{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["20"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"R\u00fat g\u1ecdn. <br\/> <br\/> $A = \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ \\dfrac{19}{1} + \\dfrac{18}{2} + \\dfrac{17}{3} + ... + \\dfrac{2}{18} + \\dfrac{1}{19} } = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Vi\u1ebft t\u1eed s\u1ed1 c\u1ee7a c\u00e1c ph\u00e2n s\u1ed1 $\\dfrac{19}{1}; \\dfrac{18}{2}; .... ; \\dfrac{2}{18}; \\dfrac{1}{19}$ th\u00e0nh hi\u1ec7u c\u1ee7a 20 v\u00e0 m\u1ed9t s\u1ed1 (V\u00ed d\u1ee5 $\\dfrac{19}{1} = \\dfrac{20 - 1}{1}$) <br\/> <b> B\u01b0\u1edbc 2: <\/b> Vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 v\u1eeba vi\u1ebft \u1edf b\u01b0\u1edbc 1 th\u00e0nh hi\u1ec7u 2 ph\u00e2n s\u1ed1 (V\u00ed d\u1ee5: $\\dfrac{20 - 1}{1} = \\dfrac{20}{1} - 1$) <br\/> <b> B\u01b0\u1edbc 3: <\/b> \u0110\u1eb7t nh\u00e2n t\u1eed chung ra ngo\u00e0i v\u00e0 r\u00fat g\u1ecdn <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align*} A &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ \\dfrac{19}{1} + \\dfrac{18}{2} + \\dfrac{17}{3} + ... + \\dfrac{2}{18} + \\dfrac{1}{19} } \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ \\dfrac{20 - 1}{1} + \\dfrac{20 - 2}{2} + \\dfrac{20 - 3}{3} + ... + \\dfrac{20 - 18}{18} + \\dfrac{20 - 19}{19} } \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ \\dfrac{20}{1} - 1 + \\dfrac{20}{2} - 1 + \\dfrac{20}{3} - 1 + ... + \\dfrac{20}{18} - 1 + \\dfrac{20}{19} - 1} \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ \\left( \\dfrac{20}{1} + \\dfrac{20}{2} + \\dfrac{20}{3} + ... + \\dfrac{20}{18} + \\dfrac{20}{19} \\right) - \\left( \\underbrace{1 + 1 + ... + 1}_{19 \\hspace{0.3cm} \\text{s\u1ed1 1}} \\right) } \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ 20 + 20 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{18} + \\dfrac{1}{19} \\right) - 19 } \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ 20 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{18} + \\dfrac{1}{19} \\right) + 20 . \\dfrac{1}{20} } \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ 20 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{18} + \\dfrac{1}{19} + \\dfrac{1}{20} \\right)} \\\\ &= \\dfrac{1}{20} \\end{align*}$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 1; 20 <\/span> <br\/><span class='basic_green'> <i> Nh\u1eadn x\u00e9t: V\u1edbi nh\u1eefng b\u00e0i r\u00fat g\u1ecdn nh\u01b0 tr\u00ean ta ch\u1ec9 c\u1ea7n x\u00e9t m\u1eabu s\u1ed1 <br\/> $\\dfrac{n + m - 1}{1} + \\dfrac{n + m - 2}{2} + \\dfrac{n + m - 3}{3} + ... + \\dfrac{2}{n + m - 2} + \\dfrac{1}{n + m - 1} \\\\ = \\dfrac{n + m}{1} - 1 + \\dfrac{n + m}{2} - 1 + \\dfrac{n + m}{3} - 1 + ... + \\dfrac{n + m}{n + m - 2} - 1 + \\dfrac{n + m }{n + m - 1} - 1 \\\\ = (n + m) . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{n + m - 1} \\right) + 1 \\\\ = (m + n) . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{n + m - 1} + \\dfrac{1}{n + m} \\right)$ <br\/> Khi \u0111\u00f3 ta r\u00fat g\u1ecdn $ \\dfrac{1}{n} + \\dfrac{1}{n + 1} + ... + \\dfrac{1}{n + m - 1} + \\dfrac{1}{n + m}$ \u1edf tr\u00ean t\u1eed v\u00e0 d\u01b0\u1edbi m\u1eabu, b\u00e0i to\u00e1n tr\u1edf n\u00ean d\u1ec5 d\u00e0ng<\/i> <\/span> "}]}],"id_ques":1691},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["100"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"R\u00fat g\u1ecdn. <br\/> <br\/> $P = \\dfrac{ \\dfrac{99}{1} + \\dfrac{98}{2} + \\dfrac{97}{3} + ... + \\dfrac{2}{98} + \\dfrac{1}{99} }{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ ","hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Vi\u1ebft t\u1eed s\u1ed1 c\u1ee7a c\u00e1c ph\u00e2n s\u1ed1 $\\dfrac{99}{1}; \\dfrac{99}{2}; .... ; \\dfrac{2}{98}; \\dfrac{2}{99}$ th\u00e0nh hi\u1ec7u c\u1ee7a 100 v\u00e0 m\u1ed9t s\u1ed1 (V\u00ed d\u1ee5 $\\dfrac{99}{1} = \\dfrac{100 - 1}{1}$) <br\/> <b> B\u01b0\u1edbc 2: <\/b> Vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 v\u1eeba vi\u1ebft \u1edf b\u01b0\u1edbc 1 th\u00e0nh hi\u1ec7u 2 ph\u00e2n s\u1ed1 (V\u00ed d\u1ee5: $\\dfrac{100 - 1}{1} = \\dfrac{100}{1} - 1$) <br\/> <b> B\u01b0\u1edbc 3: <\/b> \u0110\u1eb7t nh\u00e2n t\u1eed chung ra ngo\u00e0i v\u00e0 r\u00fat g\u1ecdn <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align*} P &= \\dfrac{ \\dfrac{99}{1} + \\dfrac{98}{2} + \\dfrac{97}{3} + ... + \\dfrac{2}{98} + \\dfrac{1}{99} }{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{ \\dfrac{100 - 1}{1} + \\dfrac{100 - 2}{2} + \\dfrac{100 - 3}{3} + ... + \\dfrac{100 - 98}{98} + \\dfrac{100 - 99}{99} }{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{ \\dfrac{100}{1} - 1 + \\dfrac{100}{2} - 1 + \\dfrac{100}{3} - 1 + ... + \\dfrac{100}{98} - 1 + \\dfrac{100}{99} - 1}{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{ \\left( \\dfrac{100}{1} + \\dfrac{100}{2} + \\dfrac{100}{3} + ... + \\dfrac{100}{98} + \\dfrac{100}{99} \\right) - \\left( \\underbrace{1 + 1 + ... + 1}_{99 \\hspace{0.3cm} \\text{s\u1ed1 1}} \\right)}{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{100 + 100 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{98} + \\dfrac{1}{99} \\right) - 99}{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{100 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{98} + \\dfrac{1}{99} \\right) + 100 . \\dfrac{1}{100}}{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{ 100 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{98} + \\dfrac{1}{99} + \\dfrac{1}{100} \\right)}{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= 100 \\end{align*}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 100 <\/span> "}]}],"id_ques":1692},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["81"],["10"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"T\u00ednh nhanh. <br\/> <br\/> $A = \\dfrac{1}{2} + \\dfrac{5}{6} + \\dfrac{11}{12} + \\dfrac{19}{20} + \\dfrac{29}{30}+ \\dfrac{41}{42} + \\dfrac{55}{56} + \\dfrac{71}{72} + \\dfrac{89}{90} = \\dfrac{ \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","hint":"Vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean th\u00e0nh hi\u1ec7u c\u1ee7a 1 v\u00e0 ph\u1ea7n b\u00f9 c\u1ee7a n\u00f3","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean th\u00e0nh hi\u1ec7u c\u1ee7a 1 v\u00e0 ph\u1ea7n b\u00f9 c\u1ee7a n\u00f3 d\u1ea1ng $\\dfrac{n - 1}{n} = 1 - \\dfrac{1}{n}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Vi\u1ebft m\u1eabu s\u1ed1 c\u1ee7a m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean th\u00e0nh t\u00edch c\u1ee7a hai s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp <br\/> <b> B\u01b0\u1edbc 3: <\/b> S\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c $\\dfrac{1}{n . (n + 1)} = \\dfrac{1}{n} - \\dfrac{1}{n + 1}$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u00ednh v\u00e0 r\u00fat g\u1ecdn <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align*} A &= \\dfrac{1}{2} + \\dfrac{5}{6} + \\dfrac{11}{12} + \\dfrac{19}{20} + \\dfrac{29}{30} + \\dfrac{41}{42} + \\dfrac{55}{56} + \\dfrac{71}{72} + \\dfrac{89}{90} \\\\ &= \\left( 1 - \\dfrac{1}{2} \\right) + \\left( 1 - \\dfrac{1}{6} \\right) + \\left( 1 - \\dfrac{1}{12} \\right) + \\left( 1 - \\dfrac{1}{20} \\right) + \\left( 1 - \\dfrac{1}{30} \\right) + \\left( 1 - \\dfrac{1}{42} \\right) \\\\ & + \\left( 1 - \\dfrac{1}{56} \\right) + \\left( 1 - \\dfrac{1}{72} \\right) + \\left( 1 - \\dfrac{1}{90} \\right) \\\\ &= 9 - \\left( \\dfrac{1}{2} + \\dfrac{1}{6} + \\dfrac{1}{12} + \\dfrac{1}{20} + \\dfrac{1}{30} + \\dfrac{1}{42} + \\dfrac{1}{56} + \\dfrac{1}{72} + \\dfrac{1}{90} \\right) \\\\ &= 9 - \\left( \\dfrac{1}{1 . 2} + \\dfrac{1}{2 . 3} + \\dfrac{1}{3 . 4} + \\dfrac{1}{4 . 5} + \\dfrac{1}{5 . 6} + \\dfrac{1}{6 . 7} + \\dfrac{1}{7 . 8} + \\dfrac{1}{8 . 9} + \\dfrac{1}{9 . 10} \\right) \\\\ &= 9 - \\left( 1 - \\dfrac{1}{2} + \\dfrac{1}{2} - \\dfrac{1}{3} + ... + \\dfrac{1}{9} - \\dfrac{1}{10} \\right) \\\\ &= 9 - \\left( 1 - \\dfrac{1}{10} \\right) \\\\ &= 9 - \\dfrac{9}{10} \\\\ &= \\dfrac{81}{10} \\end{align*}$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 81; 10 <\/span> <br\/> <span class='basic_left'><span class='basic_green'> <i> Nh\u1eadn x\u00e9t: V\u1edbi t\u1ed5ng g\u1ed3m nhi\u1ec1u ph\u00e2n s\u1ed1 d\u1ea1ng $\\dfrac{n - 1}{n}$ ta th\u01b0\u1eddng vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 \u0111\u00f3 th\u00e0nh hi\u1ec7u c\u1ee7a 1 v\u00e0 ph\u1ea7n b\u00f9 c\u1ee7a n\u00f3 $\\dfrac{n - 1}{n} = 1 - \\dfrac{1}{n}$ <br\/> Kh\u00e9o l\u00e9o vi\u1ebft n th\u00e0nh t\u00edch c\u1ee7a hai s\u1ed1 t\u1ef1 nhi\u00ean, bi\u1ebfn \u0111\u1ed1i l\u00e0m xu\u1ea5t hi\u1ec7n c\u00e1c ph\u00e2n s\u1ed1 d\u1ea1ng $\\dfrac{1}{a} - \\dfrac{1}{b} + \\dfrac{1}{b} - \\dfrac{1}{c}$ <br\/> Tri\u1ec7t ti\u00eau c\u00e1c ph\u00e2n s\u1ed1 \u0111\u1ed1i, t\u1eeb \u0111\u00f3 ta d\u1ec5 d\u00e0ng t\u00ednh \u0111\u01b0\u1ee3c gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c <\/span> "}]}],"id_ques":1693},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["39"],["200"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"T\u00ednh nhanh. <br\/> <br\/> $M = \\dfrac{3}{5 . 