\(Cho\)\(x=\left(1+\frac{1}{\sqrt{1}}\right)+\left(1+\frac{1}{\sqrt{9}}\right)+\left(\frac{1}{\sqrt{25}}\right)+\left(1+\frac{1}{\sqrt{49}}\right)+\left(1+\frac{1}{\sqrt{81}}\right)\)
\(y=\left(1+\frac{1}{\sqrt{4}}\right)+\left(1+\frac{1}{\sqrt{16}}\right)+\left(1+\frac{1}{\sqrt{36}}\right)+\left(1+\frac{1}{\sqrt{64}}\right)+\left(1+\frac{1}{\sqrt{100}}\right)\)
Tính x.y
Mn ơi, giúp mk nha, mai mk nộp òi!
=(1/1+1/3+1/5+1/7+1/9).(1/2+1/4+1/6+1/8+1/10)
=(1+1/3+1/5+1/7+1/9).137/120
=258/315.137/120
=5891/6300
Tính:
a, \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right).\left(1-\frac{1}{36}\right)\)
b, \(\left(2-\frac{3}{2}\right).\left(2-\frac{4}{3}\right).\left(2-\frac{5}{4}\right).\left(2-\frac{6}{5}\right)\)
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, \(\left(2-\frac{3}{2}\right)\cdot\left(2-\frac{4}{3}\right)\cdot\left(2-\frac{5}{4}\right)\cdot\left(2-\frac{6}{5}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1.2.3.4}{2.3.4.5}\)
\(=\frac{1}{5}\)
Tính :
\(A=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right)........\left(1-\frac{1}{196}\right).\left(1-\frac{1}{225}\right)\)
Tính:
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)\left(\frac{1}{25}-1\right)...\left(\frac{1}{121}-1\right)\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)\left(\frac{1}{25}-1\right)....\left(\frac{1}{121}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.\frac{-24}{25}....\frac{-120}{121}\)
\(=\left[\left(-1\right)\left(-1\right)\left(-1\right)\left(-1\right)....\left(-1\right)\left(10\right)\text{thừa số -1 }\right].\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{10.12}{11.11}\)
\(=\frac{1.12}{2.11}=\frac{6}{11}\)
A=\(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{100}\right)\)
B=\(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)+\left(1+\frac{1}{15}\right)......\left(1+\frac{1}{100}\right)\)
D=\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2499}{2500}\)
Tính\
\(\left(\frac{1}{16}-1\right).\left(\frac{1}{25}-1\right).\left(\frac{1}{36}-1\right)......\left(\frac{1}{100}-1\right)\))
Xin lỗi bn mk bị nhầm
\(\frac{3.5}{4^2}.\frac{4.6}{5^2}.\frac{5.7}{6^2}...\frac{9.11}{10^2}\)
= \(\frac{3.4.5...9}{4.5.6...9.10}.\frac{5.6...9.10.11}{4.5.6...9.10}=\frac{3}{10}.\frac{11}{4}=\frac{33}{40}\)
\(\left(\frac{1}{16}-1\right)\left(\frac{1}{25}-1\right)\left(\frac{1}{36}-1\right)...\left(\frac{1}{100}-1\right)\)
= \(\frac{15}{16}.\frac{24}{25}.\frac{35}{36}...\frac{99}{100}\)
= \(\frac{3.5}{4^2}.\frac{4.6}{5^2}.\frac{5.7}{6^2}...\frac{9.11}{10^2}\)
= \(\frac{3.4.5...9.10}{4.5...9.10}.\frac{5.6.7..9.10.11}{4.5.6...9.10}=3.\frac{11}{4}=\frac{33}{4}\)
Tính nhanh:
A = \(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).\left(\frac{1}{25}-1\right)......\left(\frac{1}{121}-1\right)\)
Tính các tích sau:
a)\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)
b)\(B=\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).\left(1-\frac{1}{36}\right)......\left(1-\frac{1}{1326}\right)\)
c)\(C=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{99.101}\right)\)
Tính a):\(\frac{\left(\frac{4}{9}\right)^2.\left(-\frac{9}{16}\right).\left(-1\right)^{19}}{\left(\frac{4}{25}\right)^2.\left(-\frac{25}{144}\right)^2.\left(-\frac{49}{144}\right)^2}\)
b)\(\frac{\left(\frac{2}{3}\right)^3.\left(-\frac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2.\left(-\frac{5}{12}\right)^3}\)
