{"common":{"save":0,"post_id":"961","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"701","post_id":"961","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":"","input_hint":["sqrt","x_sqrt","frac","sqr","x_"],"ques":"Cho h\u00e0m s\u1ed1 $ f(x) = \\dfrac{2}{x-1} .$ T\u00ecm $x$ sao cho $f(x) = \\dfrac{-2}{3} ?$ <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $x = $ _input_ ","explain":"Ta c\u00f3: $ f(x) = \\dfrac{-2}{3} \\\\ \\Rightarrow \\dfrac{2}{x-1} = \\dfrac{-2}{3} \\\\ \\Rightarrow \\begin{cases} x \\neq 1 \\\\ 6 = -2(x-1) \\end{cases} \\\\ \\Rightarrow \\begin{cases} x \\neq 1 \\\\ 6 = -2x + 2 \\end{cases} \\\\ \\Rightarrow \\begin{cases} x \\neq 1 \\\\ -2x = 4 \\end{cases} \\\\ \\Rightarrow \\begin{cases} x \\neq 1 \\\\ x = -2 \\end{cases} $ <br\/><br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $x = -2$ <\/span><\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:22:50"},{"id":"702","post_id":"961","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","options":{"time":14,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","t","f","f"]],"list":[{"point":10,"image":"","content":"","col_name":["Cho h\u00e0m s\u1ed1 $y = f(x) = \\dfrac{1}{4}x^{3} $","\u0110\u00fang","Sai"],"arr_ques":[" $f(-2) = -2 $"," $f(-1) = \\dfrac{-1}{4} $","$f(0) = \\dfrac{1}{2} $"," $f(3) = \\dfrac{3}{4} $"],"explain":[" V\u1edbi $x=-2$ th\u00ec $f(-2) = \\dfrac{1}{4}.(-2)^{3} = \\dfrac{1}{4}.(-8) = -2 $","<br\/>V\u1edbi $x=-1$ th\u00ec $f(-1) = \\dfrac{1}{4}.(-1)^{3} = \\dfrac{-1}{4} $","<br\/> V\u1edbi x = 0 th\u00ec $f(0) = \\dfrac{1}{4}.0^{3} = 0 $ ","<br\/> V\u1edbi x = 3 th\u00ec $f(3) = \\dfrac{1}{4}.(3)^{3} = \\dfrac{1}{4}.27= \\dfrac{27}{4} $"]}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:22:50"},{"id":"703","post_id":"961","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":[[["-8"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Cho h\u00e0m s\u1ed1 $y = f(x) = (a+2)x + 2a + 5 .$ T\u00ecm a bi\u1ebft $f(-3) = 7 ? $ <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $a = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$","explain":" Ta c\u00f3: $y = (a+2)x + 2a + 5 $ <br\/> Do \u0111\u00f3: $f(-3) $ $ = (a+2).(-3) + 2a + 5 \\\\ = -3a - 6 + 2a + 5 \\\\ = -a - 1 $ <br\/> M\u00e0 $f(-3) = 7 \\\\ \\Rightarrow -a - 1 = 7 \\\\ \\Rightarrow -a = 8 \\\\ \\Rightarrow a = -8 $ <br\/><br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-8$ <\/span><\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:22:50"}]}