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Bài làm
\(A=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right]:\left[\frac{25}{4002}.\frac{2001}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{34}+\frac{33}{34}\right]:\left[\frac{1}{2}+\frac{9}{2}\right]\)
\(=\frac{1}{5}\)
\(M=\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{2001}+\dfrac{11}{4002}\right)\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(M=\left[\left(\dfrac{4}{386}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{14}{4002}+\dfrac{11}{4002}\right)\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(M=\left(\dfrac{1}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right):\left(\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right)\)
\(M=\left(\dfrac{1}{34}+\dfrac{33}{34}\right):\left(\dfrac{1}{2}+\dfrac{9}{2}\right)\)
\(M=1:5\)
\(M=\dfrac{1}{5}\)
\(=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{1}{34}+\dfrac{33}{34}\right):\left[\dfrac{1}{2}+\dfrac{9}{2}\right]\)
=1/5
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right]\)
\(=\left[\left(\frac{4}{368}-\frac{3}{368}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{14}{4002}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right]\)
\(=\left(\frac{1}{368}.\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{4002}.\frac{2001}{25}+\frac{9}{2}\right)\)
\(=\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)
\(=1:5=\frac{1}{5}\)
a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)
b, \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)
a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền
b, - 1 hoặc + 1 vs mỗi phân số nha
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 ~
\(M=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right] \)
\(=\left(\frac{2}{17}-\frac{3}{34}+\frac{33}{34}\right):\left(\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right)\)
\(=\frac{4-3+33}{34}:\frac{14+11+225}{50}=1:5=0.2\)
khó thế tôi ko làm được
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