Giải bài toán bằng cách lập hệ phương trình

Home Lớp 9 Toán lớp 9 - Sách Kết nối tri thức Bài 3. Giải bài toán bằng cách lập hệ phương trìnhBài tập nâng cao

{"common":{"save":0,"post_id":"7749","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"7142","post_id":"7749","mon_id":"1159278","chapter_id":"1159281","question":"M\u1ed9t canô \u0111i t\u1eeb b\u1ebfn A \u0111\u1ebfn b\u1ebfn B d\u1ef1 \u0111\u1ecbnh \u0111\u1ebfn B lúc 12 gi\u1edd tr\u01b0a. N\u1ebfu ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c 20 km\/h thì s\u1ebd \u0111\u1ebfn B lúc 13 gi\u1edd. N\u1ebfu canô ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c 35 km\/h thì s\u1ebd \u0111\u1ebfn B s\u1edbm h\u01a1n 2 gi\u1edd. Tính \u0111\u1ed9 dài quãng \u0111\u01b0\u1eddng AB. <\/p>","options":["A.<\/strong> 150 km","B.<\/strong> 146 km","C.<\/strong> 160 km","D.<\/strong> 140 km"],"correct":"4","level":"3","hint":"","answer":"Ch\u1ecdn D.<\/strong> 140 km<\/span><\/p>G\u1ecdi x (km, x > 0) là \u0111\u1ed9 dài quãng \u0111\u01b0\u1eddng AB, y (gi\u1edd, y > 2) là th\u1eddi gian d\u1ef1 \u0111\u1ecbnh \u0111\u1ec3 \u0111i \u0111\u1ebfn B lúc 12 gi\u1edd tr\u01b0a.<\/p>N\u1ebfu ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c 20 km\/h thì th\u1eddi gian \u0111i h\u1ebft quãng \u0111\u01b0\u1eddng AB là:  $\\dfrac{x}{20}$<\/span> (gi\u1edd)<\/p>Vì ô tô \u0111\u1ebfn B lúc 13 gi\u1edd nên ô tô \u0111\u1ebfn ch\u1eadm h\u01a1n so v\u1edbi d\u1ef1 \u0111\u1ecbnh 1 gi\u1edd nên ta có ph\u01b0\u01a1ng trình: $\\dfrac{x}{20}=y+1$<\/span> hay $x-20y=20$<\/span> (1).<\/p>N\u1ebfu ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c 35 km\/h thì th\u1eddi gian \u0111i h\u1ebft quãng \u0111\u01b0\u1eddng AB là:  $\\dfrac{x}{35}$<\/span> (gi\u1edd) mà ô tô \u0111\u1ebfn B s\u1edbm h\u01a1n so v\u1edbi d\u1ef1 \u0111\u1ecbnh 2 gi\u1edd nên ta có ph\u01b0\u01a1ng trình $\\dfrac{x}{35}=y-2$<\/span> hay $x-35y=-70$<\/span> (2).<\/p>T\u1eeb (1) và (2) ta có h\u1ec7 ph\u01b0\u01a1ng trình $\\begin{cases}x-20y=20\\\\x-35y=-70\\end{cases}$<\/span> .<\/p>Tr\u1eeb t\u1eebng v\u1ebf hai ph\u01b0\u01a1ng trình c\u1ee7a h\u1ec7 m\u1edbi ta \u0111\u01b0\u1ee3c 15y = 90 hay y = 6.<\/p>Thay vào ph\u01b0\u01a1ng trình th\u1ee9 nh\u1ea5t ta \u0111\u01b0\u1ee3c: x – 120 = 20 hay x = 140.