(\(\frac{2016}{2}+\frac{2016}{6}+\frac{2016}{12}+...+\frac{2016}{99.100}\)) . 0,(01)
Những câu hỏi liên quan
(\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+......+\frac{1}{99.100}\) ). x =\(\frac{2016}{51}+\frac{2016}{51}+\frac{2016}{53}+....+\frac{2016}{100}\)
tìm x
\(\frac{6}{\frac{2016}{\frac{12}{2016}-}}-\frac{6}{\frac{2017}{\frac{12}{2017}}}\)tính :
Ta có: 6/2016/12/2016 - 6/2017/12/2017
= 6/12 - 6/12 = 0
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Cho \(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(N=\frac{2016}{51}+\frac{2016}{52}+...+\frac{2016}{100}\)
CMR N chia hết cho M.
Tính giá trị biểu thức :
\(A=\frac{1}{2016^{-2016}+1}+\frac{1}{2016^{-2015}+1}+..............+\frac{1}{2016^{-1}+1}+\frac{1}{2016^0+1}+\frac{1}{2016^1+1}+.............+\frac{1}{2016^{2016}+1}\)
Cho \(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}\)
\(N=\frac{2016}{51}+\frac{2016}{52}+...+\frac{2016}{100}\)
CMR; N chia hết cho M
\(\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}+1\right)\left(\frac{2105}{2016}+\frac{2016}{2017}+\frac{7}{22}\right)-\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}\right)\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{7}{22}+1\right)\)
23 tháng 8 2016 lúc 10:00
Cho \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\).Chứng minh \(\frac{x^{2016}}{a^{2016}}+\frac{y^{2016}}{b^{2016}}+\frac{z^{2016}}{c^{2016}}=\frac{x^{2016}+y^{2016}+z^{2016}}{a^{2016}+b^{2016}+c^{2016}}\)
A frac{frac{3}{4}-frac{3}{11}+frac{3}{13}}{frac{5}{4}-frac{5}{11}+frac{5}{13}}+frac{frac{1}{2}-frac{1}{3}+frac{1}{4}}{frac{5}{4}-frac{5}{6}+frac{5}{8}}B frac{2^{12}.3^5-4^6.9^2}{left(2^2.3right)^6+8^4.3^5}-frac{5^{10}.7^3-25^5.49}{left(125.7right)^3+5^9.14^3}C frac{left(a^{2016}+b^{2016}right)^{2017}}{left(c^{2016}+d^{2016}right)^{2017}} frac{left(a^{2017}-b^{2017}right)^{2016}}{left(c^{2017}-d^{2017}right)^{2016}}
Đọc tiếp
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
C = \(\frac{\left(a^{2016}+b^{2016}\right)^{2017}}{\left(c^{2016}+d^{2016}\right)^{2017}}\)= \(\frac{\left(a^{2017}-b^{2017}\right)^{2016}}{\left(c^{2017}-d^{2017}\right)^{2016}}\)
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
cs ng làm đung r
đag định lm
Cho frac{x^2}{a^2}+ frac{y^2}{b^2}+ frac{z^2}{c^2}frac{x^2+y^2+z^2}{a^2+b^2+c^2}CMR: frac{x^{2016}}{a^{2016}}+ frac{y^{2016}}{b^{2016}}+ frac{z^{2016}}{c^{2016}} frac{x^{2016}+y^{2016}+z^{2016}}{a^{2016}+b^{2016}+c^{2016}}
Đọc tiếp
Cho \(\frac{x^2}{a^2}\)+ \(\frac{y^2}{b^2}\)+ \(\frac{z^2}{c^2}\)=\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
CMR: \(\frac{x^{2016}}{a^{2016}}\)+ \(\frac{y^{2016}}{b^{2016}}\)+ \(\frac{z^{2016}}{c^{2016}}\)= \(\frac{x^{2016}+y^{2016}+z^{2016}}{a^{2016}+b^{2016}+c^{2016}}\)
Ta có
\(1\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(1\Leftrightarrow x^2+\frac{\left(b^2+c^2\right)x^2}{a^2}+y^2+\frac{\left(a^2+c^2\right)y^2}{b^2}+z^2+\frac{\left(a^2+b^2\right)z^2}{c^2}=x^2+y^2+z^2\)
\(\Leftrightarrow\frac{\left(b^2+c^2\right)x^2}{a^2}+\frac{\left(c^2+a^2\right)y^2}{b^2}+\frac{\left(a^2+b^2\right)z^2}{c^2}=0\)
Ta thấy rằng cả 3 phân số đó đều \(\ge0\)nên tổng 3 phân số sẽ \(\ge0\)
Dấu = xảy ra khi x = y = z = 0
Với x = y = z = 0 thì
\(\frac{x^{2016}}{a^{2016}}+\frac{y^{2016}}{b^{2016}}+\frac{z^{2016}}{c^{2016}}=\frac{x^{2016}+y^{2016}+z^{2016}}{a^{2016}+b^{2016}+c^{2016}}\Leftrightarrow\frac{0}{a^{2016}}+\frac{0}{b^{2016}}+\frac{0}{c^{2016}}=\frac{0+0+0}{a^{2016}+b^{2016}+c^{2016}}\)
\(\Leftrightarrow0=0\)(đúng)
\(\Rightarrow\)ĐPCM
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