Giúp mik vs ạ !! :33
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10³ + 2¹⁵
= 1000 + 32768
= 33768
Mà 33768 : 33 = 1023 (dư 9)
Em xem lại đề
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a) \(\dfrac{x-1}{5}\)=\(\dfrac{3}{y+4}\)
⇒( x - 1)( y + 4) = 3.5 = (-3)(-5) = 5.3 = (-5)(-3) = 1.15 = (-1)(-15)
x - 1 | 1 | -1 | 3 | -3 | 5 | -5 | 15 | -15 |
x
| 2 | 0 | 4 | -2 | 6 | -4 | 16 | -14 |
y + 4 | 15 | -15 | 5 | -5 | 3 | -3 | 1 | -1 |
y | 11 | -19 | 1 | -9 | -1 | -7 | -3 | -5 |
Vậy \(\left\{{}\begin{matrix}x=2\\y=11\end{matrix}\right.\); \(\left\{{}\begin{matrix}x=0\\y=-19\end{matrix}\right.\); \(\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\);.........\(\left\{{}\begin{matrix}x=-14\\y=-5\end{matrix}\right.\)
b) A= 1+\(3^2\)+\(2^2\).\(3^3\)+\(3^2\).\(3^2\)+\(3^2\).\(4^2\)+\(3^2\).\(5^2\)+\(3^2\).\(6^2\)+.....\(3^2\).\(12^2\)
=\(3^2\).(\(\dfrac{1}{3}^2\)+\(2^2\)+\(3^2\)+\(4^2\)+\(5^2\)+\(6^2\)+......+\(12^2\))
=9.(\(\dfrac{1}{9}\)+649)=5842
b) Vì \(2^2+3^2+...+12^2=649\)
\(=>3^2\left(2^2+3^2+...+12^2\right)=3^2\cdot649\)
\(=>6^2+9^2+...+36^2=5841\)
=> A = \(1+9+5841\) = 5851
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Câu 2:
a, Các từ láy: lênh khênh, rả rích
b, Tác dụng: Làm cho bài thơ trở nên phong phú, đa dạng hơn
Câu 3:
a, BPTT: Ẩn dụ
b, Tác dụng: Cho thấy ánh nắng nhẹ nhàng, dịu êm chiếu lên hai cha con khi đi trên biển.
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\(\sqrt{9-3\sqrt{5}}-\sqrt{9+3\sqrt{5}}=\dfrac{1}{\sqrt{2}}\left(\sqrt{18-6\sqrt{5}}-\sqrt{18+6\sqrt{5}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{15}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{15}+\sqrt{3}\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{15}-\sqrt{3}-\sqrt{15}-\sqrt{3}\right)=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8-2\sqrt{15}}+\sqrt{8+2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}-2\left(\sqrt{5}-1\right)\right)\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
k: Ta có: \(\sqrt{9-3\sqrt{5}}-\sqrt{9+3\sqrt{5}}\)
\(=\dfrac{\sqrt{18-6\sqrt{5}}-\sqrt{18+6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{15}-\sqrt{3}-\sqrt{15}-\sqrt{3}}{\sqrt{2}}\)
\(=-\sqrt{6}\)