<\/p>V\u1eady \u0111\u1ed9 dài quãng \u0111\u01b0\u1eddng AB là 140 km.<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"127","test":"0","date":"2024-10-03 01:41:17","option_type":"txt","len":0},{"id":"7144","post_id":"7749","mon_id":"1159278","chapter_id":"1159281","question":"Hai vòi ng\u01b0\u1edbc cùng ch\u1ea3y vào m\u1ed9t b\u1ec3 không có n\u01b0\u1edbc thì sau 1,5 gi\u1edd s\u1ebd \u0111\u1ea7y b\u1ec3. N\u1ebfu m\u1edf vòi 1 ch\u1ea3y trong 0,25 gi\u1edd r\u1ed3i khóa l\u1ea1i và m\u1edf vòi 2 ch\u1ea3y trong $\\dfrac{1}{3}$<\/span> gi\u1edd thì \u0111\u01b0\u1ee3c $\\dfrac{1}{5}$<\/span> b\u1ec3. H\u1ecfi n\u1ebfu vòi 2 ch\u1ea3y riêng thì bao lâu \u0111\u1ea7y b\u1ec3?<\/p>","options":["A.<\/strong> 2,5 gi\u1edd","B.<\/strong> 2 gi\u1edd","C.<\/strong> 3 gi\u1edd","D.<\/strong> 4 gi\u1edd"],"correct":"1","level":"3","hint":"","answer":"Ch\u1ecdn A.<\/strong> 2,5 gi\u1edd<\/span><\/p>G\u1ecdi th\u1eddi gian vòi 1, vòi 2 ch\u1ea3y m\u1ed9t mình \u0111\u1ea7y b\u1ec3 l\u1ea7n l\u01b0\u1ee3t là x, y (gi\u1edd; x, y > 0).<\/p>1 gi\u1edd vòi 1 ch\u1ea3y \u0111\u01b0\u1ee3c $\\dfrac{1}{x}$<\/span> (b\u1ec3) suy ra 0,25 gi\u1edd vòi 1 ch\u1ea3y \u0111\u01b0\u1ee3c $\\dfrac{1}{4x}$<\/span> (b\u1ec3).<\/p>1 gi\u1edd vòi 2 ch\u1ea3y \u0111\u01b0\u1ee3c $\\dfrac{1}{y}$<\/span> (b\u1ec3) suy ra $\\dfrac{1}{3}$<\/span> gi\u1edd vòi 1 ch\u1ea3y \u0111\u01b0\u1ee3c $\\dfrac{1}{3y}$<\/span> (b\u1ec3).<\/p>1 gi\u1edd c\u1ea3 2 vòi ch\u1ea3y \u0111\u01b0\u1ee3c $\\dfrac{1}{1,5}=\\dfrac{2}{3}$<\/span> (b\u1ec3) nên ta có ph\u01b0\u01a1ng trình $\\dfrac{1}{x}+\\dfrac{1}{y}=\\dfrac{2}{3}$<\/span> (1).<\/p>N\u1ebfu m\u1edf vòi 1 ch\u1ea3y trong 0,25 gi\u1edd r\u1ed3i khóa l\u1ea1i và m\u1edf vòi 2 ch\u1ea3y trong $\\dfrac{1}{3}$<\/span> gi\u1edd thì \u0111\u01b0\u1ee3c $\\dfrac{1}{5}$<\/span> b\u1ec3 nên ta có ph\u01b0\u01a1ng trình $\\dfrac{1}{4x}+\\dfrac{1}{3y}=\\dfrac{1}{5}$<\/span> (2).<\/p>T\u1eeb (1) và (2) ta có h\u1ec7 ph\u01b0\u01a1ng trình $\\begin{cases}\\dfrac{1}{x}+\\dfrac{1}{y}=\\dfrac{2}{3}\\\\\\dfrac{1}{4x}+\\dfrac{1}{3y}=\\dfrac{1}{5}\\end{cases}$<\/span> hay $\\begin{cases}\\dfrac{1}{4x}+\\dfrac{1}{4y}=\\dfrac{1}{6}\\\\\\dfrac{1}{4x}+\\dfrac{1}{3y}=\\dfrac{1}{5}\\end{cases}$<\/span><\/p>Tr\u1eeb t\u1eebng v\u1ebf hai ph\u01b0\u01a1ng trình c\u1ee7a h\u1ec7 m\u1edbi ta \u0111\u01b0\u1ee3c $-\\dfrac{1}{12y}=-\\dfrac{1}{30}$<\/span> suy ra y = 2,5.