Đa thức một biến - Bài tập dễ

Home Toán lớp 7 - Sách KNTT Bài 25: Đa thức một biếnBài tập cơ bản

{"common":{"save":0,"post_id":"5806","level":1,"total":10,"point":10,"point_extra":0},"segment":[{"id":"8449","post_id":"5806","mon_id":"1158532","chapter_id":"1158660","question":"Bi\u1ec3u th\u1ee9c nào sau \u0111ây là \u0111\u01a1n th\u1ee9c m\u1ed9t bi\u1ebfn?<\/p>","options":["A.<\/strong> x + y","B. <\/strong>3x²","C. <\/strong>x² + y","D. <\/strong>$\\dfrac{1}{x}$<\/span>"],"correct":"2","level":"1","hint":"\u0110\u01a1n th\u1ee9c m\u1ed9t bi\u1ebfn (g\u1ecdi t\u1eaft là \u0111\u01a1n th\u1ee9c) là bi\u1ec3u th\u1ee9c \u0111\u1ea1i s\u1ed1 có d\u1ea1ng tích c\u1ee7a m\u1ed9t s\u1ed1 th\u1ef1c v\u1edbi m\u1ed9t l\u0169y th\u1eeba c\u1ee7a bi\u1ebfn<\/span>, trong \u0111ó s\u1ed1 th\u1ef1c g\u1ecdi là h\u1ec7 s\u1ed1, s\u1ed1 m\u0169 c\u1ee7a l\u0169y th\u1eeba c\u1ee7a bi\u1ebfn g\u1ecdi là b\u1eadc c\u1ee7a \u0111\u01a1n th\u1ee9c.<\/p>","answer":"\u0110\u01a1n th\u1ee9c m\u1ed9t bi\u1ebfn là B. <\/strong>3x²<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"1","date":"2022-05-31 02:01:38","option_type":"math","len":0},{"id":"8452","post_id":"5806","mon_id":"1158532","chapter_id":"1158660","question":"Cho bi\u1ebft h\u1ec7 s\u1ed1 và b\u1eadc c\u1ee7a m\u1ed7i \u0111\u01a1n th\u1ee9c sau:<\/p>","options":[["$2x^5$<\/span> có","$-\\dfrac{1}{2}$<\/span> có","$4^2x$<\/span> có ","0 có"],["h\u1ec7 s\u1ed1 là $-\\dfrac{1}{2}$<\/span>, b\u1eadc là 0","h\u1ec7 s\u1ed1 là 16, b\u1eadc là 1","h\u1ec7 s\u1ed1 là 2, b\u1eadc là 5","h\u1ec7 s\u1ed1 là 0, không có b\u1eadc"],["h\u1ec7 s\u1ed1 là 2, b\u1eadc là 5","h\u1ec7 s\u1ed1 là $-\\dfrac{1}{2}$<\/span>, b\u1eadc là 0","h\u1ec7 s\u1ed1 là 16, b\u1eadc là 1","h\u1ec7 s\u1ed1 là 0, không có b\u1eadc"]],"correct":[1,2,0,3],"level":"1","hint":"\u0110\u01a1n th\u1ee9c m\u1ed9t bi\u1ebfn (g\u1ecdi t\u1eaft là \u0111\u01a1n th\u1ee9c) là bi\u1ec3u th\u1ee9c \u0111\u1ea1i s\u1ed1 có d\u1ea1ng tích c\u1ee7a m\u1ed9t s\u1ed1 th\u1ef1c v\u1edbi m\u1ed9t l\u0169y th\u1eeba c\u1ee7a bi\u1ebfn, trong \u0111ó s\u1ed1 th\u1ef1c g\u1ecdi là h\u1ec7 s\u1ed1<\/span>, s\u1ed1 m\u0169 c\u1ee7a l\u0169y th\u1eeba c\u1ee7a bi\u1ebfn g\u1ecdi là b\u1eadc c\u1ee7a \u0111\u01a1n th\u1ee9c<\/span>.<\/p>M\u1ed7i s\u1ed1 khác 0 là m\u1ed9t \u0111\u01a1n b\u1eadc 0.<\/p>S\u1ed1 0 là \u0111\u01a1n th\u1ee9c không có b\u1eadc.<\/p>","answer":"\u0110áp án là<\/p>$2x^5$<\/span> có <\/span>→ h\u1ec7 s\u1ed1 là 2, b\u1eadc là 5<\/span><\/p>$-\\dfrac{1}{2}$<\/span> có <\/span>→ h\u1ec7 s\u1ed1 là $-\\dfrac{1}{2}$<\/span>, b\u1eadc là 0<\/span><\/p>$4^2x$<\/span> có <\/span>→ h\u1ec7 s\u1ed1 là 16, b\u1eadc là 1 <\/span>(vì 4² = 16 nên h\u1ec7 s\u1ed1 là 16)<\/p>0 có <\/span>→ h\u1ec7 s\u1ed1 là 0, không có b\u1eadc<\/span><\/p>","type":"matching","extra_type":"mat_txt","time":"0","user_id":"131","test":"1","date":"2022-05-31 02:09:38"},{"id":"8454","post_id":"5806","mon_id":"1158532","chapter_id":"1158660","question":"Tính t\u1ed5ng c\u1ee7a hai \u0111\u01a1n th\u1ee9c sau:<\/p>3x³ + $\\dfrac{1}{3}$<\/span>x³<\/p>","options":["A.<\/strong> $\\dfrac{10}{3}$<\/span>x","B.<\/strong> $\\dfrac{10}{3}x^6$<\/span>","C.<\/strong> $\\dfrac{10}{3}$<\/span>","D.<\/strong> $\\dfrac{10}{3}$<\/span>x³"],"correct":"4","level":"1","hint":"\u0110\u1ec3 c\u1ed9ng hay tr\u1eeb hai \u0111\u01a1n th\u1ee9c m\u1ed9t bi\u1ebfn cùng b\u1eadc b\u1eb1ng cách c\u1ed9ng (hay tr\u1eeb) các h\u1ec7 s\u1ed1 v\u1edbi nhau và gi\u1eef nguyên l\u0169y th\u1eeba c\u1ee7a bi\u1ebfn.<\/p>","answer":"   3x³ + $\\dfrac{1}{3}$<\/span>x³<\/p>= (3 + $\\dfrac{1}{3}$<\/span>) . x³<\/p>= $\\dfrac{10}{3}$<\/span>x³<\/p>\u0110áp án \u0111úng là  D.<\/strong> $\\dfrac{10}{3}$<\/span>x³<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"1","date":"2022-05-31 02:19:31","option_type":"math","len":0}]}