Kẻ `AH, CK` vuông góc `CD`.

Xét `\DeltaADH` và `\DeltaBCK` có:

`AH =CK` 

`\hatD=\hatC`

`AD=BC` 

`=> \DeltaADH=\DeltaBCK`

`=> DH=CK=x`

Có: `CD=DH+HK+KC = x+12+x=18 => x=3` (cm)

`tanC=(BK)/(CK) <=> tan75^@ = (BK)/3 => BK =6+3\sqrt3 (cm)`

`=> S=1/2 .(AB+CD) .BK = 90+45\sqrt3 ≈ 168 (cm^2)`

c)

`y=3mx+m-2`

`<=>3mx+m-2-y=0`

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

`=> {(3x+1=0),(y+2=0):}`

`<=> {(x=-1/3),(y=-2):}`

Vậy điểm cố định mà d luôn đi qua là: `(-1/3 ; -2)`

Phương trình hoành độ giao điểm:

`mx-3=x^2`

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

(P) cắt (d) tại 2 điểm phân biệt `<=>` PT (1) có 2 nghiệm phân biệt.

`<=> \Delta >0`

`<=>m^2-3>0`

`<=> m<-\sqrt3 \vee m>\sqrt3`

Viet: `{(x_1+x_2=m),(x_1x_2=3):}`

`|x_1-x_2|=2`

`<=>(x_1-x_2)^2=4`

`<=> (x_1+x_2)^2-4x_1x_2=4`

`<=>m^2-4.3=4`

`<=>m= \pm 4` (TM)

Vậy....

`(a-b)^3 = [-(b-a)]^3 = (-1)^3 . (b-a)^3=-(b-a)^3`

a) `x^2+y^2-2x+4y+5`

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

`=(x-1)^2+(y+2)^2 >=0 forall x,y`

b) `-3x^2+2x-5`

`=-(3x^2-2x+5)`

`=-[(\sqrt3 x)^2 -2.\sqrt3 x .\sqrt3/3 + (\sqrt3/3)^2 +14/5]`

`=-(\sqrt3 x-\sqrt3/3)^2-14/5 < 0 forall x`

Minh's brother is more intelligent than anyone else in his family.

ĐK: `x>=0, x \ne 1`.

\(B=\left(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\\ =\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{x-\sqrt{x}+\sqrt{x}}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{x}\\ =\dfrac{x-\sqrt{x}+1}{x}\)

\(B=4\Leftrightarrow\dfrac{x-\sqrt{x}+1}{x}=4\\ \Leftrightarrow x-\sqrt{x}+1=4x\\ \Leftrightarrow x=\dfrac{7-\sqrt{13}}{18}\)

a) `x^2+7x+12=x^2+3x+4x+12=x(x+3)+4(x+3)=(x+3)(x+4)`

b) `a^10+a^5+1`

`=(a^5)^2 + 2.a^5 . 1/2 + (1/2)^2 + 3/4`

`=(a^5+1/2)^2+3/4`

`=>` Không thể phân tích thành nhân tử.

`24 xx 5^5 + 5^2 xx 5^3`

`=24 xx 5^5 + 5^5`

`=5^5 (24 +1)`

`=5^5 . 25`

`=5^5 .5^2`

`=5^7`