [(x+4=0),(x+1=0):}`

`<=> [(x=-4),(x=-1):}`

Vậy `S={-4;-1}`

`b) x^2 -x =0`

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

`<=> [(x=0),(x=1):}`

Vậy `S={0;1}`

`c) 3x(x-3)-x+3=0`

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

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

`<=> [(3x=1),(x-3=0):}`

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

Vậy `S={1/3;3}`

Bài 2

`a)M=x^2 -3x +10 = [x^2 - 2 * 3/2 *x + (3/2)^2 ] +10 -3/2`

`M = (x-3/2)^2 + 17/2 >= 17/2∀x`

Dấu ''='' xảy ra khi : `x-3/2 =0`

                               `=> x=3/2`

`b) P=y^2 +8y +15 = (y^2 + 2*y*4 + 4^2 ) +15 -4^2`

`P= (y+4)^2 -1 >= -1 ∀x`

Dấu ''='' xảy ra khi `y+4 =0`

                           `=> y =0-4=-4`

Bài 3

`A=12a -4a^2 +3 = - ( 4a^2 -12a -3) = -{[(2a)^2 - 2*2a*3 + 3^2] -3 -3^2}`

`A = - [(2a-3)^2 -12] = 12 - (2a-3)^2 <= 12`

dấu ''='' xảy ra khi `2a-3 =0 => 2a =3`

                             `=> a = 3/2`

`B =3a-a^2 +3  = -(a^2 -3a -3) = -{[a^2 - 2*a*3/2 + (3/2)^2 ] - 3-(3/2)^2}`

`B = -[(a-3/2)^2 - 21/4] = 21/4 -(a-3/2)^2 <= 21/4`

dấu ''='' xảy ra khi : ` a-3/2 =0  => a =3/2`

`3x^2(2x+1)-6x(1+2x)=0 `

`<=> (3x^2 -6x ) (2x+1) =0`

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

`<=> [(3x=0),(x-2=0),(2x=-1):}`

`<=> `\(\left[{}\begin{matrix}x=0\\x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy `S={0;2;-1/2}`