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Home Lớp 6 Toán lớp 6 - Sách kết nối tri thức Chia hai lũy thừa cùng cơ sốBài tập nâng cao

{"common":{"save":0,"post_id":"723","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"341","post_id":"723","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":[[2]],"list":[{"point":10,"ques":" H\u00e3y ch\u1ecdn d\u1ea5u th\u00edch h\u1ee3p thay cho d\u1ea5u ?<\/span> <br\/><center><img src='img\/5.jpg' \/><\/center> <br\/> $625^{5}$ ?<\/span> $125^{7}$","select":["A. >","B. < ","C. = "],"hint":"\u0110\u01b0a c\u1ea3 hai v\u1ebf v\u1ec1 c\u00f9ng l\u0169y th\u1eeba c\u01a1 s\u1ed1 5 \u0111\u1ec3 so s\u00e1nh","explain":"H\u01b0\u1edbng d\u1eabn: <br\/> <\/span> PH\u01af\u01a0NG PH\u00c1P CHUNG: <\/b> <br\/> Th\u01b0\u1eddng \u0111\u1ec3 so s\u00e1nh hai l\u0169y th\u1eeba ta \u0111\u01b0a v\u1ec1 d\u1ea1ng so s\u00e1nh hai l\u0169y th\u1eeba c\u00f9ng c\u01a1 s\u1ed1 ho\u1eb7c c\u00f9ng s\u1ed1 m\u0169 <br\/> + N\u1ebfu hai l\u0169y th\u1eeba c\u00f9ng c\u01a1 s\u1ed1, l\u0169y th\u1eeba n\u00e0o c\u00f3 s\u1ed1 m\u0169 l\u1edbn h\u01a1n th\u00ec l\u1edbn h\u01a1n <br\/>N\u1ebfu m > n th\u00ec $a^m > a^n$ <br\/> + N\u1ebfu hai l\u0169y th\u1eeba c\u00f9ng s\u1ed1 m\u0169 th\u00ec l\u0169y th\u1eeba n\u00e0o c\u00f3 c\u01a1 s\u1ed1 l\u1edbn h\u1edbn s\u1ebd l\u1edbn h\u01a1n <br\/> N\u1ebfu a > b th\u00ec $a^n > b^n$ <\/i> <br\/> C\u1ee5 th\u1ec3: <\/b> Nh\u1eadn th\u1ea5y 625 v\u00e0 125 \u0111\u1ec1u l\u00e0 l\u0169y th\u1eeba c\u1ee7a 5 n\u00ean ta \u0111\u01b0a v\u1ec1 c\u00f9ng l\u0169y th\u1eeba c\u01a1 s\u1ed1 5<br\/> B\u01b0\u1edbc 1: <\/b> \u0110\u01b0a $625^{5}$ v\u00e0 $125^7$ v\u1ec1 l\u0169y th\u1eeba c\u01a1 s\u1ed1 5 b\u1eb1ng c\u00e1ch l\u1ea5y 625 v\u00e0 125 chia li\u00ean ti\u1ebfp cho 5 <br\/>$625 = 5^x$ n\u00ean $625^{5} = (5^x)^{5}= 5^{5x}$ <br\/> $125 = 5^a$ n\u00ean $125^7 =(5^a)^7 = 5^{7a}$ <br\/> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $5^{5x}$ v\u00e0 $5^{7a}$ <br\/><br\/><\/span>B\u00e0i gi\u1ea3i:<\/span> <br\/> Nh\u1eadn th\u1ea5y 625 v\u00e0 125 \u0111\u1ec1u l\u00e0 c\u00e1c l\u0169y th\u1eeba c\u1ee7a 5 n\u00ean ta \u0111\u01b0a v\u1ec1 c\u00f9ng l\u0169y th\u1eeba c\u01a1 s\u1ed1 5 \u0111\u1ec3 so s\u00e1nh<br\/> $625^{5} = (5 . 5 . 5 . 5)^{5} = (5^4)^5 = 5^{5.4} = 5^{20}$ <br\/>$125^7 = (5 . 5 . 5)^7 = (5^3)^7 = 5^{3.7} = 5^{21}$ <br\/> V\u00ec $5^{20} < 5^{21}$ <br\/> N\u00ean $625^{5} < 125^{7}$<br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B. < <\/span> <br\/> L\u01b0u \u00fd: L\u0169y th\u1eeba c\u1ee7a m\u1ed9t t\u00edch b\u1eb1ng t\u00edch c\u00e1c l\u0169y th\u1eeba <br\/> $(a^m)^n = a^m.a^n (a \\neq 0; m,n \\in \\mathbb{n})$ <\/i><\/span> ","column":3}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:40"},{"id":"342","post_id":"723","mon_id":"0","chapter_id":"0","question":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","options":{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"T\u00ecm x bi\u1ebft: <br\/><center><img src='img\/4.jpg' \/><\/center> <br\/> $[(2x - 4)+9]^3 = 3^6$ <br\/> V\u1eady x = _input_","hint":"Bi\u1ebfn \u0111\u1ed5i VP v\u1ec1 l\u0169y th\u1eeba b\u1eadc 3 c\u1ee7a m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean sau \u0111\u00f3 cho 2 c\u01a1 s\u1ed1 b\u1eb1ng nhau \u0111\u1ec3 t\u00ecm