Hai bài toán về phân số - Bài tập nâng cao số 2

Home Lớp 6 Toán lớp 6 - Sách kết nối tri thức Bài 27: Hai bài toán về phân sốBài tập nâng cao

{"common":{"save":0,"post_id":"4789","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"8145","post_id":"4789","mon_id":"1158113","chapter_id":"1158119","question":"M\u1ed9t h\u1ed9p bánh có 75 chi\u1ebfc. Các chi\u1ebfc bánh trong h\u1ed9p gi\u1ed1ng nhau nh\u01b0ng \u0111\u01b0\u1ee3c b\u1ecdc b\u1eb1ng hai lo\u1ea1i gi\u1ea5y có màu khác nhau: tr\u1eafng và \u0111en. S\u1ed1 bánh b\u1ecdc trong gi\u1ea5y tr\u1eafng chi\u1ebfm  $\\dfrac{8}{15}$<\/span>  s\u1ed1 bánh trong h\u1ed9p. Di\u1ec7n tích m\u1ed7i t\u1edd gi\u1ea5y b\u1ecdc m\u1ed9t chi\u1ebfc bánh \u0111\u1ec1u b\u1eb1ng 100 ${cm}^2$<\/span>. Tính di\u1ec7n tích m\u1ed7i lo\u1ea1i gi\u1ea5y (tr\u1eafng và \u0111en) c\u1ea7n dùng \u0111\u1ec3 b\u1ecdc bánh trong h\u1ed9p bánh \u0111ó.<\/p>","options":["4000 ${cm}^2$<\/span> gi\u1ea5y tr\u1eafng và 3500 ${cm}^2$<\/span> gi\u1ea5y \u0111en","3500 ${cm}^2$<\/span> gi\u1ea5y tr\u1eafng và 4000 ${cm}^2$<\/span> gi\u1ea5y \u0111en"],"correct":"1","level":"3","hint":"Dùng quy t\u1eafc: "Mu\u1ed1n tìm $\\dfrac{m}{n}$<\/span> c\u1ee7a m\u1ed9t s\u1ed1  a  cho tr\u01b0\u1edbc ta tính  a . $\\dfrac{m}{n}$<\/span> ." \u0111\u1ec3 tính s\u1ed1 bánh m\u1ed7i lo\u1ea1i r\u1ed3i tính di\u1ec7n tích m\u1ed7i lo\u1ea1i gi\u1ea5y.<\/p>","answer":"S\u1ed1 chi\u1ebfc bánh b\u1ecdc gi\u1ea5y màu tr\u1eafng là     $\\dfrac{8}{15}$<\/span> . 75 = 40 (chi\u1ebfc).<\/p>S\u1ed1 chi\u1ebfc bánh b\u1ecdc gi\u1ea5y màu \u0111en  là      75 - 40 = 35  (chi\u1ebfc).<\/p>Di\u1ec7n tích gi\u1ea5y gói màu tr\u1eafng là       40 . 100 = 4000  (${cm}^2$<\/span>)<\/p>Di\u1ec7n tích gi\u1ea5y gói màu \u0111en là          35 . 100 = 3500 (${cm}^2$<\/span>)<\/p>\u0110áp án \u0111úng là   4000 <\/strong>${cm}^2$<\/span> gi\u1ea5y tr\u1eafng và 3500 <\/strong>${cm}^2$<\/span> gi\u1ea5y \u0111en<\/strong><\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"1","date":"2021-10-03 11:58:57","option_type":"math","len":0},{"id":"8149","post_id":"4789","mon_id":"1158113","chapter_id":"1158119","question":"M\u1ed9t xe ô tô và m\u1ed9t xe máy kh\u1edfi hành cùng lúc theo hai h\u01b0\u1edbng ng\u01b0\u1ee3c nhau: ô tô \u0111i t\u1eeb A \u0111\u1ebfn B, xe máy \u0111i t\u1eeb B \u0111\u1ebfn A. Sau 7 gi\u1edd hai xe ch\u01b0a g\u1eb7p nhau và còn cách nhau 130 km. Bi\u1ebft r\u1eb1ng \u0111\u1ec3 \u0111i h\u1ebft quãng \u0111\u01b0\u1eddng AB, ô tô ph\u1ea3i \u0111i trong 15 gi\u1edd và xe máy ph\u1ea3i \u0111i trong 18 gi\u1edd liên t\u1ee5c v\u1edbi v\u1eadn t\u1ed1c không \u0111\u1ed5i. Tính chi\u1ec1u dài quãng \u0111\u01b0\u1eddng AB.