Rút gọn :
A = \(\frac{2^3+3^2-5^2}{5^2-3^2+2^3}\)
B = \(\frac{2^3.3^2-3^2.4^2}{3^2.4^3-2^5.3^4}\)
Tính:
a,A=\(\dfrac{12^{15}.3^4-4^5.3^9}{27^3.2^{10}-32^3.3^9}\)
b. B= \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^3.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{99}{49^2.50^2}\)
\(A=\dfrac{12^{15}\cdot3^4-4^5\cdot3^9}{27^3\cdot2^{10}-32^3\cdot3^9}\\ =\dfrac{\left(2^2\cdot3\right)^{15}\cdot3^4-\left(2^2\right)^5\cdot3^9}{\left(3^3\right)^3\cdot2^{10}-\left(2^5\right)^3\cdot3^9}\\ =\dfrac{2^{30}\cdot3^{15}\cdot3^4-2^{10}\cdot3^9}{3^9\cdot2^{10}-2^{15}\cdot3^9}\\ =\dfrac{3^9\cdot2^{10}\left(2^{20}\cdot3^{10}\right)}{3^9\cdot2^{10}\left(1-2^5\right)}\\ =\dfrac{\left(2^2\right)^{10}\cdot3^{10}}{1-32}\\ =\dfrac{\left(2^2\cdot3\right)^{10}}{-31}\\ =\dfrac{-12^{10}}{31}\)
\(B=\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{99}{49^2\cdot50^2}\\ =\dfrac{2^2-1^2}{1^2\cdot2^2}+\dfrac{3^2-2^2}{2^2\cdot3^2}+...+\dfrac{50^2-49^2}{49^2\cdot50^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{49^2}-\dfrac{1}{50^2}\\ =1-\dfrac{1}{2500}\\ =\dfrac{2499}{2500}\)
\(\frac{5^4-5^3}{100}\)
\(\frac{3^3+3^2+3}{39}\)
\(\frac{5.3^7-5.3^5}{5.3^5-5.3^3}\)
\(\frac{2^{15}+2^{14}+2^{13}}{2^{13}+2^{12}+2^{11}}\)
đề bài yêu cầu là rút gọn các bạn nhớ giúp mình nha
5^4-3/100=1/20
3^3+2+1/3*13=3^5/13
5.3^7-5/5.3^5-3=1
2^15+14+13/2^13+12+11=2^6
Rút gọn biểu thức
a)\(\frac{\left(\frac{2}{3}\right)^3.\left(-\frac{3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(-\frac{5}{12}\right)^2}\)
b)\(6^6+6^3.3^3+3^6\)/-73
A=10^2-(5^2.4-4^3.3)+2^3
Ai tick mk lên 30 -> 40 điểm mk tick cho cả tháng
chứng minh rằng:
a) A= \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)<1
b)B=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\text{(đpcm) }\)
rút gọn:
a)\(\frac{4^3.1^5}{9^2.8^2}\) b)\(\frac{25^2.2^2.4^1}{3^3.2^3.224}\) c)\(\frac{25^3.5^6.5^2}{9^4.10^2}\)
\(a.\frac{4^3.1^5}{9^2.8^2}=\frac{2^6.1}{3^4.2^6}=\frac{1}{81}\)
\(b.\frac{25^2.2^2.4^1}{3^3.2^3.224}=\frac{5^4.2^4}{3^3.2^8.7}=\frac{5^4}{3^3.2^4.7}=\frac{625}{3024}\)
\(c.\frac{25^3.5^6.5^2}{9^4.10^2}=\frac{5^{14}}{3^8.2^2.5^2}=\frac{5^{10}}{3^8.2^2}\)
Thu gọn :
A = \(\frac{2^4.3^3+2^3.3^4}{2^5.3^3-2^4.3^2}\)
\(A=\frac{2^4.3^3+2^3.3^4}{2^5.3^3-2^4.3^2}\)
\(A=\frac{2^3.3^3\left(2+3\right)}{2^4.3^2\left(2.3-1\right)}\)
\(A=\frac{2^3.3^3.5}{2^4.3^2.5}\)
\(A=\frac{3}{2}\)
\(A=\frac{2^4.3^3+2^3.3^4}{2^5.3^3-2^4.3^2}\)
\(A=\frac{2^3.3^3\left(2+3\right)}{2^4.3^2\left(2.3-1\right)}\)
\(A=\frac{2^3.3^3.5}{2^4.3^2.5}\)
\(A=\frac{3}{2}\)
Chướng minh rằng:
a, \(\frac{1}{1^2.2^2}\)+$\frac{5}{2^2.3^2}$+$\frac{5}{3^2.4^2}$+...+$\frac{5}{9^2.10^2}$ <1
b, \(\frac{1}{3}\)+\(\frac{2}{3^2}\)+$\frac{3}{3^3}$+$\frac{4}{3^4}$+...+$\frac{100}{3^100}$ <\(\frac{3}{4}\)
rút gọn r quy đồng mẫu các ps
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}va\frac{3^4.5-3^6}{3^4.13+3^4}\)