<\/p>V\u1eady n\u1ebfu vòi 2 ch\u1ea3y riêng thì m\u1ea5t 2,5 gi\u1edd \u0111\u1ec3 \u0111\u1ea7y b\u1ec3.<\/p>","type":"choose","extra_type":"shape1","time":"0","user_id":"127","test":"0","date":"2024-10-03 01:50:45","option_type":"txt","len":0},{"id":"7145","post_id":"7749","mon_id":"1159278","chapter_id":"1159281","question":"Gi\u1ea3i bài toán c\u1ed5 sau \u0111ây:<\/p>Quýt, cam m\u01b0\u1eddi b\u1ea3y qu\u1ea3 t\u01b0\u01a1i \u0110em chia cho m\u1ed9t tr\u0103m ng\u01b0\u1eddi cùng vui Chia ba m\u1ed7i qu\u1ea3 quýt r\u1ed3i Còn cam m\u1ed7i qu\u1ea3 chia m\u01b0\u1eddi th\u1eadt xinh Tr\u0103m ng\u01b0\u1eddi tr\u0103m mi\u1ebfng ngon lành Quýt cam m\u1ed7i lo\u1ea1i tính rành là bao?<\/span><\/strong><\/p>Câu tr\u1ea3 l\u1eddi nào sau \u0111ây \u0111úng<\/strong>?<\/p>","options":["A.<\/strong> Cam: 9 qu\u1ea3; Quýt: 8 qu\u1ea3.","B.<\/strong> Cam: 5 qu\u1ea3; Quýt: 12 qu\u1ea3.","C.<\/strong> Cam: 10 qu\u1ea3; Quýt: 7 qu\u1ea3.","D.<\/strong> Cam: 7 qu\u1ea3; Quýt: 10 qu\u1ea3."],"correct":"4","level":"3","hint":"","answer":"Ch\u1ecdn D.<\/strong> Cam: 7 qu\u1ea3; Quýt: 10 qu\u1ea3.<\/span><\/p>G\u1ecdi s\u1ed1 cam là x, s\u1ed1 quýt là y (x, y ∈ N* ; x < 17, y < 17).<\/p>Quýt, cam 17 qu\u1ea3 t\u01b0\u01a1i ⇒ x + y = 17.<\/p>M\u1ed7i qu\u1ea3 quýt chia ba ⇒ Có 3y mi\u1ebfng quýt<\/p>Chia m\u01b0\u1eddi m\u1ed7i qu\u1ea3 cam ⇒ Có 10x mi\u1ebfng cam<\/p>T\u1ed5ng s\u1ed1 mi\u1ebfng tròn 100 ⇒ 10x + 3y = 100.<\/p>T\u1eeb (1) và (2) ta có h\u1ec7 ph\u01b0\u01a1ng trình $\\begin{cases}x+y=17\\\\10x+3y=100\\end{cases}$<\/span> <\/p>Nhân hai v\u1ebf ph\u01b0\u01a1ng trình th\u1ee9 nh\u1ea5t v\u1edbi 3 ta \u0111\u01b0\u1ee3c: $\\begin{cases}3x+3y=51\\\\10x+3y=100\\end{cases}$<\/span><\/p>Tr\u1eeb t\u1eebng v\u1ebf hai ph\u01b0\u01a1ng trình c\u1ee7a h\u1ec7 m\u1edbi ta \u0111\u01b0\u1ee3c 7x = 49 hay x = 7.<\/p>Thay vào ph\u01b0\u01a1ng trình th\u1ee9 nh\u1ea5t ta \u0111\u01b0\u1ee3c 7 + y = 17 hay y = 10.<\/p>V\u1eady có 7 qu\u1ea3 cam và 10 qu\u1ea3 quýt.<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"127","test":"0","date":"2024-10-03 01:59:17","option_type":"txt","len":2}]}