Giups vs ạ
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Bài 6:
\(A=3+\left(3^2+3^3\right)+\left(3^4+3^5\right)+...+\left(3^{2020}+3^{2021}\right)\\ A=3+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^{2021}\left(1+3\right)\\ A=3+4\left(3^2+3^4+...+3^{2021}\right)⋮̸4\left(3⋮̸4\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{4}x+3y=6\\\dfrac{2}{3}x+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{12}x=0\\\dfrac{1}{4}x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
Bài 2:
b: \(\Leftrightarrow x^2=10\)
hay \(x=\pm\sqrt{10}\)
1. how far is it from here to the nearest post office?
2 how far is it from new york to california?
3. how far is it from your house to your family store
4 how far is it from beijing to berlin
5 how far is it from your hotel to the beach
1 How far is it from here to the nearest post office?
2 How far is it from New Yorn to California?
3 How far is it from your house to the your family store?
4 How far is it from Beijing to Berlin?
5 How far is it from your hotel to the beach?
Giải ra hộ mình nhé
Với \(a\ge0;a\ne1\)
\(A=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2}{\left(a-1\right)^2}=1\)
-> chọn B