\(3S=3+3^2+3^3+...+3^{10}\\ \Rightarrow3S-S=3+3^2+...+3^{10}-1-3-3^2-...-3^9\\ \Rightarrow2S=3^{10}-1\\ \Rightarrow S=\dfrac{3^{10}-1}{2}\)

Ta có \(S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^8+3^9\right)\)

\(S=\left(1+3\right)+3^2\left(1+3\right)+...+3^8\left(1+3\right)\\ S=\left(1+3\right)\left(1+3^2+...+3^8\right)=4\left(1+3^2+...+3^8\right)⋮4\)

Gọi cd,cr lần lượt là \(a,b(cm;a,b>0)\)

Ta có \(\dfrac{a}{7}=\dfrac{b}{5}\). Đặt \(\dfrac{a}{7}=\dfrac{b}{5}=k\Rightarrow a=7k;b=5k\) 

Ta có \(ab=315\Rightarrow35k^2=315\Rightarrow k^2=9\Rightarrow k=3\left(k>0\right)\)

\(\Rightarrow\left[{}\begin{matrix}a=21\\b=15\end{matrix}\right.\)

Vậy 2 cạnh hcn là 21 cm và 15 cm

uầy bạn hỏi y chang bạn bên hoidap :v

\(n_{BaCl_2}=\dfrac{52}{208}=0,25(mol)\\ BaCl_2+H_2SO_4\to BaSO_4\downarrow+2HCl\\ \Rightarrow n_{BaSO_4}=0,25(mol);n_{HCl}=0,5(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{0,5.36,5}{52+150-0,25.233}.100\%=12,696\%\)

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{y-x}{7-5}=\dfrac{48}{2}=24\\ \Rightarrow\left\{{}\begin{matrix}x=120\\y=168\\z=48\end{matrix}\right.\)

\(\left(2x-y\right)^3=\left(2x\right)^3-3\left(2x\right)^2y+3\cdot2x\cdot y^2-y^3\\ =8x^3-12x^2y+6xy^2-y^3\)

Chọn \(12x^2y\)

\(a,\left\{{}\begin{matrix}BM=MC\\ME=MD\\\widehat{BMD}=\widehat{CME}\left(đđ\right)\end{matrix}\right.\Rightarrow\Delta BDM=\Delta CEM\left(c.g.c\right)\\ b,\Delta BDM=\Delta CEM\\ \Rightarrow\widehat{BDM}=\widehat{CEM}\\ \text{mà 2 góc này ở vị trí slt nên }CE\text{//}BD\\ \text{Mà }BD\bot AB\Rightarrow CE\bot AB\)

Áp dụng định lí PTG, ta có: \(BC=\sqrt{AB^2+AC^2}=2\sqrt{61}\left(cm\right)\)

Vì AM là trung tuyến ứng cạnh huyền BC nên \(AM=\dfrac{1}{2}BC=\sqrt{61}\left(cm\right)\)

\(=\dfrac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}+\dfrac{10\left(1+\sqrt{6}\right)}{-5}+\sqrt{\left(\sqrt{6}-1\right)^2}\\ =\sqrt{6}-2-2\sqrt{6}+\sqrt{6}-1=-3\)