{"segment":[{"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}]}],"id_ques":241},{"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"]}]}],"id_ques":242},{"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}]}],"id_ques":243},{"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 $A = \\dfrac{1}{2^{1}} + \\dfrac{1}{2^{2}} + \\dfrac{1}{2^{3}} + ... + \\dfrac{1}{2^{49}} + \\dfrac{1}{2^{50}} $ v\u1edbi $1,$ th\u00ec:","select":["A. $ A > 1 $ ","B. $ A = 1 $ ","C. $ A < 1 $ "],"hint":" T\u00ednh $ A = 2A - A , $ r\u1ed3i so s\u00e1nh","explain":" <span class='basic_left'> Ta c\u00f3 $2A = 2\\left(\\dfrac{1}{2^{1}} + \\dfrac{1}{2^{2}} + \\dfrac{1}{2^{3}} + ... + \\dfrac{1}{2^{49}} + \\dfrac{1}{2^{50}}\\right) \\\\ = \\left(1 + \\dfrac{1}{2^{1}} + \\dfrac{1}{2^{2}} + \\dfrac{1}{2^{3}} + ... + \\dfrac{1}{2^{49}}\\right) $ <br\/> $\\Rightarrow 2A -A= \\left(1 + \\dfrac{1}{2^{1}} + \\dfrac{1}{2^{2}} + \\dfrac{1}{2^{3}} + ... + \\dfrac{1}{2^{49}}\\right) - \\left(\\dfrac{1}{2^{1}} + \\dfrac{1}{2^{2}} + \\dfrac{1}{2^{3}} + ... + \\dfrac{1}{2^{49}} + \\dfrac{1}{2^{50}} \\right) $ <br\/> $ A = 1 + \\dfrac{1}{2^{1}} + \\dfrac{1}{2^{2}} + \\dfrac{1}{2^{3}} + ... + \\dfrac{1}{2^{49}} - \\dfrac{1}{2^{1}} - \\dfrac{1}{2^{2}} - \\dfrac{1}{2^{3}} - ... - \\dfrac{1}{2^{49}} -\\dfrac{1}{2^{50}} $ <br\/> $ A = 1 - \\dfrac{1}{2^{50}} <1 $ <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/span><\/span> ","column":3}]}],"id_ques":244},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c <br\/> $ K = \\left[\\dfrac{1}{100} - 1^{2}\\right] . \\left[\\dfrac{1}{100} - \\left(\\dfrac{1}{2}\\right)^{2}\\right]. \\left[\\dfrac{1}{100} - \\left(\\dfrac{1}{3}\\right)^{2}\\right]...\\left[\\dfrac{1}{100} - \\left(\\dfrac{1}{20}\\right)^{2}\\right] \\\\$ = _input_","hint":"T\u00ednh gi\u00e1 tr\u1ecb b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a ph\u00e2n s\u1ed1 r\u1ed3i quan s\u00e1t. ","explain":" Ta c\u00f3: $ K = \\left[\\dfrac{1}{100} - 1^{2}\\right] . \\left[\\dfrac{1}{100} - \\left(\\dfrac{1}{2}\\right)^{2}\\right]. \\left[\\dfrac{1}{100} - \\left(\\dfrac{1}{3}\\right)^{2}\\right]...\\left[\\dfrac{1}{100} - \\left(\\dfrac{1}{20}\\right)^{2}\\right] \\\\ = \\left[\\dfrac{1}{100} - 1^{2}\\right] . \\left[\\dfrac{1}{100} - \\dfrac{1}{4}\\right] . \\left[\\dfrac{1}{100} - \\dfrac{1}{9}\\right]... \\left[\\dfrac{1}{100} - \\dfrac{1}{100}\\right]... \\left[\\dfrac{1}{100} - \\dfrac{1}{400}\\right] $ <br\/> Nh\u1eadn x\u00e9t: C\u00f3 th\u1eeba s\u1ed1 $ \\left[\\dfrac{1}{100} - \\dfrac{1}{100}\\right] = 0 $ <br\/> Suy ra t\u00edch $K = 0$ <br\/> <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $0$.