Câu 1:
a) Áp dụng BĐT Δ
$⇒10+10=20$
$⇒$ Không thể tạo ra 1 tam giác
b) $MN<MP$
$⇒\widehat{P}<\widehat{N}$
Ta có:
$\widehat{P}+\widehat{PMH}=90^o$
$\widehat{N}+\widehat{NMH}=90^o$
mà $\widehat{P}<\widehat{N}$
$⇒\widehat{PMH}>\widehat{NMH}$
Bài 2:
$P(x)=2x^3-3x+x^5-4x^3+4x-x^5+x^2-2$
$P(x)=(x^5-x^5)+(2x^3-4x^3)+x^2+(-3x+4x)-2$
$P(x)=-2x^3+x^2-x-2$
$⇒$ Bậc: $3$
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$Q(x)=x^4-2x^2+3x+1+2x^2$
$Q(x)=x^4+(2x^2-2x^2)+3x+1$
$Q(x)=x^4+3x+1$
$⇒$ Bậc: $4$