O______________________A____________________________B__________x

vì A nằm giữa O và B

=>OA+AB=OB

=>5+8=OB

=>OB=13 cm

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+16\right)\)\(=\dfrac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)\(=\dfrac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)\)\(=\dfrac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\dfrac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=\dfrac{1}{2}\left(5^{32}-1\right)\)

Số mol :

\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3mol\)

3\(H_2\) + \(Fe_2O_3\) --> 2\(Fe\) +3 \(H_2O\)

\(\Rightarrow n_{H_2}=\dfrac{0,3.3}{2}=0,45mol\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,45.22,4=10,08\)

Ta có :

\(100a\left(a+1\right)+bc\)

\(=100a^2+100a+bc\)

\(=100a^2+10.10a+bc\)

\(=100a^2+10a\left(b+c\right)+bc\)

\(=100a^2+10ab+10ac+bc\)

\(=10a\left(10a+b\right)+c\left(10a+b\right)\)

\(=\left(10a+b\right)\left(10a+c\right)\)

\(\Rightarrowđpcm\)

1, Ta có :

\(\widehat{A}+\widehat{D}=180^o\Rightarrow\widehat{D}=180^o-130^o=50^o\)

\(\Rightarrow\widehat{D}=360^o-\widehat{A}-\widehat{B}-\widehat{C}=110^o\)

2,

\(c,2x^2+y^2+4x-2y+3=2\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=2\left(x+1\right)^2+\left(y-1\right)^2\)\(d,2x^2+y^2-6x+2xy+9=\left(x^2-6x+9\right)+\left(x^2+2xy+y^2\right)=\left(x-3\right)^2+\left(x+y\right)^2\)

C1:985              c2:6                          c3:15               c4:16                                c5:10                    c6:5

c7:3021                           c8:20                                c9:    8                           c10:10

 45dm²  8cm² = 0, 4508 m²

\(x+2\sqrt{2}x^2+2x^3=0\Leftrightarrow x\left(2x^2+2.\sqrt{2}x+1\right)=0\Leftrightarrow x\left(\sqrt{2}x+1\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{2}x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{2}x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{\sqrt{2}}\end{matrix}\right.\)