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\(=1+\frac{3}{2}+2+\frac{5}{2}+3+\frac{7}{2}+...+8+\frac{17}{2}\)
\(=\left(1+2+...+8\right)+\left(\frac{3}{2}+\frac{5}{2}+...+\frac{17}{2}\right)=36+\frac{80}{2}=36+40=76\)
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)
\(B=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{16}.\frac{\left(1+16\right).16}{2}\)
\(B=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+\frac{1}{4}.\frac{5.4}{2}+...+\frac{1}{16}.\frac{17.16}{2}\)
\(B=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(B=\frac{1}{2}.\left(2+3+4+5+...+17\right)\)
\(B=\frac{1}{2}.\frac{\left(2+17\right).16}{2}=19.4=76\)
A = 1 + 1/2.(1+2) + 1/3.(1+2+3) + 1/4.(1+2+3+4) + ...+ 1/16.(1+2+....+16)
A = 1 + 1/2.3 + 1/3.6 + 1/4.10 + ...+ 1/16.136
A = 1 + 3/2 + 4/2 + 5/2 + ....+ 17/2
A = 1 + (3+4+5+...+17)/2
A = 1 + 150/2
A = 1 + 75
A = 76
A = 1 + 1/2.(1+2) + 1/3.(1+2+3) + 1/4.(1+2+3+4) + ...+ 1/16.(1+2+....+16)
A = 1 + 1/2.3 + 1/3.6 + 1/4.10 + ...+ 1/16.136
A = 1 + 3/2 + 4/2 + 5/2 + ....+ 17/2
A = 1 + (3+4+5+...+17)/2
A = 1 + 150/2
A = 1 + 75
A = 76
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{16}.\frac{16.17}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(=\frac{\frac{17.18}{2}-1}{2}=76\)
x. (x^2)^3 = x^5
x^7 ≠ x^5
Nếu,
x^7 - x^5 = 0
mủ lẻ nên phương trình có 3 nghiệm
Đáp số:
x = -1
hoặc
x = 0
hoặc
x = 1
a, \(\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot\left(1-\frac{1}{25}\right)\cdot\left(1-\frac{1}{36}\right)\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\frac{4.6}{5.5}\cdot\frac{5.7}{6.6}\)
\(=\frac{1.2.3.4.5}{2.3.4.5.6}\cdot\frac{3.4.5.6.7}{2.3.4.5.6}=\frac{1}{6}\cdot\frac{7}{2}\)
\(=\frac{7}{12}\)
b, \(\left(2-\frac{3}{2}\right)\cdot\left(2-\frac{4}{3}\right)\cdot\left(2-\frac{5}{4}\right)\cdot\left(2-\frac{6}{5}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1.2.3.4}{2.3.4.5}\)
\(=\frac{1}{5}\)
Ta có công thức : 1 + 2 + 3 + ... + n = \(\frac{n\left(n+1\right)}{2}\)
Do đó
P = \(1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+16}{16}\)
\(P=1+\frac{2.3}{2.2}+\frac{3.4}{2.3}+\frac{4.5}{2.4}+...+\frac{16.17}{2.16}\)
\(P=1+\frac{1}{2}\left(3+4+5+...+17\right)\)
\(P=1+\frac{1}{2}.\frac{\left(17-3+1\right)\left(3+17\right)}{2}=76\)
Xét thừa số tổng quát:
\(\frac{1+2+3+...+n}{n}=\frac{n\left(n+1\right)}{2n}=\frac{n+1}{2}\)
Thay vào bài toán:
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}=\frac{2+3+...+17}{2}=76\)
Đặt \(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(A=1+\frac{1+2}{2}+\frac{1+2+3}{3}+\frac{1+2+3+4}{4}+...+\frac{1+2+3+...+16}{16}\)
\(A=1+\frac{2\left(2+1\right):2}{2}+\frac{3\left(3+1\right):2}{3}+\frac{4\left(4+1\right):2}{4}+...+\frac{16\left(16+1\right):2}{16}\)
\(A=1+\frac{2+1}{2}+\frac{3+1}{2}+\frac{4+1}{2}+...+\frac{16+1}{2}\)
\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(A=\frac{2+3+4+5+...+17}{2}\)
\(A=\frac{152}{2}\)
\(A=76\)