<\/span><\/span> "}]}],"id_ques":245},{"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":" T\u00ecm s\u1ed1 nguy\u00ean $n$ bi\u1ebft r\u1eb1ng: $\\dfrac{81}{(-3)^{n}} = -243$ ","select":["A. $ n = 9 $ ","B. $ n = 1 $ ","C. $ n = -1 $ ","D. $ n = -9 $"],"hint":" $ \\dfrac{81}{(-3)^{n}} = \\dfrac{(-3)^{4}}{(-3)^{n}} $ ","explain":"Ta c\u00f3: $\\dfrac{81}{(-3)^{n}} = -243 \\\\ \\Rightarrow \\dfrac{(-3)^{4}}{(-3)^{n}} = (-3)^{5} \\\\ \\Rightarrow (-3)^{4-n} = (-3)^{5} \\\\ \\Rightarrow 4 - n = 5 \\Rightarrow n = -1 $ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/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> <br\/>","column":2}]}],"id_ques":246},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"],["3"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c: $ A = \\dfrac{2.8^{4}.27^{2}+4.6^{9}}{2^{7}.6^{7}+2^{7}.40.9^{4}} $ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":" Ph\u00e2n t\u00edch c\u00e1c s\u1ed1 ra th\u1eeba s\u1ed1 nguy\u00ean t\u1ed1 r\u1ed3i v\u1eadn d\u1ee5ng c\u00e1c c\u00f4ng th\u1ee9c l\u0169y th\u1eeba \u0111\u1ec3 r\u00fat g\u1ecdn","explain":" Ta c\u00f3: $ A = \\dfrac{2.8^{4}.27^{2}+4.6^{9}}{2^{7}.6^{7}+2^{7}.40.9^{4}} $ <br\/> $ = \\dfrac{2.(2^{3})^{4}.(3^{3})^{2} + 2^{2}.(2.3)^{9}}{2^{7}.(2.3)^{7} + 2^{7}.(2^{3}.5).(3^{2})^{4}} $ <br\/> $ = \\dfrac{2.2^{12}.3^{6} + 2^{2}.2^{9}.3^{9}}{2^{7}.2^{7}.3^{7} + 2^{7}.2^{3}.5.3^{8}} \\\\ = \\dfrac{2^{13}.3^{6} + 2^{11}.3^{9}}{2^{14}.3^{7} + 2^{10}.3^{8}.5} $ <br\/> $ = \\dfrac{2^{11}.3^{6}(2^{2} + 3^{3})}{2^{10}.3^{7}(2^{4} + 3.5)} $ <br\/> $ = \\dfrac{2.31}{3.31} = \\dfrac{2}{3}$ <br\/> <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $2$ v\u00e0 $3$ <\/span><\/span>"}]}],"id_ques":247},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4"],["5"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c $ B = \\dfrac{4^{6}.9^{5}+6^{9}.120}{8^{4}.3^{12}-6^{11}} $= <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":" Ph\u00e2n t\u00edch c\u00e1c s\u1ed1 ra th\u1eeba s\u1ed1 nguy\u00ean t\u1ed1 r\u1ed3i v\u1eadn d\u1ee5ng c\u00e1c c\u00f4ng th\u1ee9c l\u0169y th\u1eeba \u0111\u1ec3 r\u00fat g\u1ecdn","explain":" Ta c\u00f3: $ B = \\dfrac{4^{6}.9^{5}+6^{9}.120}{8^{4}.3^{12}-6^{11}} \\\\ = \\dfrac{2^{12}.3^{10} + 2^{9}.3^{9}.2^{3}.3.5}{2^{12}.3^{12}-2^{11}.3^{11}} \\\\ = \\dfrac{2^{12}.3^{10} + 2^{12}.3^{10}.5}{2^{11}.3^{11}(6-1)} \\\\ = \\dfrac{2^{12}.3^{10}(1+5)}{2^{11}.3^{11}(6-1)} \\\\ = \\dfrac{2.6}{3.5} = \\dfrac{12}{15} = \\dfrac{4}{5}$ <br\/> <br\/> <span class='basic_pink'> Gi\u00e1 tr\u1ecb c\u1ee7a B l\u00e0 $\\dfrac{4}{5}$. <\/span><\/span>"}]}],"id_ques":248},{"time":4,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" Cho c\u00e1c s\u1ed1 h\u1eefu t\u1ec9 $x, y, z$ th\u1ecfa m\u00e3n: $xy = \\dfrac{2}{3}; yz = 0,6$ v\u00e0 $zx = 0,625$ <br\/> Gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0: ","select":["A. $x \\in \\left\\{ \\dfrac{5}{6}; \\dfrac{-5}{6} \\right\\}$ ","B. $x \\in \\left\\{ \\dfrac{4}{5}; \\dfrac{-4}{5} \\right\\}$ ","C. $ x = \\dfrac{-5}{6} $ ","D. $ x = \\dfrac{5}{6} $"],"hint":" Nh\u00e2n ba \u0111\u1eb3ng th\u1ee9c v\u1edbi nhau ta \u0111\u01b0\u1ee3c: $(xyz)^{2} = 0,25 = (0,5)^{2} $","explain":" <span class='basic_left'> C\u00f3: $xy = \\dfrac{2}{3}; yz = 0,6$ v\u00e0 $zx = 0,625$ <br\/> Nh\u00e2n ba \u0111\u1eb3ng th\u1ee9c v\u1edbi nhau ta \u0111\u01b0\u1ee3c: <br\/> $xy.yz.zx = \\dfrac{2}{3} . 0,6 . 0,625 \\\\ \\Rightarrow (xyz)^{2} = 0,25 = (0,5)^{2} \\\\ \\Rightarrow xyz = \\pm 0,5$ <br\/> Tr\u01b0\u1eddng h\u1ee3p 1: $xyz = 0,5\\Rightarrow$ $\\begin{cases} xyz = 0,5 \\\\ yz = 0,6 \\end{cases}$$ \\Rightarrow x = 0,5 : 0,6 = \\dfrac{5}{6}$ <br\/> Tr\u01b0\u1eddng h\u1ee3p 2: $xyz = -0,5\\Rightarrow$ $\\begin{cases} xyz = -0,5\\\\ yz = 0,6 \\end{cases} $ $\\Rightarrow x = -0,5 : 0,6 = \\dfrac{-5}{6} $ <br\/> V\u1eady gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0: $x \\in \\left\\{ \\dfrac{5}{6}; \\dfrac{-5}{6} \\right\\}$ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":249},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-3"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean $x$ bi\u1ebft: $ 2^{x-1} + 5.2^{x-2} = \\dfrac{7}{32} $ <br\/> <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $ x = $ _input_","explain":" Ta c\u00f3: $ 2^{x-1} + 5.2^{x-2} = \\dfrac{7}{32} \\\\ \\Rightarrow 2^{x-2+1} + 5.2^{x-2} = \\dfrac{7}{32} \\\\ \\Rightarrow 2^{x-2}.2^{1} + 5.2^{x-2} = \\dfrac{7}{32} \\\\ \\Rightarrow 2^{x-2}(2 + 5) = \\dfrac{7}{32} \\\\ \\Rightarrow 2^{x-2} = \\dfrac{7}{32} : 7 = \\dfrac{1}{32} = \\dfrac{1}{2^{5}} \\\\ \\Rightarrow 2^{5} . 2^{x-2} = 1 \\\\ \\Rightarrow 2^{5+x-2} = 2^{0} \\\\ \\Rightarrow 2^{x+3} = 2^{0} \\\\ \\Rightarrow x + 3 = 0 \\Rightarrow x = -3 $ <br\/> <br\/> <span class='basic_pink'> Gi\u00e1 tr\u1ecb c\u1ee7a $x$ b\u1eb1ng $-3.$ <\/span><\/span> "}]}],"id_ques":250}],"lesson":{"save":0,"level":3}}