{"common":{"save":0,"post_id":"7814","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"7626","post_id":"7814","mon_id":"1159278","chapter_id":"1159327","question":"<p>R&uacute;t g\u1ecdn A =&nbsp;<span class=\"math-tex\">$\\sqrt[3]{x^3-12x^2+48x-64}-x$<\/span>.<\/p>","options":["<strong>A.<\/strong> 4","<strong>B.<\/strong> &ndash;4","<strong>C.<\/strong> 2","<strong>D.<\/strong> &ndash;2"],"correct":"2","level":"3","hint":"","answer":"<p>Ch\u1ecdn&nbsp;<span style=\"color:#16a085;\"><strong>B.<\/strong>&nbsp;&ndash;4.<\/span><\/p><p>A =&nbsp;<span class=\"math-tex\">$\\sqrt[3]{x^3-12x^2+48x-64}-x$<\/span><\/p><p>=&nbsp;<span class=\"math-tex\">$\\sqrt[3]{(x-4)^3}-x$<\/span><\/p><p>=&nbsp;<span class=\"math-tex\">$x-4-x$<\/span><\/p><p>=&nbsp;&ndash;4.<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"127","test":"0","date":"2024-10-12 00:51:14","option_type":"txt","len":1},{"id":"7627","post_id":"7814","mon_id":"1159278","chapter_id":"1159327","question":"<p>T&iacute;nh A =&nbsp;<span class=\"math-tex\">$\\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}$<\/span>.<\/p>","options":["<strong>A.<\/strong> 2","<strong>B.<\/strong> 1","<strong>C.<\/strong> &ndash;1","<strong>D.<\/strong> &ndash;2"],"correct":"2","level":"3","hint":"<p>&Aacute;p d\u1ee5ng&nbsp;<span class=\"math-tex\">$(A+B)^3=A^3+3A^2B+3AB^2+B^3=A^3+B^3+3AB(A+B)$<\/span>.<\/p>","answer":"<p>Ch\u1ecdn&nbsp;<span style=\"color:#16a085;\"><strong>B.<\/strong> 1.<\/span><\/p><p>A&sup3; =&nbsp;<span class=\"math-tex\">$\\bigg(\\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}\\bigg)^3$<\/span><\/p><p>=&nbsp;<span class=\"math-tex\">$2+\\sqrt{5}+2-\\sqrt{5}+3\\sqrt[3]{(2+\\sqrt{5})(2-\\sqrt{5})}.\\left(\\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}\\right)$<\/span><\/p><p>=&nbsp;<span class=\"math-tex\">$4-3A$<\/span>.<\/p><p>Khi \u0111&oacute;&nbsp;<span class=\"math-tex\">$A^3+3A-4=0$<\/span><\/p><p><span class=\"math-tex\">$(A-1)(A^2+A+4)=0$<\/span><\/p><p>Hay A = 1 (v&igrave;&nbsp;<span class=\"math-tex\">$A^2+A+4=\\bigg(A+\\dfrac{1}{2}\\bigg)^2+\\dfrac{15}{4}>0;\\forall A\\in R$<\/span>).<\/p><p>V\u1eady A = 1.<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"127","test":"0","date":"2024-10-12 00:57:21","option_type":"txt","len":1},{"id":"7628","post_id":"7814","mon_id":"1159278","chapter_id":"1159327","question":"<p>T&iacute;nh B =&nbsp;<span class=\"math-tex\">$\\sqrt[3]{3+\\sqrt{9+\\dfrac{125}{27}}}-\\sqrt[3]{-3+\\sqrt{9+\\dfrac{125}{27}}}$<\/span>.<\/p>","options":["<strong>A.<\/strong> 1","<strong>B.<\/strong> 0","<strong>C.<\/strong> 2","<strong>D.<\/strong> &ndash;1"],"correct":"1","level":"3","hint":"","answer":"<p>Ch\u1ecdn&nbsp;<span style=\"color:#16a085;\"><strong>A.<\/strong>&nbsp;1.<\/span><\/p><p>\u0110\u1eb7t&nbsp;<span class=\"math-tex\">$a=\\sqrt[3]{3+\\sqrt{9+\\dfrac{125}{27}}};b=\\sqrt[3]{-3+\\sqrt{9+\\dfrac{125}{27}}}$<\/span>&nbsp;. Khi \u0111&oacute; B =&nbsp;<span class=\"math-tex\">$a-b$<\/span>.<\/p><p><span class=\"math-tex\">$a^3-b^3=$<\/span><span class=\"math-tex\">$\\left(\\sqrt[3]{3+\\sqrt{9+\\dfrac{125}{27}}}\\right)^3-\\left(\\sqrt[3]{-3+\\sqrt{9+\\dfrac{125}{27}}}\\right)^3$<\/span><\/p><p>=&nbsp;<span class=\"math-tex\">$3+\\sqrt{9+\\dfrac{125}{27}}-\\bigg(-3+\\sqrt{9+\\dfrac{125}{27}}\\bigg)$<\/span>&nbsp;= 6.<\/p><p><span class=\"math-tex\">$ab$<\/span>&nbsp;=&nbsp;<span class=\"math-tex\">$\\left(\\sqrt[3]{3+\\sqrt{9+\\dfrac{125}{27}}}\\right)\\left(\\sqrt[3]{-3+\\sqrt{9+\\dfrac{125}{27}}}\\right)$<\/span><\/p><p>=&nbsp;<span class=\"math-tex\">$\\sqrt[3]{\\bigg(3+\\sqrt{9+\\dfrac{125}{27}}\\bigg)\\bigg(-3+\\sqrt{9+\\dfrac{125}{27}}\\bigg)}$<\/span><\/p><p>=&nbsp;<span class=\"math-tex\">$\\sqrt[3]{\\bigg(\\sqrt{9+\\dfrac{125}{27}}\\bigg)^2-3^2}$<\/span><\/p><p>=&nbsp;<span class=\"math-tex\">$\\sqrt[3]{9+\\dfrac{125}{27}-9}$<\/span><\/p><p>=&nbsp;<span class=\"math-tex\">$\\dfrac{5}{3}$<\/span>.<\/p><p>V\u1eady&nbsp;<span class=\"math-tex\">$B^3=a^3-b^3-3ab(a-b)$<\/span><\/p><p><span class=\"math-tex\">$B^3=6-5B$<\/span><\/p><p><span class=\"math-tex\">$B^3+5B-6=0$<\/span><\/p><p><span class=\"math-tex\">$(B-1)(B^2+B+6)=0$<\/span><\/p><p><span class=\"math-tex\">$B=1$<\/span>&nbsp;(v&igrave;&nbsp;<span class=\"math-tex\">$B^2+B+6=\\bigg(B+\\dfrac{1}{2}\\bigg)^2+\\dfrac{23}{4}>0;\\forall B\\in R$<\/span>).<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"127","test":"0","date":"2024-10-12 01:08:45","option_type":"txt","len":1}]}