A=[(2/193-3/386).193/17+33/34]:[(7/2001+11/4002).2001/25+9/2]
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\(M=\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{2001}+\dfrac{11}{4002}\right)\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(M=\left[\left(\dfrac{4}{386}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{14}{4002}+\dfrac{11}{4002}\right)\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(M=\left(\dfrac{1}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right):\left(\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right)\)
\(M=\left(\dfrac{1}{34}+\dfrac{33}{34}\right):\left(\dfrac{1}{2}+\dfrac{9}{2}\right)\)
\(M=1:5\)
\(M=\dfrac{1}{5}\)
\(=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{1}{34}+\dfrac{33}{34}\right):\left[\dfrac{1}{2}+\dfrac{9}{2}\right]\)
=1/5
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right]\)
\(=\left[\left(\frac{4}{368}-\frac{3}{368}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{14}{4002}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right]\)
\(=\left(\frac{1}{368}.\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{4002}.\frac{2001}{25}+\frac{9}{2}\right)\)
\(=\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)
\(=1:5=\frac{1}{5}\)
\(M=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right] \)
\(=\left(\frac{2}{17}-\frac{3}{34}+\frac{33}{34}\right):\left(\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right)\)
\(=\frac{4-3+33}{34}:\frac{14+11+225}{50}=1:5=0.2\)
\(M=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{14+11}{4002}-\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{1}{17}\cdot\dfrac{193}{386}+\dfrac{33}{34}\right):\left[\dfrac{25}{4002}-\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=1:\dfrac{625-2001\cdot4002+9\cdot50525}{100050}\)
\(=-\dfrac{100050}{7552652}\)
Ta có : [ ( 2/193 - 3/386 ) * 193/17 + 32/34 ] : [ ( 7/2001 + 11/4002 ) * 2001/25 + 9/2 ] .
=> [ 2/193 * 193/17 - 3/386 * 193/17 + 32/34 ] : [ 7/2001 * 2001/25 + 11/4002 * 2001/25 + 9/2 ] .
=> [ 2/17 - 3/34 + 32/34 ] : [ 7/25 + 11/50 + 9/2 ] .
=> [ 4/34 - 3/34 + 32/34 ] : [ 14/50 + 11/50 + 225/50 ] .
=> 33/34 : 5 .
=> 33/34 * 1/5 .
=> 33/170 .
Tính bằng cách thuận tiện nhất:
\(=\left[\left(\frac{2}{193}\cdot\frac{193}{17}\right)-\left(\frac{3}{386}\cdot\frac{193}{17}\right)+\frac{32}{34}\right]:\left[\left(\frac{7}{2001}\cdot\frac{2001}{25}\right)+\left(\frac{11}{4002}\cdot\frac{2001}{25}\right)+\frac{9}{2}\right]\)
\(=\left[\left(\frac{2}{17}-\frac{3}{17}\right)+\frac{32}{34}\right]:\left[\left(\frac{7}{25}+\frac{11}{50}\right)+\frac{9}{2}\right]\)
\(=\left(-\frac{1}{17}+\frac{32}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)
\(=\frac{15}{17}+5\)
\(=\frac{100}{17}\)
~ học tốt ~
ta có
\(M=[(\dfrac{2}{193}-\dfrac{3}{386}).\dfrac{193}{17}+\dfrac{33}{34}]:[(\dfrac{7}{2001}+\dfrac{11}{4002}).\dfrac{2001}{25}+\dfrac{9}{2}]\)
\(\Rightarrow\)\(M=[\dfrac{1}{386}.\dfrac{193}{17}+\dfrac{33}{34}]:[\dfrac{25}{4002}.\dfrac{2001}{25}+\dfrac{9}{2}]\)
\(\Rightarrow\)\(M=[\dfrac{1}{34}+\dfrac{33}{34}]:[\dfrac{1}{2}+\dfrac{9}{2}]\)
\(\Rightarrow\)\(M=1:5\)
\(\Rightarrow M=\dfrac{1}{5}\)
Bài làm
\(A=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right]:\left[\frac{25}{4002}.\frac{2001}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{34}+\frac{33}{34}\right]:\left[\frac{1}{2}+\frac{9}{2}\right]\)
\(=\frac{1}{5}\)