8} + \\dfrac{11}{8 . 19} + \\dfrac{12}{19 . 31} + \\dfrac{70}{31 . 101} + \\dfrac{99}{101 . 200} = \\dfrac{ \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","hint":"Vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean th\u00e0nh hi\u1ec7u c\u1ee7a hai ph\u00e2n s\u1ed1 d\u1ea1ng $\\dfrac{1}{n} - \\dfrac{1}{n + a}$ ","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Nh\u1eadn th\u1ea5y m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean \u0111\u1ec1u c\u00f3 d\u1ea1ng $\\dfrac{a}{n . (n + a)}$ do \u0111\u00f3: <br\/> <b> B\u01b0\u1edbc 1: <\/b> Vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean th\u00e0nh hi\u1ec7u c\u1ee7a hai ph\u00e2n s\u1ed1 d\u1ea1ng $\\dfrac{1}{n} - \\dfrac{1}{n + a}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Tri\u1ec7t ti\u00eau c\u00e1c ph\u00e2n s\u1ed1 \u0111\u1ed1i v\u00e0 t\u00ednh <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align*} M &= \\dfrac{3}{5 . 8} + \\dfrac{11}{8 . 19} + \\dfrac{12}{19 . 31} + \\dfrac{70}{31 . 101} + \\dfrac{99}{101 . 200} \\\\ &= \\dfrac{8 - 5}{5 . 8} + \\dfrac{19 - 8}{8 . 19} + \\dfrac{31 - 19}{19 . 31} + \\dfrac{101 - 31}{31 . 101} + \\dfrac{200 - 101}{101 . 200} \\\\ &= \\dfrac{1}{5} - \\dfrac{1}{8} + \\dfrac{1}{8} - \\dfrac{1}{19} + \\dfrac{1}{19} - \\dfrac{1}{31} + \\dfrac{1}{31} - \\dfrac{1}{101} + \\dfrac{1}{101} - \\dfrac{1}{200} \\\\ &= \\dfrac{1}{5} - \\dfrac{1}{200} \\\\ &= \\dfrac{39}{200} \\end{align*}$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 39; 200 <\/span> "}]}],"id_ques":1694},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"T\u00ecm x bi\u1ebft: <br\/><br\/> $2 . \\left( \\dfrac{1}{3}x - \\dfrac{1}{2} \\right) - \\dfrac{2}{3} . (2x - 3) = \\dfrac{-7}{6}$ ","select":["$A. x = \\dfrac{13}{4} $ ","$B. x = \\dfrac{-13}{4}$","$C. x = \\dfrac{1}{4} $","$D. x = \\dfrac{-1}{4} $"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Nh\u00e2n ph\u00e1 ngo\u1eb7c <br\/> <b> B\u01b0\u1edbc 2: <\/b> Chuy\u1ec3n c\u00e1c s\u1ed1 h\u1ea1ng kh\u00f4ng ch\u1ee9a x sang v\u1ebf ph\u1ea3i <br\/> <b> B\u01b0\u1edbc 3: <\/b> Bi\u1ebfn \u0111\u1ed5i v\u00e0 t\u00ecm x <\/span> <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> $ \\begin{align*} 2 . \\left( \\dfrac{1}{3}x - \\dfrac{1}{2} \\right) - \\dfrac{2}{3} . (2x - 3) = \\dfrac{-7}{6} \\\\ \\dfrac{2}{3}x - 1 - \\dfrac{4}{3}x + 2 &= \\dfrac{-7}{6} \\\\ \\dfrac{2}{3}x - \\dfrac{4}{3}x &= \\dfrac{-7}{6} + 1 - 2 \\\\ \\dfrac{-2}{3}x &= \\dfrac{-7}{6} - 1 \\\\ \\dfrac{-2}{3}x &= \\dfrac{-13}{6} \\\\ x &= \\dfrac{-13}{6} : \\dfrac{-2}{3} \\\\ x &= \\dfrac{13}{4} \\end{align*}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: $A. x = \\dfrac{13}{4}$ <\/span> ","column":2}]}],"id_ques":1695},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["25"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"N\u1ebfu l\u00e0m m\u1ed9t m\u00ecnh ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t c\u00f3 th\u1ec3 ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c trong 25 