\(\frac{101}{19}\)x \(\frac{61}{218}\)- \(\frac{101}{218}\)x \(\frac{42}{19}\)+ \(\frac{117}{218}\)
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Tính hợp lý:a) A frac{101}{19}. frac{61}{218}- frac{101}{218}. frac{42}{19}+ frac{117}{218}b) B left(frac{5}{2011^2}+frac{7}{2012^2}-frac{9}{2013^2}right). left(frac{4}{5}-frac{3}{4}-frac{1}{20}right)
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Tính hợp lý:
a) A= \(\frac{101}{19}\). \(\frac{61}{218}\)\(-\) \(\frac{101}{218}\). \(\frac{42}{19}\)+ \(\frac{117}{218}\)
b) B= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right)\). \(\left(\frac{4}{5}-\frac{3}{4}-\frac{1}{20}\right)\)
a) A = \(\frac{101}{19}.\) \(\frac{61}{218}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)
= \(\frac{101}{218}.\frac{61}{19}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)
=\(\frac{101}{218}.\left(\frac{61}{19}-\frac{42}{19}\right)+\frac{117}{218}\)
=\(\frac{101}{218}.\frac{19}{19}+\frac{117}{218}\)
=\(\frac{101}{218}.1+\frac{117}{218}\)
=\(\frac{101}{218}+\frac{117}{218}\)
=\(\frac{218}{218}\)\(=1\)
b) B = \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).\left(\frac{4}{5}-\frac{3}{4}-\frac{1}{20}\right)\)
= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right)\)\(.\left(\frac{1}{20}-\frac{1}{20}\right)\)
= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).0\)
= \(0\)
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tính hợp lý
A=101/19*61/218-101/218*42/19+117/218
Tính hợp lí :
\(B=\frac{20+\frac{19}{3}+\frac{19}{101}}{7+\frac{7}{13}+\frac{7}{19}+\frac{7}{101}}\)
So sánh 2 phân số sau :
\(\frac{217}{18}\)và \(\frac{218}{19}\)
Ta có : \(\frac{217}{18}-1=\frac{199}{18}\)
\(\frac{218}{19}-1=\frac{199}{19}\)
Do \(\frac{199}{18}>\frac{199}{19}\Rightarrow\frac{217}{18}< \frac{218}{19}\)
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\(\frac{217}{18}=12+\frac{1}{8}>11+1>11+\frac{9}{19}=\frac{218}{19}\)
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217/18 - 1 = 199/18 ; 218/19 - 1 = 199/19
199/18 > 199/19 nên 217/18 > 217/19
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Xem thêm câu trả lời
\(\frac{10}{11}x\frac{12}{13}:\frac{50}{51}-\frac{19}{20}x\frac{12}{13}:\frac{101}{102}+\frac{99}{100}\)
so sánh các phân số sau theo cách thuận tiện:
\(\frac{13}{19};\frac{47}{53};\frac{-13}{-20}\)
\(\frac{125}{131};\frac{117}{109};\frac{101}{93}\)
Dễ ợt! Đại ca toàn đây
giải phương trình
a,\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{9\cdot10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
b,\(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+\frac{40}{39}\)
c,(x-20)+(x-19)+(x-18)+...+100+101=101
a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)
\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)
=>-9/10=-9/10(luôn đúng)
b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)
=>347x+780=1552
=>347x=772
hay x=772/347
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\(\times\frac{218}{\frac{##}{+\frac{####}{\frac{###}{####}}}}\)tim # m c o
giải phương trìnha,(frac{1}{1cdot2}+frac{1}{2cdot3}+frac{1}{3cdot4}+frac{1}{9cdot10})(x-1)+frac{1}{10}xx-frac{9}{10}b,frac{x+1}{1}+frac{2x+3}{3}+frac{3x+5}{5}+frac{20x+39}{39}22+frac{4}{3}+frac{6}{5}+frac{40}{39}c,(x-10)+(x-19)+(x-18)+...+100+101101d,(frac{1}{1cdot51}+frac{1}{2cdot52}+frac{1}{3cdot53}+...+frac{1}{10cdot60})xfrac{1}{1cdot11}+frac{1}{1cdot12}+frac{1}{1cdot13}
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giải phương trình
a,(\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{9\cdot10}\))(x-1)+\(\frac{1}{10}x\)=\(x-\frac{9}{10}\)
b,\(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+\frac{40}{39}\)
c,(x-10)+(x-19)+(x-18)+...+100+101=101
d,(\(\frac{1}{1\cdot51}+\frac{1}{2\cdot52}+\frac{1}{3\cdot53}+...+\frac{1}{10\cdot60}\))x=\(\frac{1}{1\cdot11}+\frac{1}{1\cdot12}+\frac{1}{1\cdot13}\)
