a)\(P\left(x\right)+Q\left(x\right)=x^4-x^3+3x^2-1+3x\)

b)\(H\left(x\right)=Q\left(x\right)+2x^4+2\)

\(H\left(x\right)=-2x^4-x^2+x-2+2x^4+2\)

\(H\left(x\right)=-x^2+x\)

c) cho H(x) = 0

\(=>-x^2+x=0\Leftrightarrow x\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

1) xét hai tam giác AHB và AHC ta có

AB = AC (gt)

\(\widehat{BAH}=\widehat{CAH}\left(gt\right)\)

AH chung 

=> ΔAHB = ΔAHC

2)Xét hai tam giác vuông AHE và AHF ta có

AH chung

AE = AF (theo hình vẽ)

=>  ΔAHE = ΔAHF ( c.h - c.g.v)

3)ta có \(\left\{{}\begin{matrix}\text{ ΔAHE = ΔAHF ( c.h - c.g.v)}\\\text{ ΔAHB = ΔAHC}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}EH=FH\\BH=HC;\end{matrix}\right.\)

Xét hai tam giác vuông HEB và HFC

\(BH=HC\)

HE = HF

=> ΔHEB = ΔHFC ( c.h - c.g.v)

\(A=4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1010.1011}\right)\)

\(A=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1010}-\dfrac{1}{1011}\right)\)

\(A=4\left(1-\dfrac{1}{1010}\right)=4\left(\dfrac{1011}{1011}-\dfrac{1}{1011}\right)=\dfrac{4040}{1011}\)

\(=>\left|2x-1\right|=3\dfrac{1}{2}:\dfrac{21}{22}=\dfrac{7}{2}:\dfrac{21}{22}=\dfrac{11}{3}\)

\(=>\left[{}\begin{matrix}2x-1=\dfrac{11}{3}\\2x-1=-\dfrac{11}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{14}{3}\\2x=-\dfrac{8}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Phải thêm cả kẻ \(BH\perp AC\) chứ nhỉ

nêu ko thì \(\widehat{ABH}\) ko thể bằng 30 độ

\(\dfrac{a}{b}=\dfrac{b}{c}\Leftrightarrow b^2=ac\)

\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(c+a\right)}=\dfrac{a}{c}\)

Chu vi hình thoi

5 x 4 =20 cm

Diện tích hình thoi

\(\dfrac{8\times6}{2}=24\left(cm^2\right)\)