<\/p>Quãng \u0111\u01b0\u1eddng AB dài<\/strong> <input class=\"tim_x input_number\" tpl=\"dien_so\" type=\"text\"\/> km<\/strong>.<\/p>","options":"","correct":["[\"900\"]"],"level":"3","hint":"","answer":"M\u1ed7i gi\u1edd ô tô \u0111i \u0111\u01b0\u1ee3c       $\\dfrac{1}{15}$<\/span>  quãng \u0111\u01b0\u1eddng.<\/p>M\u1ed7i gi\u1edd xe máy \u0111i \u0111\u01b0\u1ee3c  $\\dfrac{1}{18}$<\/span> quãng \u0111\u01b0\u1eddng.<\/p>M\u1ed7i gi\u1edd hai xe cùng \u0111i \u0111\u01b0\u1ee3c   $\\dfrac{1}{15}+\\dfrac{1}{18}=\\dfrac{6}{90}+\\dfrac{5}{90}$<\/span>  = $\\dfrac{11}{90}$<\/span>  (quãng \u0111\u01b0\u1eddng).<\/p>Sau 7 gi\u1edd hai xe cùng \u0111i \u0111\u01b0\u1ee3c 7 . $\\dfrac{11}{90}$<\/span> =$\\dfrac{77}{90}$<\/span>  (quãng \u0111\u01b0\u1eddng)<\/p>Sau 7 gi\u1edd hai xe còn cách nhau 1  -  $\\dfrac{77}{90}$<\/span>   =   $\\dfrac{90}{90}-\\dfrac{77}{90}=$<\/span>   $\\dfrac{13}{90}$<\/span>  (quãng \u0111\u01b0\u1eddng), \u1ee9ng v\u1edbi 130 km.<\/p>V\u1eady quãng \u0111\u01b0\u1eddng AB dài  130 :  $\\dfrac{13}{90}$<\/span>   = 130 . $\\dfrac{90}{13}$<\/span> = 900 (km)<\/strong><\/p>S\u1ed1 c\u1ea7n \u0111i\u1ec1n là  900<\/strong>.<\/p>","type":"blank","extra_type":"","time":"0","user_id":"131","test":"1","date":"2021-10-03 12:06:05"},{"id":"8153","post_id":"4789","mon_id":"1158113","chapter_id":"1158119","question":"S\u1ed1 nào trong ba s\u1ed1  S, Q, R  l\u1edbn h\u01a1n 1, bi\u1ebft r\u1eb1ng:<\/p>   $\\dfrac{5}{7}$<\/span> c\u1ee7a S là $\\dfrac{1}{3}$<\/span><\/p>   $\\dfrac{2}{9}$<\/span> c\u1ee7a Q là $\\dfrac{3}{8}$<\/span><\/p>   $\\dfrac{9}{4}$<\/span> c\u1ee7a R là $\\dfrac{5}{3}$<\/span> <\/p>","options":["Q","R","S"],"correct":"1","level":"3","hint":"Mu\u1ed1n tìm m\u1ed9t s\u1ed1 khi bi\u1ebft $\\dfrac{m}{n}$<\/span> c\u1ee7a s\u1ed1 \u0111ó là a, ta tính a : $\\dfrac{m}{n}$<\/span><\/p>","answer":"Ta có:<\/p>   $\\dfrac{5}{7}$<\/span> c\u1ee7a S là $\\dfrac{1}{3}$<\/span>  nên  S = $\\dfrac{1}{3}$<\/span> :  $\\dfrac{5}{7}$<\/span> = $\\dfrac{1}{3}$<\/span> . $\\dfrac{7}{5}$<\/span> = $\\dfrac{7}{15}$<\/span>  < 1<\/p>   $\\dfrac{2}{9}$<\/span> c\u1ee7a Q là $\\dfrac{3}{8}$<\/span>  nên Q = $\\dfrac{3}{8}$<\/span> :  $\\dfrac{2}{9}$<\/span> = $\\dfrac{3}{8}$<\/span> . $\\dfrac{9}{2}$<\/span>  = $\\dfrac{27}{16}$<\/span>  > 1<\/p>   $\\dfrac{9}{4}$<\/span> c\u1ee7a R là $\\dfrac{5}{3}$<\/span>  nên  R = $\\dfrac{5}{3}$<\/span> : $\\dfrac{9}{4}$<\/span> = $\\dfrac{5}{3}$<\/span> .  $\\dfrac{4}{9}$<\/span> = $\\dfrac{20}{27}$<\/span>  < 1<\/p>V\u1eady   Q > 1<\/strong><\/span><\/p>\u0110áp án \u0111úng là  Q<\/strong><\/span><\/p>","type":"choose","extra_type":"star","time":"0","user_id":"131","test":"3","date":"2021-10-03 12:20:38","option_type":"txt","len":0}]}