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#)Giải :
a)\(2^6.5^6=\left(2.5\right)^6=10^6\)
b)\(8^2.5^2=\left(8.5\right)^2=40^2\)
c)\(4^3.5^3=\left(4.5\right)^3=20^3\)
d)\(5^2.6^2.3^2=\left(5.6.2\right)^2=60^2\)
e)\(\frac{625^5}{25^8}=\frac{\left(25^2\right)^5}{25^8}=\frac{25^{10}}{25^8}=25^2\)
g)\(\frac{3^9}{7}.\frac{7^9}{3}=\frac{\left(3.7\right)^9}{7.3}=\frac{21^9}{21}=21^8\)
\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{1}{18}\)
\(b,2^8:2^5+3^3.2-12\)
\(=2^3+9.2-12\)
\(=8+18-12\)
\(=26-12\)
\(=14\)
Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!
Sửa lại câu b
\(=2^3+27.2-12\)
\(=8+54-12\)
\(=62-12\)
\(=50\)
Đặt \(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(\Rightarrow A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(\Rightarrow A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(\Rightarrow A=\frac{1}{1^2}-\frac{1}{10^2}\)
\(\Rightarrow A=1-\frac{1}{10^2}\)
Mà \(1-\frac{1}{10^2}< 1.\)
\(\Rightarrow A< 1\left(đpcm\right).\)
Chúc bạn học tốt!
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+....+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+....+\frac{1}{9^2}-\frac{1}{10^2}=\frac{1}{1^2}-\frac{1}{10^2}<1\)
=>đpcm
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\text{(đpcm) }\)
Ta có :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\)\(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=\)\(\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+\frac{4^2}{3^2.4^2}-\frac{3^2}{3^2.4^2}+...+\frac{10^2}{9^2.10^2}-\frac{9^2}{9^2.10^2}\)
\(=\)\(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=\)\(1-\frac{1}{10^2}\)
\(=\)\(\frac{100-1}{100}\)
\(=\)\(\frac{99}{100}\)
Chúc bạn học tốt ~
=3/1.4+5/4.9+7/9.16+......+19/81.100
=(1/1-1/4)+(1/4-1/9)+........+(1/81-1/100)
=1-1/100
=99/100<1(đpcm)
Xét số bất kì a. Ta sẽ chứng mỉnh (a + 1)2 - a2 = 2a + 1.
Thật vậy, ta có (a + 1)2 - a2 = a(a + 1) + (a + 1) - a2 = (a2 + a) + (a + 1) = 2a + 1 (đpcm).
Áp dụng ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=\frac{1}{1^2}-\frac{1}{10^2}< 1\left(đpcm\right)\)
\(a.\frac{4^3.1^5}{9^2.8^2}=\frac{2^6.1}{3^4.2^6}=\frac{1}{81}\)
\(b.\frac{25^2.2^2.4^1}{3^3.2^3.224}=\frac{5^4.2^4}{3^3.2^8.7}=\frac{5^4}{3^3.2^4.7}=\frac{625}{3024}\)
\(c.\frac{25^3.5^6.5^2}{9^4.10^2}=\frac{5^{14}}{3^8.2^2.5^2}=\frac{5^{10}}{3^8.2^2}\)