\(\left|-1\frac{1}{2}\right|+25\%+\frac{1}{4}\)
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D= [(1-1/2)(1-1/3)...(1-1/25)]:[(1+1/2)(1+1/3)...(1+1/25)]
D= [1/2. 2/3. ... . 24/25]: [3/2. 4/3. ... . 26/25]
D= 1/25 : 2/26
D= 1/25 . 26/2= 13/25
Vậy D= 13/25
\(D=\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{25}\right)\right]\)\(:\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{25}\right)\right]\)
\(D=\left[\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{24}{25}\right]:\left[\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{26}{25}\right]\)
\(D=\frac{1.2.3...24}{2.3.4...25}:\frac{3.4.5...26}{2.3.4...25}\)
\(D=\frac{1}{25}:13\)
\(D=\frac{1}{325}\)
\(a,A=\left[\frac{4}{11}.\left(\frac{1}{25}\right)^0+\frac{7}{22}.2\right]^{2010}-\left(\frac{1}{2^2}:\frac{8^2}{4^4}\right)^{2009}\)
\(A=\left(\frac{4}{11}.1+\frac{7}{11}\right)^{2010}-\left(\frac{1}{2^2}.2^2\right)^{2009}\)
\(A=1-1=0\)
\(b,B=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)
\(B=\frac{0,8:1}{\frac{3}{5}}+\frac{\left(1\right):\frac{4}{7}}{\left(\frac{59}{9}-\frac{13}{4}\right).36}\)
\(B=0,8.\frac{5}{3}+\frac{\frac{7}{4}}{\frac{119}{36}.36}\)
\(B=\frac{4}{3}+\frac{7}{4}.\frac{1}{119}\)
\(B=\frac{4}{3}+\frac{1}{68}=\frac{275}{204}\)
Ta có: \(\frac{0,8:\left(\frac{4}{5}\cdot1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)\cdot2\frac{2}{17}}+\frac{\left(1,2\cdot0,5\right)}{\frac{4}{5}}\)
\(=\frac{\frac{4}{5}:\left(\frac{4}{5}\cdot\frac{5}{4}\right)}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right)\cdot\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right)\cdot\frac{36}{17}}+\frac{6}{5}\cdot\frac{1}{2}\cdot\frac{5}{4}\)
\(=\frac{\frac{4}{5}}{\frac{3}{5}}+\frac{\frac{7}{4}}{\frac{119}{36}\cdot\frac{36}{17}}+\frac{3}{4}\)
\(=\frac{4}{5}\cdot\frac{5}{3}+\frac{7}{4}\cdot\frac{1}{7}+\frac{3}{4}=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}=\frac{7}{3}\)
x. (x^2)^3 = x^5
x^7 ≠ x^5
Nếu,
x^7 - x^5 = 0
mủ lẻ nên phương trình có 3 nghiệm
Đáp số:
x = -1
hoặc
x = 0
hoặc
x = 1
a, \(\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot\left(1-\frac{1}{25}\right)\cdot\left(1-\frac{1}{36}\right)\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\frac{4.6}{5.5}\cdot\frac{5.7}{6.6}\)
\(=\frac{1.2.3.4.5}{2.3.4.5.6}\cdot\frac{3.4.5.6.7}{2.3.4.5.6}=\frac{1}{6}\cdot\frac{7}{2}\)
\(=\frac{7}{12}\)
b, \(\left(2-\frac{3}{2}\right)\cdot\left(2-\frac{4}{3}\right)\cdot\left(2-\frac{5}{4}\right)\cdot\left(2-\frac{6}{5}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1.2.3.4}{2.3.4.5}\)
\(=\frac{1}{5}\)
\(=\frac{\frac{6}{5}:\left(\frac{6}{5}-\frac{5}{4}\right)}{\frac{8}{25}+\frac{2}{25}}+\frac{\frac{2027}{25}:\frac{9}{4}}{\left(\frac{38}{7}-\frac{9}{4}\right):\frac{267}{56}}\)
\(=\frac{\frac{6}{5}:\left(\frac{-1}{20}\right)}{\frac{2}{5}}+\frac{\frac{8180}{225}}{\frac{89}{28}:\frac{167}{56}}\)
\(=\frac{-12}{5}:\frac{2}{5}+\frac{8180}{225}:\frac{178}{167}\)
\(=-1+...\)ra số to vcl
Đề sai à ???
\(25\cdot\left(-\frac{1}{3}\right)^3+\frac{1}{5}-2\cdot\left(-\frac{1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\frac{-1}{27}+\frac{1}{5}-2\cdot\frac{\left(-1\right)^2}{2^2}-\frac{1}{2}\)
\(=\frac{-25}{27}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=\frac{-98}{135}-\frac{1}{2}-\frac{1}{2}=\frac{-233}{135}\)
\(\left(-2\right)\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
\(=\left(-2\right)\left(\frac{3}{4}-\frac{25}{100}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)
\(=\left(-2\right)\left(\frac{75}{100}-\frac{25}{100}\right):\left(\frac{27}{12}-\frac{14}{12}\right)\)
\(=\left(-2\right)\cdot\frac{50}{100}:\frac{13}{12}\)
\(=\left(-2\right)\cdot\frac{1}{2}:\frac{13}{12}\)
\(=-1\cdot\frac{12}{13}=-\frac{12}{13}\)
\(\left|-1\frac{1}{2}\right|+25\%+\frac{1}{4}\)
\(=\left|-\frac{3}{2}\right|+\frac{1}{4}+\frac{1}{4}\)
\(=\frac{3}{2}+\frac{1}{2}\)
\(=2\)