ng\u00e0y, ng\u01b0\u1eddi th\u1ee9 hai ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c trong 20 ng\u00e0y, ng\u01b0\u1eddi th\u1ee9 ba ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c trong 24 ng\u00e0y. C\u1ea3 ba ng\u01b0\u1eddi c\u00f9ng l\u00e0m trong hai ng\u00e0y sau \u0111\u00f3 ng\u01b0\u1eddi th\u1ee9 3 l\u00e0m ti\u1ebfp 6 ng\u00e0y, r\u1ed3i ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t tr\u1edf l\u1ea1i c\u00f9ng ng\u01b0\u1eddi th\u1ee9 t\u01b0, c\u1ea3 3 ng\u01b0\u1eddi c\u00f9ng l\u00e0m th\u00eam 4 ng\u00e0y n\u1eefa th\u00ec xong. H\u1ecfi n\u1ebfu ng\u01b0\u1eddi th\u1ee9 t\u01b0 l\u00e0m m\u1ed9t m\u00ecnh th\u00ec trong bao l\u00e2u s\u1ebd ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c \u0111\u00f3? <br\/> \u0110\u00e1p \u00e1n l\u00e0: _input_ ng\u00e0y ","hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> G\u1ecdi s\u1ed1 ng\u00e0y \u0111\u1ec3 ng\u01b0\u1eddi th\u1ee9 t\u01b0 ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c l\u00e0 x <br\/> T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng c\u00f4ng vi\u1ec7c m\u1ed7i ng\u01b0\u1eddi l\u00e0m \u0111\u01b0\u1ee3c trong 1 ng\u00e0y <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng c\u00f4ng vi\u1ec7c c\u1ea3 3 (th\u1ee9 nh\u1ea5t, th\u1ee9 hai, th\u1ee9 ba) l\u00e0m \u0111\u01b0\u1ee3c trong 2 ng\u00e0y (1) <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng c\u00f4ng vi\u1ec7c ng\u01b0\u1eddi th\u1ee9 3 l\u00e0m \u0111\u01b0\u1ee3c trong 6 ng\u00e0y (2) <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng c\u00f4ng vi\u1ec7c ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t, ng\u01b0\u1eddi th\u1ee9 ba v\u00e0 ng\u01b0\u1eddi th\u1ee9 t\u01b0 l\u00e0m trong 4 ng\u00e0y (3) <br\/> <b> B\u01b0\u1edbc 5: <\/b> V\u00ec kh\u1ed1i l\u01b0\u1ee3ng c\u00f4ng vi\u1ec7c l\u00e0 1 c\u00f4ng vi\u1ec7c n\u00ean ta c\u00f3 bi\u1ec3u th\u1ee9c <br\/> (1) + (2) + (3) = 1 t\u1eeb \u0111\u00f3 ta t\u00ecm \u0111\u01b0\u1ee3c x <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> G\u1ecdi s\u1ed1 ng\u00e0y \u0111\u1ec3 ng\u01b0\u1eddi th\u1ee9 t\u01b0 l\u00e0m 1 m\u00ecnh ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c \u0111\u00f3 l\u00e0 x (x > 0) <br\/> Kh\u1ed1i l\u01b0\u1ee3ng c\u00f4ng vi\u1ec7c ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t, ng\u01b0\u1eddi th\u1ee9 2, ng\u01b0\u1eddi th\u1ee9 ba, ng\u01b0\u1eddi th\u1ee9 t\u01b0 l\u00e0m \u0111\u01b0\u1ee3c trong 1 ng\u00e0y l\u1ea7n l\u01b0\u1ee3t l\u00e0: $\\dfrac{1}{25}; \\dfrac{1}{20}; \\dfrac{1}{24}; \\dfrac{1}{x}$ <br\/> Sau 2 ng\u00e0y ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t, ng\u01b0\u1eddi th\u1ee9 hai v\u00e0 ng\u01b0\u1eddi th\u1ee9 ba l\u00e0m \u0111\u01b0\u1ee3c l\u00e0: <br\/> $\\left( \\dfrac{1}{25} + \\dfrac{1}{20} + \\dfrac{1}{24} \\right) . 2 = \\dfrac{79}{300}$ (c\u00f4ng vi\u1ec7c) <br\/> Sau 6 ng\u00e0y ng\u01b0\u1eddi th\u1ee9 3 l\u00e0m \u0111\u01b0\u1ee3c l\u00e0: <br\/> $\\dfrac{1}{24} . 