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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
\(1\frac{1}{3}\)x \(1\frac{1}{8}\)x \(1\frac{1}{15}\)x .......x \(1\frac{1}{99}\)= \(\frac{4}{3}\)x \(\frac{9}{8}\)x\(\frac{16}{15}\)x..........x \(\frac{100}{99}\)
=\(\frac{2.2}{1.3}\)x\(\frac{3.3}{2.4}\)x\(\frac{4.4}{3.5}\)x.......x \(\frac{10.10}{9.11}\)= \(\frac{2.3.4....10}{1.2.3.....9}\)x \(\frac{2.3.4....10}{3.4.5....11}\)
= \(\frac{10}{1}\)x\(\frac{2}{11}\)=\(\frac{20}{11}\)
4/3 x 9/8 x16/15 x ..... x 100/99
=2x2/1x3 x 3x3/2x4 x 4x4/3x5 x ... x 10x10/9x11
=2x3x4x....x10/1x2x3x..x9 x 2x3x4x...x10/3x4x5x...x11
=10 x 2/11=20/11
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