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1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
1) \(+2x+3y⋮17\)
\(\Rightarrow26x+39y⋮17\)
\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)
Mà \(17x+34y⋮17\)
\(\Rightarrow9x+5y⋮17\)
\(+9x+5y⋮17\)
\(\Rightarrow36x+20y⋮17\)
\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)
Mà \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
A=1+\(\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+\frac{1}{4}\cdot\frac{4\cdot5}{2}+....+\frac{1}{100}+\frac{100\cdot101}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)
\(=1+\left(\frac{101\cdot2}{2}-3\right)\cdot\frac{1}{2}=1+98\cdot\frac{1}{2}=49+1=50\)
a. \(\frac{20^5.5^{10}}{100^5}\)
\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)
\(=\frac{20^5.25^5}{100^5}\)
\(=\frac{500^5}{100^5}\)
\(=\left(\frac{500}{100}\right)^5\)
\(=5^5=3125\)
b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)
\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)
\(=3^5.\frac{1}{0,3}\)
\(=810\)
c. \(\frac{6^3+3.6^2+3^3}{-13}\)
\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)
\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\frac{3^3.13}{-13}\)
\(=\left(-3\right)^3\)
\(=-27\)
\(A=\dfrac{-1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow\dfrac{1}{3}A=\dfrac{-1}{3^2}+\dfrac{1}{3^3}-\dfrac{1}{3^4}+...-\dfrac{1}{3^{100}}+\dfrac{1}{3^{101}}\)
Cộng vế với vế:
\(A+\dfrac{1}{3}A=\dfrac{-1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{100}}+\dfrac{1}{3^{101}}\)
\(\Rightarrow\dfrac{4}{3}A=\dfrac{-1}{3}+\dfrac{1}{3^{101}}\)
\(\Rightarrow A=\dfrac{1}{4}\left(\dfrac{1}{3^{100}}-1\right)\)
Do \(\dfrac{1}{3^{100}}< \dfrac{1}{3}< 1\Rightarrow A< 0\)
\(\Rightarrow\left|A\right|=-A=-\dfrac{1}{4}\left(\dfrac{1}{3^{100}}-1\right)=\dfrac{1}{4}\left(1-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow B=4\left|A\right|+\dfrac{1}{3^{100}}=1-\dfrac{1}{3^{100}}+\dfrac{1}{3^{100}}=1\)