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A=1+\(\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+\frac{1}{4}\cdot\frac{4\cdot5}{2}+....+\frac{1}{100}+\frac{100\cdot101}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)
\(=1+\left(\frac{101\cdot2}{2}-3\right)\cdot\frac{1}{2}=1+98\cdot\frac{1}{2}=49+1=50\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
Bài 2:
a, 1/3 + 1/2 : x = -4
=> 1/2 : x = -4 - 1/3
=> 1/2 : x = -13/3
=> x = 1/2 ; -13/3
=> x = -3/26
Vậy x = -3 / 26
Bài 2:
b, x2 - 4x = 0
=> x.(x - 4) =0
=> x=0 hoặc x - 4 = 0
x - 4= 0 => x=4
Vậy x=0 và x=4
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{100}\left(1+2+...+100\right)\)
\(=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{100}\cdot\frac{100.101}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)
\(=\frac{1}{2}\left(2+3+...+101\right)=\frac{1}{2}\cdot\frac{100.103}{2}=25.103=2575\)