x","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> PH\u01af\u01a0NG PH\u00c1P CHUNG: <\/b><br\/> Ta bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c \u1edf hai v\u1ebf v\u1ec1 c\u00f9ng c\u01a1 s\u1ed1 ho\u1eb7c c\u00f9ng s\u1ed1 m\u0169 <br\/> +) N\u1ebfu VT v\u00e0 VP c\u00f3 c\u00f9ng c\u01a1 s\u1ed1 th\u00ec cho s\u1ed1 m\u0169 \u1edf 2 v\u1ebf b\u1eb1ng nhau \u0111\u1ec3 t\u00ecm x <br\/> ($a^x = a^y$ khi v\u00e0 ch\u1ec9 khi x = y)<br\/> +) N\u1ebfu VT v\u00e0 VP c\u00f3 c\u00f9ng s\u1ed1 m\u0169 th\u00ec ta cho c\u01a1 s\u1ed1 \u1edf 2 v\u1ebf b\u1eb1ng nhau r\u1ed3i t\u00ecm x <br\/> ($a^m = b^m$ khi v\u00e0 ch\u1ec9 khi a = b) <\/i> <br\/> C\u1ee5 th\u1ec3: <\/b> <br\/> B\u01b0\u1edbc 1: <\/b> Bi\u1ebfn \u0111\u1ed5i $3^6$ v\u1ec1 l\u0169y th\u1eeba b\u1eadc 3 <br\/> B\u01b0\u1edbc 2: <\/b> V\u00ec c\u1ea3 VP v\u00e0 VT \u0111\u1ec1u l\u00e0 l\u0169y th\u1eeba b\u1eadc 3 <br\/> Cho hai c\u01a1 s\u1ed1 b\u1eb1ng nhau \u0111\u1ec3 t\u00ecm x theo c\u00f4ng th\u1ee9c tr\u00ean<\/span> <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> $ \\begin{align*} \\text{Ta c\u00f3: } [(2x - 4)+9]^3 &= 3^6 \\\\ [(2x - 4)+9]^3 &= 3^{2.3} \\\\ [(2x - 4)+9]^3 &= (3^2)^3 \\\\ [(2x - 4)+9]^3 &= 9^3 \\end{align*}$ <br\/> $ \\begin{align*} \\text{Suy ra: } (2x - 4)+9 &= 9 \\\\ 2x - 4 + 9 &= 9 \\\\ 2x + 5 &= 9 \\\\ 2x &= 9 - 5 \\\\ x &= 2 \\end{align*}$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 2<\/span> <br\/> L\u01b0u \u00fd: <\/i> $(a^m)^n = a^{m.n} (a \\neq 0; m, n \\in \\mathbb{N})$<\/span>"}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:40"},{"id":"343","post_id":"723","mon_id":"0","chapter_id":"0","question":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","options":{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["625"]]],"list":[{"point":10,"width":60,"type_input":"","ques":"T\u00ednh. <br\/><center><img src='img\/2.jpg' \/><\/center> <br\/> $5^{10} . 25^4 . 125^2 : 625^5$ = _input_","hint":"Bi\u1ebfn \u0111\u1ed5i t\u1ea5t c\u1ea3 v\u1ec1 l\u0169y th\u1eeba c\u01a1 s\u1ed1 5 sau \u0111\u00f3 \u00e1p d\u1ee5ng quy t\u1eafc nh\u00e2n v\u00e0 chia hai l\u0169y th\u1eeba c\u00f9ng c\u01a1 s\u1ed1","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> * Quy t\u1eafc: <\/b> <br\/> +) $a^m.a^n = a^{m + n}$ <br\/> +) $a^m:a^n = a^{m - n}$ <br\/> +) $a^m.b^m = (a.b)^n$ <br\/> +) $(a^m)^n = a^{m.n}$ <br\/> +) $a^m:b^m = (a: b)^m$ <br\/> V\u1edbi $a, b \\neq 0; m, n \\in \\mathbb{N}$ <br\/> C\u1ee5 th\u1ec3: <\/b> <br\/> B\u01b0\u1edbc 1: <\/b> Bi\u1ebfn \u0111\u1ed5i $25^4$, $125^2$; $625^5$ v\u1ec1 c\u00f9ng l\u0169y th\u1eeba c\u01a1 s\u1ed1 5 <br\/> B\u01b0\u1edbc 2: <\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh nh\u00e2n v\u00e0 chia theo quy t\u1eafc nh\u00e2n, chia hai l\u0169y th\u1eeba c\u00f9ng c\u01a1 s\u1ed1 <\/span> <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> $ \\begin{align*} \\text{Ta c\u00f3: } & 5^{10} . 25^4 . 125^2 : 625^5 \\\\ =& 5^{10}. (5 . 5)^4 . (5 . 5 . 5)^2 : (5 . 5 . 5 . 5)^5 \\\\ =& 5^{10}. (5^2)^4 . (5^3)^2 : (5^4)^5 \\\\ =& 5^{10} . 5^8 . 5^6 : 5^{20} \\\\ =& 5^{10 + 8 + 6} : 5^{20} \\\\ =& 5^{24} : 5^{20} \\\\ =& 5^{24 - 20} \\\\ =& 5^4 \\\\ =& 625 \\end{align*}$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 625 <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:40"}]}