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\(\frac{11}{15}+\frac{23}{16}-\left(5+\frac{27}{16}-\frac{19}{15}\right)+\left(\frac{-1}{2}\right)^2\)
\(=\frac{11}{15}+\frac{23}{16}-5-\frac{27}{16}+\frac{19}{15}+\frac{1}{4}\)
\(=\left(\frac{11}{15}+\frac{19}{15}\right)+\left(\frac{23}{16}-\frac{27}{16}\right)-5+\frac{1}{4}\)
\(=\frac{30}{15}-\frac{4}{16}-5+\frac{1}{4}\)
\(=2-\frac{1}{4}-5+\frac{1}{4}\)
\(=-3\)
học tốt ngôlãmtân
\(\frac{11}{15}+\frac{23}{16}-\left(5+\frac{27}{16}-\frac{19}{15}\right)+\left(-\frac{1}{2}\right)^2\) = \(\frac{11}{15}+\frac{23}{16}-5-\frac{27}{16}+\frac{19}{15}+\frac{1}{4}\)
\(=2-\frac{1}{4}-5+\frac{1}{4}=2-5=-3\)
Kb với mình nha!
\(5\frac{5}{27}+\frac{7}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)
\(=\left(5\frac{5}{27}-\frac{5}{27}\right)+\left(\frac{7}{23}+\frac{16}{23}\right)+0,5\)
\(=5+\frac{23}{23}+0,5=5,5+1=6,5\)
i) \(\frac{25^9}{5^{16}}-5^3:5\)
\(=\frac{\left(5^2\right)^9}{5^{16}}-5^2\)
\(=\frac{5^{18}}{5^{16}}-25\)
\(=5^2-25\)
\(=25-25\)
\(=0.\)
k) \(\frac{5}{7}-\left|\frac{2}{-7}\right|\)
\(=\frac{5}{7}-\left|\frac{-2}{7}\right|\)
\(=\frac{5}{7}-\frac{2}{7}\)
\(=\frac{3}{7}.\)
l) \(\frac{3^6.3^4}{9^3}\)
\(=\frac{3^{6+4}}{\left(3^2\right)^3}\)
\(=\frac{3^{10}}{3^6}\)
\(=3^4.\)
\(=81.\)
Chúc bạn học tốt!
1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!
