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\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+\frac{15}{16.31}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}\)
\(=1-\frac{1}{31}\)
\(=\frac{30}{31}\)
Dựa vào công thức được chứng minh:
(Em có thể chứng minh lại)
Ta có:
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}\)
\(=1-\frac{1}{31}\)
\(=\frac{30}{31}\)
Chúc em học tốt^^
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a, \(4^7.3^4.9^6:6^{13}\)
\(=\left(2^{14}.3^4.3^{12}\right):\left(2^{13}.3^{13}\right)\)
\(=2^{14}:2^{13}.3^{16}:3^{13}\)
\(=2.3^3=54\)
b, \(2^3.3^2-5^{16}:25^7\)
\(=72-5^{16}:5^{14}\)
\(=72-5^2=47\)
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4^7.3^4.9^6:6^13=4^7.3^4.(3^2)^6:6^13
=4^7.3^16:3^13.2^13
=(2^2)^7.3^16:3^13.2^13
=2^14.3^16:3^13.2^13
=2.3^3
=54
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a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)
\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)
\(P=1-\dfrac{1}{56}\)
\(P=\dfrac{55}{56}\)
b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)
\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=3\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}\)
\(A=\dfrac{297}{100}\)
c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(B=1-\dfrac{1}{103}\)
\(B=\dfrac{102}{103}\)
d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)
\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}.\dfrac{102}{103}\)
\(C=\dfrac{170}{103}\)
e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)
\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}.\dfrac{104}{105}\)
\(D=\dfrac{26}{15}\)
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a)=-4.25.-8.-125.-3
=-100.1000.-3
=300000
b)=-8.-125.25.-4.-9
=1000.-100.-9
=900000
c)=-4.-25.-9.7
=100.-63
=-6300
d) làm theo cách thường(ko chắc)
\(C=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+\frac{15}{16.31}\) (số 9 mk viết thêm cho đề nó hợp lý)
\(=\frac{4-1}{1.4}+\frac{9-4}{4.9}+\frac{16-9}{9.16}+\frac{31-16}{16.31}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}\)
\(=1-\frac{1}{31}=\frac{30}{31}\)