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63 - 33 = (6 - 3)3
63 . 33 = (6 . 3)3
63 : 33 = (6 : 3)3
63 + 33 = (6 + 3)3
Bạn có thể viết rõ đề lại không, mình không hiểu cho lắm
\(A=1+6+6^2+6^4+...+6^{100}\)
\(\Rightarrow6A=6+6^2+6^4+...+6^{100}+6^{101}\)
\(\Rightarrow6A-A=\left(6+6^2+6^4+....+6^{102}\right)-\left(1+6+6^2+6^4+...+6^{100}\right)\)
\(\Rightarrow5A=6^{101}-1\)
\(\Rightarrow A=\frac{6^{101}-1}{5}\)
Đặt \(A=5+5^2+5^3+....+5^{199}+5^{200}\)
\(\Leftrightarrow5A=5\left(5+5^2+5^3+....+5^{199}+5^{200}\right)\)
\(\Leftrightarrow5A=5^2+5^3+5^4+....+5^{200}+5^{201}\)
\(\Leftrightarrow5A-A=\left(5^2+5^3+5^4+....+5^{200}+5^{201}\right)-\left(5+5^2+5^3+....+5^{199}+5^{200}\right)\)
\(\Leftrightarrow4A=5^{201}-5\)
\(\Leftrightarrow A=\frac{5^{201}-5}{4}\)
B=\(1+3^2+3^4+...+3^{100}\)
9B=\(3^2+3^4+...+3^{100}\)
9B-B=\(\left(3^2+3^4+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
8B=\(3^{102}-1\)
B=\(\left(3^{102}-1\right):8\)
C=\(1+5^3+5^6+...+5^{99}\)
125C=\(5^3+5^6+5^9+...+5^{102}\)
125C-C=\(\left(5^3+5^6+5^9+...+5^{102}\right)-\left(1+5^3+5^6+...+5^{99}\right)\)
124C=\(5^{102}-1\)
C=\(\left(5^{102}-1\right):124\)
\(6+6^2+\cdot\cdot\cdot+6^{10}\)
\(=6\cdot\left(1+6\right)+6^3\cdot\left(1+6\right)+\cdot\cdot\cdot+6^9\cdot\left(1+6\right)\)
\(=6\cdot7+6^3\cdot7+\cdot\cdot\cdot+6^9\cdot7\)
\(=7\cdot\left(6+6^3+\cdot\cdot\cdot+6^9\right)⋮7\)
\(\Rightarrow6+6^2+\cdot\cdot\cdot\cdot+6^{10}⋮7\)
( 1 + 1 +1 ) ! = 6
2 + 2+ 2 = 6
3.3-3 = 6
5+(5:5) = 6
6+6-6 = 6
7- ( 7 : 7) = 6
2 + 2 + 2 = 6
3 x 3 - 3 = 6
5 : 5 + 5 = 6
6 x 6 : 6 = 6
7 - 7 : 7 = 6
a: \(A=\dfrac{2}{3}-4\cdot\dfrac{5}{4}=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
b: \(B=\dfrac{-2+5}{6}\cdot11-7=\dfrac{11}{2}-7=-\dfrac{3}{2}\)
c: \(=1+\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{27}{4}+5-1\)
\(=\dfrac{1}{3}+\dfrac{27}{4}+5=\dfrac{145}{12}\)
d: \(D=\dfrac{2}{3}\cdot\dfrac{2}{5}-\dfrac{1}{5}\cdot\dfrac{1}{6}=\dfrac{4}{15}-\dfrac{1}{30}=\dfrac{7}{30}\)
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-3+(3+6)
=-3+9
=6
-3 + ( 3 + 6 )
= -3 + 9
= 6
nhaaaaa