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\(\frac{437^2-363^2}{537^2-463^2}\)
\(=\frac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)
\(=\frac{74.800}{74.1000}\)
\(=\frac{80}{1000}=\frac{2}{25}\)
\(\frac{437^2-363^2}{537^2-463^2}\)
\(=\frac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)( Áp dụng hằng đẳng thức \(A^2-B^2=\left(A-B\right)\left(A+B\right)\))
\(=\frac{74\cdot800}{74\cdot1000}\)
\(=\frac{4}{5}\)
\(a.\)
\(A=5\dfrac{4}{23}.27\dfrac{3}{47}+5\dfrac{4}{23}.\left(-4\dfrac{3}{47}\right)\)
\(A=5\dfrac{4}{23}\left(27\dfrac{3}{47}-4\dfrac{3}{47}\right)\)
\(A=5\dfrac{4}{23}\left(27-4\right)\)
\(A=5\dfrac{4}{23}.23\)
\(A=119\)
\(b.\)
\(B=2^3+3.1-2^{-2}.4+\left(-2^2:\dfrac{1}{2}\right).8\)
\(B=2^3+3-\dfrac{1}{4}.4+\left(-8\right).8\)
\(B=2^3+3-1-64\)
\(B=-54\)
Ta có: \(A=\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\)
\(A>\dfrac{1}{100.101}+\dfrac{1}{101.102}+\dfrac{1}{102.103}+\dfrac{1}{103.104}+\dfrac{1}{104.105}\)\(A>\dfrac{1}{100}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{102}+\dfrac{1}{102}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{104}+\dfrac{1}{104}-\dfrac{1}{105}\)\(A>\dfrac{1}{100}-\dfrac{1}{105}\)
\(A>\dfrac{1}{2100}\)
Mà \(B=\dfrac{1}{2^2.3.5^2.7}\)=\(\dfrac{1}{2100}\)
=> \(A>B\)
Vậy \(A>B\)
\(a,A=\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2009}+\dfrac{1}{2008}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ A=1+\dfrac{1}{2010}=\dfrac{2011}{2010}\)
\(b,B=\left(-124\right)\left(63-37\right)+\dfrac{17}{66}\left(-66\right)=-124\cdot26+17=-3224+17=-3207\)
=>x(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9)+99/x=-3/7
=>8/9x+99/x=-3/7
\(\Leftrightarrow\dfrac{8x}{9}+\dfrac{99}{x}=\dfrac{-3}{7}\)
\(\Leftrightarrow\dfrac{8x^2+99\cdot9}{9x}=\dfrac{-3}{7}\)
\(\Leftrightarrow-56x^2-6237=27x\)
hay \(x\in\varnothing\)
\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
hay \(\dfrac{a}{b}\in\left\{\dfrac{\sqrt{7}}{3};-\dfrac{\sqrt{7}}{3}\right\}\)
\(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
\(\Leftrightarrow4.\left(3a^2-b^2\right)=3\left(a^2+b^2\right)\)
\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow12a^2-3a^2=3b^2+4b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
\(\text{hoặc }\dfrac{a}{b}=\pm\dfrac{\sqrt{7}}{3}\)
a. \(-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=\dfrac{-180}{7}+\dfrac{-30}{7}+\left(-105\right).\dfrac{1}{105}\)
\(=\dfrac{-180}{7}+\dfrac{-30}{7}+\left(-1\right)\)
\(=-31\)
b. \(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^6.3^6.2^9}=\dfrac{2^{15}.3^6.3^2}{2^{15}.3^6}=3^2=9\)
a: \(\dfrac{63^2-47^2}{215^2-105^2}=\dfrac{\left(63-47\right)\cdot\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)
\(=\dfrac{16\cdot110}{110\cdot320}=\dfrac{16}{320}=\dfrac{1}{20}\)
b: \(\dfrac{437^2-363^2}{537^2-463^2}=\dfrac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)
\(=\dfrac{74\cdot800}{74\cdot1000}=\dfrac{800}{1000}=\dfrac{4}{5}\)