bài 2: rút gọn biểu thức sau:
(\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(11^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)...\left(12^4+\frac{1}{4}\right)}\)
Những câu hỏi liên quan
Rút gọn biểu thức A = \(\left(2-1\frac{1}{4}\right)\left(2-1\frac{1}{9}\right)\left(2-1\frac{1}{16}\right)...\left(2-1\frac{1}{400}\right)\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{399}{400}\Rightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{19.21}{20.20}\Rightarrow\frac{1.2.3...19}{2.3.4...20}.\frac{3.4.5...21}{2.3.4...20}\) \(\Rightarrow A=\frac{1}{20}.\frac{21}{2}=\frac{21}{40}\)
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\(\frac{\left(1^2+\frac{1}{4}\right)\left(3^2+\frac{1}{4}\right)\left(5^2+\frac{1}{4}\right)..........\left(29^2+\frac{1}{4}\right)}{\left(2^2+\frac{1}{4}\right)\left(4^2+\frac{1}{4}\right)\left(6^2+\frac{1}{4}\right)...........\left(30^2+\frac{1}{4}\right)}\)
tính giá trị biểu thức trên
Tính giá trị biểu thức \(M=\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(2014^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(2013^4+\frac{1}{4}\right)}\) .
Tổng quát: a^4+1/4=(a^2+1/2)^2-a^2=(a^2+1/2-a)(a^2+1/2+a)=[(a-1/2)^2+1/4][(a^1/2)^2+1/4]=[(a-0,5)^2+0,25][(a+0,5)^2+0,25]
Tử số của M=[(2-0,5)^2+0,25][(2+0,5)^2+0,25][(4-0,5)^2+0,25][(4+0,5)^2+0,25][(6-0,5)^2+0,25][(6+0,5)^2+0,25]....[(2014-0,5)^2+0,25][(2014+0,5)^2+0,25]
=(1,5^2+0,25)(2,5^2+0,25)(3,5^2+0,25)(4,5^2+0,25)(5,5^2+0,25)(6,5^2+0,25)....(2013,5^2+0,25)(2014,5^2+0,25)
Mẫu số của M=[(1-0,5)^2+0,25][(1+0,5)^2+0,25][(3-0,5)^2+0,25][(3+0,5)^2+0,25][(5-0,5)^2+0,25][(5+0,5)^2+0,25]....[(2013-0,5)^2+0,25][(2013+0,5)^2+0,25]
=(0,5^2+0,25)(1,5^2+0,25)(2,5^2+0,25)(3,5^2+0,25)(4,5^2+0,25)(5,5^2+0,25)....(2012,5^2+0,25)(2013,5^2+0,25)
Vậy M=(2014,5^2+0,25)/(0,5^2+0,25)
Còn bao nhiêu bạn tính tiếp nhá
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Xem thêm câu trả lời
Tính giá trị biểu thức \(M=\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(2014^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(2013^4+\frac{1}{4}\right)}\) .
Xét số hạng tổng quát:
\(k^4+\frac{1}{4}=\left(k^4+2\cdot\frac{1}{2}\cdot k^2+\frac{1}{4}\right)-k^2\)
= \(\left(k^2+\frac{1}{2}\right)^2-k^2\)= \(\left(k^2-k+\frac{1}{2}\right)\left(k^2+k+\frac{1}{2}\right)\)
Thay k từ 1 đến 2014 , ta được
M=
\(\frac{\left(2+\frac{1}{2}\right)\left(6+\frac{1.}{2}\right)...\left(4054182+\frac{1}{2}\right)\left(4058210+\frac{1}{2}\right)}{\frac{1}{2}\cdot\left(2+\frac{1}{2}\right)...\left(4050156+\frac{1}{2}\right)\left(4054182+\frac{1}{2}\right)}\)=\(\frac{4058210+\frac{1}{2}}{\frac{1}{2}}=8116421\)
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rút gọn biểu thức( không tính cái gạch ở cuối nha)\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4x^2-4}{5}\)
\(=\left(\frac{x^2+2x+1+6-x^2+x-3x+3}{2\left(x-1\right)\left(x+1\right)}\right)\frac{4\left(x^2-1\right)}{5}\)
\(=\frac{10}{2\left(x-1\right)
\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
Vậy giá trị của biểu thức là 4
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23 tháng 8 2016 lúc 10:43
Tính : \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)\left(7^4+\frac{1}{4}\right)\left(9^4+\frac{1}{4}\right)\left(11^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)\left(8^4+\frac{1}{4}\right)\left(10^4+\frac{1}{4}\right)\left(12^4+\frac{1}{4}\right)}\)
Tính A=\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(11^4+\frac{1}{4}\right)}{\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(12^4+\frac{1}{4}\right)}{ }}\)
Xét số hạng tổng quát:
\(k^4+\frac{1}{4}=\left(k^4+2\cdot\frac{1}{2}\cdot k^2+\frac{1}{4}\right)-k^2\)=\(\left(k^2+\frac{1}{2}\right)^2-k^2\)
= \(\left(k^2+\frac{1}{2}-k\right)\left(k^2+\frac{1}{2}+k\right)\)
Thay k từ 1 đến 12 ta được:
A=\(\frac{\frac{1}{2}\cdot\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(110+\frac{1}{2}\right)\left(132+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)...\left(132+\frac{1}{2}\right)\left(152+\frac{1}{2}\right)}\)=\(\frac{\frac{1}{2}}{152+\frac{1}{2}}=\frac{1}{305}\)
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Vì cộng thêm k2 trong ngoặc nên phải trừ đi k2
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Tính giá trị của biểu thức:Tfrac{4}{2.4}+frac{4}{4.6}+frac{4}{6.8}+...+frac{4}{2008.2010}Afrac{1}{1.3}+frac{1}{3.5}+frac{1}{5.7}+...+frac{1}{2007.2009}+frac{1}{2009.2011}Cleft(1-frac{1}{2}right)left(1-frac{1}{3}right)left(1-frac{1}{4}right)...left(1-frac{1}{1000}right)Sfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}Bleft(1+frac{1}{2}right)left(1+frac{1}{3}right)left(1+frac{1}{4}right)...left(1+frac{1}{100}right)Dleft(1-frac{1}{17}right)left(1-frac{2}{17}right)left(1-frac{3}{17}right)...left(1-f...
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Tính giá trị của biểu thức:
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)
\(S=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{100}\right)\)
\(D=\left(1-\frac{1}{17}\right)\left(1-\frac{2}{17}\right)\left(1-\frac{3}{17}\right)...\left(1-\frac{27}{17}\right)\)
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
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\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
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\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(C=\frac{1.2.3...99}{2.3.4...100}\)
\(\Rightarrow C=\frac{1}{100}\)
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left(frac{-5}{12}+frac{7}{4}-frac{3}{8}right)-left[4frac{1}{2}-7frac{1}{3}right]-left(frac{1}{4}-frac{5}{2}right)left[2frac{1}{4}-5frac{3}{2}right]-left(frac{3}{10}-1right)-5frac{1}{2}+left(frac{1}{3}-frac{5}{6}right)frac{4}{7}-left(3frac{2}{5}-1frac{1}{2}right)-frac{5}{21}+left[3frac{1}{2}-4frac{2}{3}right]frac{1}{8}-1frac{3}{4}+left(frac{7}{8}-3frac{7}{2}+frac{3}{4}right)-left[frac{7}{4}-frac{5}{8}right]left(frac{3}{5}-2frac{1}{10}+frac{11}{20}right)-left[frac{-3}{4}+1frac{7}{2}right]left[-2fr...
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\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
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