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A=[(1+2+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x (2,4x42 - 21x4,8)] / 1+1/2+1/3+...+1/100
= [(1+2+3+...+100) x (1/2 - 1/3 - 1/4-1/5) x (2,4x2x21 - 21x2x 4,8)] / 1+1/2+1/3+...+1/100
=[(1+2+3+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x 0] / 1+1/2+1/3+...+1/100
=0 / 1+1/2+1/3+...+1/100 = 0
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
Dễ thấy 6,3 . 12 - 21 . 3,6 = 63 . 1,2 - 63 . 1,2 = 0
Do đó biểu thức trên bằng 0
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)
\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)
\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\left[\frac{2}{3}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)
~ Hok tốt ~
$A=\dfrac{\left(\dfrac15-\dfrac27\right)\cdot\dfrac34-\dfrac34\cdot\left(\dfrac13-\dfrac27\right)}{\dfrac15\cdot\dfrac27-\dfrac13\cdot\left(\dfrac27+\dfrac39\right)+\dfrac39+\dfrac15}$
$=\dfrac{\dfrac34\left[\left(\dfrac15-\dfrac27\right)-\left(\dfrac13-\dfrac27\right)\right]}{\dfrac2{35}-\dfrac13\left(\dfrac27+\dfrac13\right)+\dfrac13+\dfrac15}$
$=\dfrac{\dfrac34\left(\dfrac15-\dfrac13\right)}{\dfrac2{35}-\dfrac13\cdot\dfrac{13}{21}+\dfrac13+\dfrac15}$
$=\dfrac{\dfrac34\cdot\left(-\dfrac2{15}\right)}{\dfrac2{35}-\dfrac{13}{63}+\dfrac{8}{15}}$
$=\dfrac{-\dfrac1{10}}{\dfrac{18-65+168}{315}}$
$=\dfrac{-\dfrac1{10}}{\dfrac{121}{315}}$
$=-\dfrac1{10}\cdot\dfrac{315}{121}$
$=-\dfrac{63}{242}.$
$\textbf{a)}$
$\left(-\dfrac34+\dfrac27\right):\dfrac27+\left(-\dfrac14+\dfrac57\right):\dfrac23$
$=\left(-\dfrac{13}{28}\right)\cdot\dfrac72+\dfrac{13}{28}\cdot\dfrac32$
$=-\dfrac{13}{8}+\dfrac{39}{56}$
$=-\dfrac{13}{14}.$
$\textbf{b)}$
$\left(-\dfrac13\right)^2\cdot\dfrac4{11}+\dfrac7{11}\cdot\left(-\dfrac13\right)^2$
$=\dfrac19\left(\dfrac4{11}+\dfrac7{11}\right)$
$=\dfrac19.$
$A=\left(\dfrac14-1\right)\left(\dfrac19-1\right)\left(\dfrac1{16}-1\right)\cdots\left(\dfrac1{100}-1\right)\left(\dfrac1{121}-1\right)$
$=\left(-\dfrac34\right)\left(-\dfrac89\right)\left(-\dfrac{15}{16}\right)\cdots\left(-\dfrac{99}{100}\right)\left(-\dfrac{120}{121}\right)$
$=(-1)^{10}\cdot\dfrac34\cdot\dfrac89\cdot\dfrac{15}{16}\cdots\dfrac{99}{100}\cdot\dfrac{120}{121}$
$=\dfrac34\cdot\dfrac89\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdots\dfrac{99}{100}\cdot\dfrac{120}{121}$
$=\dfrac{3\cdot8\cdot15\cdot24\cdots99\cdot120}{4\cdot9\cdot16\cdot25\cdots100\cdot121}$
$=\dfrac{(1\cdot2\cdot3\cdots10)\,(3\cdot4\cdot5\cdots12)}{(2\cdot3\cdot4\cdots11)^2}$
$=\dfrac{1\cdot12}{2\cdot11}$
$=\dfrac6{11}.$
Cảm ơn các bạn
Ta thấy biểu thức trong ngoặc thứ ba của tử số bằng 0
\(\Rightarrow\)tử số phân số trên bằng 0
\(\Rightarrow\) phân số trên bằng 0
\(=\frac{\left(1+2+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(=\frac{0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{100}}\)
\(=0\)
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