{"common":{"save":0,"post_id":"970","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"241","post_id":"970","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","options":{"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":" Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $ A = \\dfrac{(-5)^{3}.(-0,9)^{2}}{\\left(1\\dfrac{1}{2}\\right)^{4}.\\left(-3\\dfrac{1}{3}\\right)^{3}.(-1)^{7}} $ l\u00e0:","select":["A. $ \\dfrac{-27}{50} $ ","B. $ \\dfrac{27}{50} $ ","C. $ \\dfrac{-50}{27} $ ","D. $ \\dfrac{50}{27} $ "],"hint":" S\u1eed d\u1ee5ng quy t\u1eafc b\u1ecf d\u1ea5u \u00e2m c\u1ee7a c\u01a1 s\u1ed1: <br\/> V\u1edbi $a > 0$ ta c\u00f3: $\\quad (-a)^{2n} = a^{2n}; (-a)^{2n+1} = -a^{2n+1}$ ","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b>* \u0110\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a m\u1ed9t bi\u1ec3u th\u1ee9c c\u00f3 ch\u1ee9a l\u0169y th\u1eeba, ta l\u00e0m theo b\u1ed1n b\u01b0\u1edbc sau \u0111\u00e2y:<\/b> <br\/> <u> B\u01b0\u1edbc 1:<\/u> Kh\u1eed d\u1ea5u \u00e2m c\u1ee7a s\u1ed1 m\u0169 (n\u1ebfu c\u00f3). <br\/> <u> B\u01b0\u1edbc 2:<\/u> Kh\u1eed d\u1ea5u \u00e2m c\u1ee7a c\u01a1 s\u1ed1 (n\u1ebfu c\u00f3). <br\/> <u> B\u01b0\u1edbc 3:<\/u> Bi\u1ebfn \u0111\u1ed5i \u0111\u1ec3 \u0111\u01b0a c\u00e1c l\u0169y th\u1eeba v\u1ec1 c\u01a1 s\u1ed1 nguy\u00ean t\u1ed1. <br\/> <u> B\u01b0\u1edbc 4:<\/u> R\u00fat g\u1ecdn. <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <br\/> S\u1eed d\u1ee5ng quy t\u1eafc b\u1ecf d\u1ea5u \u00e2m c\u1ee7a c\u01a1 s\u1ed1: <br\/> V\u1edbi $a > 0$ ta c\u00f3: $\\quad (-a)^{2n} = a^{2n}; (-a)^{2n+1} = -a^{2n+1}$ <br\/> $A = \\dfrac{(-5)^{3}.(-0,9)^{2}}{\\left(1\\dfrac{1}{2}\\right)^{4}.\\left(-3\\dfrac{1}{3}\\right)^{3}.(-1)^{7}} $<br\/>$ = \\left(-5^{3}.\\left(\\dfrac{9}{10}\\right)^{2}\\right): \\left[\\left(\\dfrac{3}{2}\\right)^{4}.\\left(\\dfrac{10}{3}\\right)^{3}\\right] $<br\/>$ = -\\left(5^{3}.\\dfrac{9^{2}}{10^{2}}\\right):\\left(\\dfrac{3^{4}}{2^{4}}.\\dfrac{10^{3}}{3^{3}}\\right) = -\\left(5^{3}.\\dfrac{(3^{2})^{2}}{(2.5)^{2}}\\right):\\left(\\dfrac{3^{4}}{2^{4}}.\\dfrac{(2.5)^{3}}{3^{3}}\\right) $<br\/>$ = -\\left(\\dfrac{5^{3}.3^{4}}{2^{2}.5^{2}}\\right):\\left(\\dfrac{3.2^{3}.5^{3}}{2^{4}}\\right) = -\\left(\\dfrac{5.3^{4}}{2^{2}}\\right):\\left(\\dfrac{3.5^{3}}{2}\\right) $<br\/>$= -\\dfrac{5.3^{4}}{2^{2}}.\\dfrac{2}{3.5^{3}} = -\\dfrac{3^{3}}{2.5^{2}} = \\dfrac{-27}{50}$ <br\/> <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A. <\/span><\/span> <span class='basic_left'> <i> L\u01b0u \u00fd: <\/i> $\\bullet$ L\u0169y th\u1eeba b\u1eadc ch\u1eb5n c\u1ee7a m\u1ed9t s\u1ed1 \u00e2m l\u00e0 m\u1ed9t s\u1ed1 d\u01b0\u01a1ng. <br\/> $\\qquad \\quad \\bullet$ L\u0169y th\u1eeba b\u1eadc l\u1ebb c\u1ee7a m\u1ed9t s\u1ed1 \u00e2m l\u00e0 m\u1ed9t s\u1ed1 \u00e2m. <\/span>","column":2}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:22:44"},{"id":"242","post_id":"970","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","options":{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["f","f","f","t"]],"list":[{"point":10,"image":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":[" $(-5)^{2} . (-5)^{3} = (-5)^{6} $ "," $ (0,2)^{10} : (0,2)^{5} = (0,2)^{2} $ "," $ \\dfrac{(8)^{10}}{(4)^{8}} = \\left(\\dfrac{8}{4}\\right)^{10-8} = (2)^{2} $ ","$ (0,75)^{3} : 0,75 = (0,75)^{2}$"],"hint":"\u00c1p d\u1ee5ng c\u00e1c c\u00f4ng th\u1ee9c: <br\/> $ x^{m} . x^{n} = x^{m+n}; \\qquad x^{m} : x^{n} = x^{m-n}; \\quad \\left(\\dfrac{x}{y}\\right)^{n} = \\dfrac{x^{n}}{y^{n}} (y \\neq 0) $","explain":["Ta c\u00f3: $(-5)^{2} . (-5)^{3} = (-5)^{2+3} = (-5)^{5}. $ Suy ra kh\u1eb3ng \u0111\u1ecbnh kh\u00f4ng \u0111\u00fang","<br\/> Ta c\u00f3:$(0,2)^{10} : (0,2)^{5} = (0,2)^{10-5} = (0,2)^{5} . $ Suy ra kh\u1eb3ng \u0111\u1ecbnh kh\u00f4ng \u0111\u00fang ","<br\/>Ta c\u00f3: $\\dfrac{(8)^{10}}{(4)^{8}} = \\dfrac{(2^{3})^{10}}{(2^{2})^{8}} = \\dfrac{2^{30}}{2^{16}} = 2^{30-16} = 2^{14} .$ Suy ra kh\u1eb3ng \u0111\u1ecbnh kh\u00f4ng \u0111\u00fang","<br\/> Ta c\u00f3: $ (0,75)^{3} : 0,75 = (0,75)^{3-1} = (0,75)^{2}. $ Suy ra kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang"]}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:22:44"},{"id":"243","post_id":"970","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","options":{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" So s\u00e1nh gi\u00e1 tr\u1ecb c\u1ee7a $B = \\dfrac{1}{3^{1}} + \\dfrac{1}{3^{2}} + \\dfrac{1}{3^{3}} + ... + \\dfrac{1}{3^{99}} + \\dfrac{1}{3^{100}} $ v\u1edbi $\\dfrac{1}{2},$ ta \u0111\u01b0\u1ee3c:","select":["A. $ B > \\dfrac{1}{2} $ ","B. $ B = \\dfrac{1}{2} $ ","C.$ B < \\dfrac{1}{2} $ "],"hint":" T\u00ednh $ 2B = 3B - B ,$ r\u1ed3i so s\u00e1nh v\u1edbi $\\dfrac{1}{2}$ ","explain":"<span class='basic_left'> Ta c\u00f3 $ 3B = \\left(\\dfrac{1}{3^{1}} + \\dfrac{1}{3^{2}} + \\dfrac{1}{3^{3}} + ... + \\dfrac{1}{3^{99}} + \\dfrac{1}{3^{100}} \\right) \\\\ = 1 + \\dfrac{1}{3^{1}} + \\dfrac{1}{3^{2}} + \\dfrac{1}{3^{3}} + ... + \\dfrac{1}{3^{98}} + \\dfrac{1}{3^{99}}$ <br\/> $\\Rightarrow 3B-B = \\left ( 1 + \\dfrac{1}{3^{1}} + \\dfrac{1}{3^{2}} + \\dfrac{1}{3^{3}} + ... + \\dfrac{1}{3^{98}} + \\dfrac{1}{3^{99}}\\right) - \\left ( \\dfrac{1}{3^{1}} + \\dfrac{1}{3^{2}} + \\dfrac{1}{3^{3}} + ... + \\dfrac{1}{3^{99}} + \\dfrac{1}{3^{100}} \\right) $ <br\/> $ 2B = 1 + \\dfrac{1}{3^{1}} + \\dfrac{1}{3^{2}} + \\dfrac{1}{3^{3}} + ... + \\dfrac{1}{3^{98}} + \\dfrac{1}{3^{99}} - \\dfrac{1}{3^{1}} - \\dfrac{1}{3^{2}} - \\dfrac{1}{3^{3}} - ... - \\dfrac{1}{3^{99}} - \\dfrac{1}{3^{100}} $ <br\/> $ 2B = 1 - \\dfrac{1}{3^{100}} $ <br\/> Suy ra : $B = \\dfrac{1}{2} - \\dfrac{1}{2.3^{100}} < \\dfrac{1}{2} $ <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/span><\/span> ","column":3}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:22:44"}]}