Tính A= \(\frac{\left(2000x4+1995+2001x1995\right)}{\left(1995x495+1995x5+1995x3\right)}\)
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\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)
\(\Leftrightarrow a^2b+a^2c+b^2a+b^2c+abc+abc+bc^2+ac^2=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\Leftrightarrow...\)
\(P=0\)
\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
\(=\frac{5.18-10.27+15.36}{5.2.18.2-10.2.27.2+15.2.36.2}\)
\(=\frac{5.18-10.27+15.36}{5.8.2.2-10.27.2.2+15.36.2.2}\)
\(=\frac{1}{2.2-2.2+2.2}\)
\(=\frac{1}{2.2}=\frac{1}{4}\)
Trả lời
120-(-0,5).(-40).(-5).(-0,2).20.0,25/5+10+10+1995
=120-[(-0,5).(0,2)].[(-40).0,25].[20.(-5)]/2020
=120-0,1.(-10).-100/2020
=120-101/2020
=120-101/2020
=19/2020
\(=\frac{-120+\frac{1}{2}.\left(-40\right).\left(-5\right).\frac{-1}{5}.20.\frac{1}{4}}{5+20.1+1995}\)
\(=\frac{-120+1.\left(-1\right).-5.1.5}{5+1995}\)
\(=\frac{120.-1.1.-5.1.5}{2000}\)
\(=\frac{-120.1\left(-5+5\right)}{2000}\)
\(=0\)
(\(\frac{121}{122}\). \(\frac{123}{125}\)) .(\(\frac{1995}{1996}\).\(\frac{17}{16}\)- \(\frac{21}{25}\):\(\frac{16}{17}\)).(\(\frac{42}{30}\).\(\frac{75}{23}\)-\(\frac{19}{23}\).\(\frac{210}{38}\))
= (121/122 . 123/125) . ( 1995/1996 . 17/16 - 21/25 : 16/ 17) . 0
= 0
ai vay