6 = \\dfrac{1}{4}$ (c\u00f4ng vi\u1ec7c) <br\/> Sau 4 ng\u00e0y ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t, ng\u01b0\u1eddi th\u1ee9 ba v\u00e0 ng\u01b0\u1eddi th\u1ee9 t\u01b0 l\u00e0m \u0111\u01b0\u1ee3c l\u00e0: <br\/> $\\left( \\dfrac{1}{25} + \\dfrac{1}{24} + \\dfrac{1}{x} \\right) . 4 = \\dfrac{49}{150} + \\dfrac{4}{x}$ (c\u00f4ng vi\u1ec7c) <br\/> V\u00ec c\u00f4ng vi\u1ec7c \u0111\u00e3 ho\u00e0n th\u00e0nh n\u00ean ta c\u00f3: <br\/> $ \\begin{align*} \\dfrac{79}{300} + \\dfrac{1}{4} + \\dfrac{49}{150} + \\dfrac{4}{x} &= 1 \\\\ \\dfrac{79}{300} + \\dfrac{75}{300} + \\dfrac{98}{300} + \\dfrac{4}{x} &= \\dfrac{300}{300} \\\\ \\dfrac{252}{300} + \\dfrac{4}{x} &= \\dfrac{300}{300} \\\\ \\dfrac{4}{x} &= \\dfrac{300 - 252}{300} \\\\ \\dfrac{4}{x} &= \\dfrac{4}{25} \\end{align*} $ <br\/> $\\Rightarrow$ x = 25 <br\/> <span class='basic_pink'> V\u1eady ng\u01b0\u1eddi th\u1ee9 t\u01b0 l\u00e0m 1 m\u00ecnh m\u1ea5t 25 ng\u00e0y th\u00ec ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c <\/span> "}]}],"id_ques":1696},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"],["84"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["frac"],"ques":"Cho hai ph\u00e2n s\u1ed1 $\\dfrac{5}{12}$ v\u00e0 $\\dfrac{15}{21}$. T\u00ecm ph\u00e2n s\u1ed1 l\u1edbn nh\u1ea5t sao cho khi chia m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean cho ph\u00e2n s\u1ed1 \u0111\u00f3 th\u00ec k\u1ebft qu\u1ea3 l\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean. <br\/> \u0110\u00e1p \u00e1n l\u00e0: <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> G\u1ecdi ph\u00e2n s\u1ed1 l\u1edbn nh\u1ea5t c\u1ea7n t\u00ecm l\u00e0 $\\dfrac{a}{b}$ <br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\dfrac{5}{12} : \\dfrac{a}{b}$, t\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 th\u01b0\u01a1ng tr\u00ean l\u00e0 s\u1ed1 nguy\u00ean (1) <br\/> (\u0110\u1ec3 m\u1ed9t ph\u00e2n s\u1ed1 l\u00e0 s\u1ed1 nguy\u00ean th\u00ec t\u1eed s\u1ed1 ph\u1ea3i chia h\u1ebft cho m\u1eabu s\u1ed1) <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\dfrac{15}{21} : \\dfrac{a}{b}$, t\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 th\u01b0\u01a1ng \u0111\u00f3 l\u00e0 s\u1ed1 nguy\u00ean (2) <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 $\\dfrac{a}{b} \\in \\mathbb{Z}$ l\u00e0 a = \u01afCLN; b = BCNN <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> G\u1ecdi ph\u00e2n s\u1ed1 l\u1edbn nh\u1ea5t c\u1ea7n t\u00ecm l\u00e0 $\\dfrac{a}{b}$ $(a, b \\in \\mathbb{N^*}; (a, b) = 1)$ <br\/> Ta c\u00f3: $\\dfrac{5}{12} : \\dfrac{a}{b} = \\dfrac{5b}{12a}$ <br\/> \u0110\u1ec3 $\\dfrac{5b}{12a} \\in \\mathbb{Z}$ th\u00ec 5b $ \\vdots $ 12a $\\Rightarrow$ 5 $\\vdots$ a v\u00e0 b $\\vdots$ 12 (1) <br\/> $\\dfrac{15}{21} : \\dfrac{a}{b} = \\dfrac{15b}{21a}$. \u0110\u1ec3 $\\dfrac{15b}{21a} \\in \\mathbb{Z}$ th\u00ec 15 $\\vdots$ a v\u00e0 b $\\vdots$ 21 (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ \u0110\u1ec3 $\\dfrac{a}{b}$ l\u1edbn nh\u1ea5t th\u00ec ph\u1ea3i c\u00f3: <br\/> a = \u01afCLN(5, 15) = 5 ; b = BCNN(12, 21) = 84 <br\/> $\\Rightarrow$ $\\dfrac{a}{b} = \\dfrac{5}{84}$ <br\/> Th\u1eed l\u1ea1i: $\\dfrac{5}{12} : \\dfrac{5}{84} = 7$; $\\dfrac{15}{21} : \\dfrac{5}{84} = 12$ (th\u1ecfa m\u00e3n)"}]}],"id_ques":1697},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-2"],["5"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["frac"],"ques":" T\u00ednh. <br\/><br\/> $\\dfrac{1}{1 - \\dfrac{1}{1 - \\dfrac{1}{2}}} + \\dfrac{ 1}{1 + \\dfrac{1}{1 + \\dfrac{1}{2}}}$ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"T\u00ednh ng\u01b0\u1ee3c t\u1eeb cu\u1ed1i l\u00ean","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh t\u1ea7ng d\u01b0\u1edbi c\u00f9ng c\u1ee7a m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh t\u1ea7ng th\u1ee9 2 c\u1ee7a m\u1ed7i ph\u00e2n s\u1ed1 \u0111\u00f3 <br\/> <b> B\u01b0\u1edbc 3: <\/b> Quy \u0111\u1ed3ng r\u1ed3i t\u00ednh t\u1ed5ng hai ph\u00e2n s\u1ed1 \u0111\u00f3 <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align*} & \\dfrac{1}{1 - \\dfrac{1}{1 - \\dfrac{1}{2}}} + \\dfrac{ 1}{1 + \\dfrac{1}{1 + \\dfrac{1}{2}}} = \\dfrac{1}{1 - \\dfrac{1}{ \\dfrac{2}{2} - \\dfrac{1}{2}}} + \\dfrac{1}{1 + \\dfrac{1}{\\dfrac{2}{2} + \\dfrac{1}{2}}} = \\dfrac{1}{1 - \\dfrac{1}{\\dfrac{1}{2}}} + \\dfrac{1}{1 + \\dfrac{1}{\\dfrac{3}{2}}} \\\\\\\\ &= \\dfrac{1}{1 - 2} + \\dfrac{1}{1 + \\dfrac{2}{3}} = -1 + \\dfrac{1}{\\dfrac{3}{3} + \\dfrac{2}{3}} = -1 + \\dfrac{1}{\\dfrac{5}{3}} = -1 + \\dfrac{3}{5} = \\dfrac{-5}{5} + \\dfrac{3}{5} = \\dfrac{-2}{5} \\end{align*}$<br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: Ta c\u00f3 $\\dfrac{1}{\\dfrac{a}{b}} = \\dfrac{b}{a}$ <br\/> V\u1edbi nh\u1eefng ph\u00e2n s\u1ed1 c\u00f3 nhi\u1ec1u t\u1ea7ng nh\u01b0 tr\u00ean \u0111\u1ec3 r\u00fat g\u1ecdn ta t\u00ednh l\u1ea7n l\u01b0\u1ee3t t\u1eeb d\u01b0\u1edbi cu\u1ed1i l\u00ean tr\u00ean <\/i> <\/span> "}]}],"id_ques":1698},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["8"],["13"]]],"list":[{"point":10,"width":50,"ques":" R\u00fat g\u1ecdn ph\u00e2n s\u1ed1. <br\/><br\/> $\\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{1}{2}}}}}$ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"T\u00ednh ng\u01b0\u1ee3c t\u1eeb cu\u1ed1i l\u00ean","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> L\u01b0u \u00fd: Ta c\u00f3 $\\dfrac{1}{\\dfrac{a}{b}} = \\dfrac{b}{a}$ <\/span> <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align*} & \\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{1}{2}}}}} = \\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{1}{\\dfrac{3}{2}}}}} = \\dfrac{1}{1 +\\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{2}{3}}}} \\\\\\\\ &= \\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{1}{ \\dfrac{5}{3}}}} = \\dfrac{1}{1 + \\dfrac{1}{1 + \\dfrac{3}{5}}} = \\dfrac{1}{1 + \\dfrac{1}{\\dfrac{8}{5}}} = \\dfrac{1}{1 + \\dfrac{5}{8}} = \\dfrac{1}{\\dfrac{13}{8}} = \\dfrac{8}{13} \\end{align*}$"}]}],"id_ques":1699},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Ba v\u00f2i n\u01b0\u1edbc I, II, III n\u1ebfu ch\u1ea3y