a: Tọa độ giao điểm là:

\(\left\{{}\begin{matrix}x^2-x-1=x-1\\y=x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x-2\right)=0\\y=x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{0;2\right\}\\y\in\left\{-1;1\right\}\end{matrix}\right.\)

GIÚP EM VỚI Ạ! EM ĐANG CẦN GẤP Ạ

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a: A={6;7;8;9;10;11;12;13;14}

A={x∈N|5<x<15}

A có 14-6+1=9 phần tử

b: Tổng của A là:

(14+6)*9/2=90

\(a.\)

\(n_{Na_2CO_3}=\dfrac{10.6}{106}=0.1\left(mol\right)\)

\(n_{Na}=0.1\cdot2=0.2\left(mol\right)\)

\(\Rightarrow\%Na=\dfrac{0.2\cdot23}{10.6}\cdot100\%=43.39\%\)

\(n_C=0.2\left(mol\right)\)

\(\Rightarrow\%C=\dfrac{0.1\cdot12}{10.6}\cdot100\%=11.32\%\)

\(n_O=0.2\cdot3=0.6\left(mol\right)\)

\(\Rightarrow\%O=\dfrac{0.6\cdot16}{10.6}\cdot100\%=45.29\%\)

\(b.\)

\(n_{Ca_3\left(PO_4\right)_2}=\dfrac{620}{310}=2\left(mol\right)\)

\(n_{Ca}=3\cdot2=6\left(mol\right)\)

\(\Rightarrow\%Ca=\dfrac{6\cdot40}{620}\cdot100\%=38.71\%\)

\(n_P=2\cdot2=4\left(mol\right)\)

\(\Rightarrow\%P=\dfrac{4\cdot31}{620}\cdot100\%=20\%\)

\(n_O=2\cdot8=16\left(mol\right)\)

\(\Rightarrow\%O=\dfrac{16\cdot16}{620}\cdot100\%=41.29\%\)

có \(\frac{a}{b}=\frac{c}{d}\)

\(=\orbr{\begin{cases}\frac{a+c}{b+d}\\\frac{a-c}{b-d}\end{cases}}\)

\(\Rightarrow\frac{a+c}{b+d}=\frac{a-c}{b-d}\left(=\orbr{\begin{cases}\frac{a}{b}\\\frac{c}{d}\end{cases}}\right)\)

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