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63+3x62+33/13
=63+31x62+33/13
=62x31x(6+32)/13
=1620/13
Em nghĩ là thế !
= (2.3)^3+ 3. ( 2.3)^2 + 3^3 / -13
= 2^3 .3^3 + 3. 3^2 . 2^2 + 3^3 / -13
= 3^3. ( 2^3 + 2^2 + 1) /-13
= 27.13/-13
= -27
\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{6^2\left(6+3\right)+3^2\cdot3}{-13}=\frac{36\cdot9+9\cdot3}{-13}=\frac{39\cdot9}{-13}=-3\cdot9=-27\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy \(x=-2004\)
a, \(\dfrac{20^5.5^{10}}{100^5}=\dfrac{20^5.5^{10}}{\left(20.5\right)^5}=\dfrac{20^5.5^{10}}{20^5.5^5}=5^5\)
b,\(\dfrac{\left(0,9\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^6}=\dfrac{3^5}{\left(0,3\right)}\)
\(\frac{2^7.3^6}{6^5.8^2}\)= \(\frac{2^7.3^6}{3^5.2^5.2^6}\)=\(\frac{2^7.3^6}{3^5.2^{11}}\)=\(\frac{3}{2^4}\)=\(\frac{3}{16}\)
đó là ý kiến của mình
\(a.\frac{2^7\times9^3}{6^5\times8^2}\)
\(=\frac{3}{16}\)
\(b.\frac{6^3+3\times6^2+3^3}{-13}\)
\(-\frac{333}{13}\)
\(\frac{4^2\times4^3}{2^{10}}=\frac{\left(2^2\right)^2\times\left(2^2\right)^3}{2^{10}}=\frac{2^4\times2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\times3\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\right)^5\times3^5}{\left(0,2\right)^6}=\frac{3^5}{0,2}=\frac{243}{0,2}=243:\left(0,2\right)=243\times5=1215\)
\(\frac{2^7\times9^3}{6^5\times8^2}=\frac{2^7\times\left(3^2\right)^3}{\left(2\times3\right)^5\times\left(2^3\right)^2}=\frac{2^7\times3^6}{2^5\times3^5\times2^6}=\frac{3}{2^4}=\frac{3}{16}\)
\(\frac{6^3+3\times6^2+3^3}{-13}=\frac{\left(2\times3\right)^3+3\times6^2+3^3}{-13}=\frac{3\left(2^3\times3^2+6^2+3^2\right)}{-13}=\frac{3\times117}{-13}=\frac{351}{-13}=-27\)
\(\frac{\left(-0,25\right)^{-5}.9^4.\left(-2\right)^{-3}-2^{-2}.6^9}{2^9.3^6+6^6.40}\)
\(=\frac{\left(-4\right)^5.\left(3^2\right)^4.\left(-2\right)^{-3}-2^{-2}.\left(3.2\right)^9}{2^9.3^6+\left(2.3\right)^6.2^3.5}\)
\(=\frac{-\left(2^2\right)^5.3^8.\left(-2\right)^{-3}-2^{-2}.3^9.2^9}{2^9.3^6+2^6.3^6.2^3.5}\)
\(=\frac{-2^{10}3^8.\left(-2\right)^{-3}-2^7.3^9}{2^9.3^6+2^9.3^6.5}\)
\(=\frac{2^73^8.-2^7.3^9}{2^9.3^6+2^9.3^6.5}\)
\(=\frac{2^7.3^8.\left(1-3\right)}{2^9.3^6.\left(1+5\right)}\)
\(=\frac{3^2.\left(-2\right)}{2^2.6}\)
\(=\frac{-3}{4}\)
d)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(8+4+1\right)}{-13}=\frac{3^3\cdot13}{-13}=-27\)
giang làm a,b,c rồi nên làm d thôi
lười quá, hehe ^_^
a) \(\frac{4^2\cdot4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
b)\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\right)^5.3^5}{\left(0,2\right)^6}=\frac{3^5}{0,2}=\frac{243.5}{1}=1215\)
c)\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^5}{2^{11}.3^5}=\frac{1}{2^4}=\frac{1}{16}\)