m\u1ed9t m\u00ecnh v\u00e0o b\u1ec3 c\u1ea1n th\u00ec ch\u1ea3y \u0111\u1ea7y b\u1ec3 theo th\u1ee9 t\u1ef1 4 gi\u1edd, 6 gi\u1edd, 9 gi\u1edd. L\u00fac \u0111\u1ea7u ng\u01b0\u1eddi ta m\u1edf v\u00f2i I v\u00e0 v\u00f2i II trong 1 gi\u1edd 30 ph\u00fat, sau \u0111\u00f3 \u0111\u00f3ng v\u00f2i I r\u1ed3i m\u1edf ti\u1ebfp v\u00f2i III c\u00f9ng ch\u1ea3y v\u1edbi v\u00f2i II cho \u0111\u1ebfn khi \u0111\u1ea7y b\u1ec3. H\u1ecfi v\u00f2i III ch\u1ea3y trong bao l\u00e2u? <br\/> \u0110\u00e1p \u00e1n l\u00e0: ","select":["A. 1 gi\u1edd 15 ph\u00fat ","B. 1 gi\u1edd 20 ph\u00fat","C. 1 gi\u1edd 30 ph\u00fat ","D. 1 gi\u1edd 21 ph\u00fat "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> <b> B\u01b0\u1edbc 1: <\/b> \u0110\u1ed5i ra \u0111\u01a1n v\u1ecb ph\u00fat <br\/> T\u00ednh l\u01b0\u1ee3ng n\u01b0\u1edbc m\u1ed7i v\u00f2i ch\u1ea3y \u0111\u01b0\u1ee3c trong 1 ph\u00fat <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh l\u01b0\u1ee3ng n\u01b0\u1edbc v\u00f2i I v\u00e0 II ch\u1ea3y \u0111\u01b0\u1ee3c trong 1 gi\u1edd 30 ph\u00fat (1) <br\/> <b> B\u01b0\u1edbc 3: <\/b> N\u1ebfu g\u1ecdi th\u1eddi gian v\u00f2i III ch\u1ea3y l\u00e0 x ph\u00fat <br\/> T\u00ednh l\u01b0\u1ee3ng n\u01b0\u1edbc v\u00f2i II v\u00e0 III ch\u1ea3y trong th\u1eddi gian x ph\u00fat (2) <br\/> <b> B\u01b0\u1edbc 4: <\/b> V\u00ec c\u00f3 1 b\u1ec3 n\u01b0\u1edbc n\u00ean ta c\u00f3: (1) + (2) = 1 <br\/> Bi\u1ebfn \u0111\u1ed5i v\u00e0 t\u00ecm x d\u1ef1a v\u00e0o bi\u1ec3u th\u1ee9c tr\u00ean <\/span> <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <br\/> \u0110\u1ed1i: 9 gi\u1edd = 540 ph\u00fat; 6 gi\u1edd = 360 ph\u00fat; 4 gi\u1edd = 240 ph\u00fat; 1gi\u1edd 30 ph\u00fat = 90 ph\u00fat <br\/> L\u01b0\u1ee3ng n\u01b0\u1edbc v\u00f2i I ch\u1ea3y \u0111\u01b0\u1ee3c trong 1 ph\u00fat l\u00e0: <br\/> $1 : 240 = \\dfrac{1}{240}$ (b\u1ec3) <br\/> L\u01b0\u1ee3ng n\u01b0\u1edbc v\u00f2i II ch\u1ea3y \u0111\u01b0\u1ee3c trong 1 ph\u00fat l\u00e0: <br\/> $1 : 360 = \\dfrac{1}{360}$ (b\u1ec3) <br\/> L\u01b0\u1ee3ng n\u01b0\u1edbc v\u00f2i III ch\u1ea3y \u0111\u01b0\u1ee3c trong 1 ph\u00fat l\u00e0: <br\/> $1 : 540 = \\dfrac{1}{540}$ (b\u1ec3) <br\/> L\u01b0\u1ee3ng n\u01b0\u1edbc v\u00f2i I v\u00e0 v\u00f2i II ch\u1ea3y \u0111\u01b0\u1ee3c trong 1 gi\u1edd 30 ph\u00fat l\u00e0: <br\/> $\\left( \\dfrac{1}{240} + \\dfrac{1}{360} \\right) . 90 = \\dfrac{5}{8}$ (b\u1ec3) <br\/> G\u1ecdi th\u1eddi gian v\u00f2i III ch\u1ea3y \u0111\u1ebfn khi \u0111\u1ea7y b\u1ec3 l\u00e0 x (ph\u00fat) <br\/> L\u01b0\u1ee3ng n\u01b0\u1edbc v\u00f2i II v\u00e0 v\u00f2i III ch\u1ea3y trong x ph\u00fat l\u00e0: <br\/> $\\left( \\dfrac{1}{360} + \\dfrac{1}{540} \\right) . x = \\dfrac{x}{216}$ (b\u1ec3) <br\/> Theo \u0111\u1ec1 b\u00e0i ta c\u00f3: <br\/> $\\dfrac{5}{8} + \\dfrac{x}{216} = 1$ <br\/> $\\Rightarrow$ x = 81 <br\/> Th\u1eddi gian v\u00f2i III ch\u1ea3y l\u00e0 81 ph\u00fat = 1 gi\u1edd 21 ph\u00fat <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":2}]}],"id_ques":1700}],"lesson":{"save":0,"level":3}}