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a)\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}......\frac{99}{100}\)
\(=\frac{1.2.3.4.....99}{2.3.4.5.6.....100}\)
\(=\frac{1}{100}\)
b) Tương tự như câu a
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{100}{99}\)
\(1\frac{1}{3}\times1\frac{1}{8}\times1\frac{1}{15}\times1\frac{1}{24}\times...\times1\frac{1}{99}\)
\(=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times\frac{25}{24}\times...\times\frac{100}{99}\)
\(=\frac{2\times2}{1\times3}\times\frac{3\times3}{2\times4}\times\frac{4\times4}{3\times5}\times\frac{5\times5}{4\times6}\times...\times\frac{10\times10}{9\times11}\)
\(=\frac{2}{1}\times\frac{10}{11}\)
\(=\frac{20}{11}\)
Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
a, (x+2)+(x+4)+(x+6)+...+(x+100)=6000
(x+x+x+...+x)+(2+4+6+...+100)=6000
50.x+2550=6000
50.x=6000-2550
50.x=3450
x=3450:50
x=69
b, 1+2+3+4+...+x=15
10+...+x=15
x=15-10
x=5
Nho **** cho minh nha
Ta có: (x+x+x+...+x) + (2+4+6+...+100) = 6000
Ta thấy vế phải có: (100-2):2+1=50(số hạng)
Tổng của vế phải: [(2+100)*50]:2=2550
\(\Rightarrow\)có 50 số x
\(\Rightarrow\)50*x + 2550 = 6000
\(\Rightarrow\)50*x=6000-2550
\(\Rightarrow\)50*x=3450
\(\Rightarrow\)x=3450:50
\(\Rightarrow\)x=69
Vậy x=69
Mình đúng nè, nhớ k nha
\(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\)
=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\)
=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)
2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)
\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\)
\(\dfrac{35}{12}-\dfrac{7}{5}\)
\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)
4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)
(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\)
6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)
7 + \(\dfrac{103}{28}\)
\(\dfrac{299}{28}\)
3/4 × 8/9 × 15/16 × ... × 99/100
= 1×3/2×2 × 2×4/3×3 × 3×5/4×4 × ... × 9×11/19×10
= 1×2×3×...×9/2×3×4×...×10 × 3×4×5×...×11/2×3×4×...×10
= 1/10 × 11/2
= 11/20
bạn làm rõ giúp mình nhé