Giup to voi

T

thaiha_98

Bài 1:
$x^4-7x^3+12x^2+4x-16=0$
\Leftrightarrow $x^4-2x^3-5x^3+10x^2+2x^2-4x+8x-16=0$
\Leftrightarrow $x^3(x-2) - 5x^2(x-2) + 2x(x-2) + 8(x-2)=0$
\Leftrightarrow $(x-2)(x^3-5x^2+2x+8)=0$
\Leftrightarrow $(x-2)(x^3-4x^2-x^2+4x-2x+8)=0$
\Leftrightarrow $(x-2)[x^2(x-4)-x(x-4)-2(x-4)]=0$
\Leftrightarrow $(x-2)(x-4)(x^2-x-2)=0$
\Leftrightarrow $(x-2)(x-4)(x+1)=0$
\Leftrightarrow $x=2; x=4$ hoặc $x=-1$
 
P

phucvo29

Bài 2

Bạn tự vẽ hình nha

Ta có:[TEX] EB = AB = \frac{BC}{2}[/TEX]
[TEX]\Rightarrow \triangle{BAE}[/TEX] cân tại B
[TEX]\Rightarrow \widehat{BAE} = \widehat{BEA}[/TEX]
Ta có : [TEX]\widehat{BEA} = 180^0 - \widehat{AEC} = 180^0 - 143^0 = 37^0[/TEX]
[TEX]\triangle{ABE}[/TEX] có : [TEX]\widehat{ABE} = 180^0 - 2\widehat{BEA}[/TEX]
[TEX]= 180^0 - 2.37^0 = 106^0[/TEX]
Ta có [TEX]AB // BC[/TEX] ( hbh ABCD)
[TEX]\Rightarrow \widehat{ABE} + \widehat{BCD} = 180^0[/TEX]
[TEX]\Rightarrow \widehat{BCD} = 180^0 - 106^0 = 74^0[/TEX]
 
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