{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $3\\sqrt{2x}-5\\sqrt{8x}+7\\sqrt{18x}=28$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${ _input_}","hint":"\u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c.<br\/><b>B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng: V\u1edbi $B\\ge 0$ ta c\u00f3 $\\sqrt{{{A}^{2}}B}=\\left| A \\right|\\sqrt{B}\\,$$=\\left\\{ \\begin{align} & \\begin{matrix} \\,\\,\\,A\\sqrt{B} & \\text{n\u1ebfu}\\,\\,\\, A\\ge0 \\\\\\end{matrix} \\\\ & \\begin{matrix} -A\\sqrt{B} & \\text{n\u1ebfu} \\,\\,\\,A < 0 \\\\\\end{matrix} \\\\ \\end{align} \\right.$ <br\/><b>B\u01b0\u1edbc 3:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ge 0$ <br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,3\\sqrt{2x}-5\\sqrt{8x}+7\\sqrt{18x}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=28\\\\ & \\Leftrightarrow 3\\sqrt{2x}-5\\sqrt{4.2x}+7\\sqrt{9.2x}=28 \\\\ & \\Leftrightarrow 3\\sqrt{2x}-5.2\\sqrt{2x}+7.3\\sqrt{2x}=28 \\\\ & \\Leftrightarrow 3\\sqrt{2x}-10\\sqrt{2x}+21\\sqrt{2x}\\,\\,\\,=28 \\\\ & \\Leftrightarrow 14\\sqrt{2x}\\,=28 \\\\ & \\Leftrightarrow \\sqrt{2x}=2 \\\\ & \\Leftrightarrow 2x=4 \\\\&\\Leftrightarrow x= 2 \\,\\, \\text{(th\u1ecfa m\u00e3n)}\\\\\\end{align}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 2 \\right\\}$ <br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2$ <\/span>"}]}],"id_ques":561},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["4"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"T\u00ednh: $\\dfrac{1}{\\sqrt{5}-2}-\\dfrac{1}{\\sqrt{5}+2}=$ _input_ ","hint":"\u00c1p d\u1ee5ng quy t\u1eafc: Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>\u00c1p d\u1ee5ng quy t\u1eafc: Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu:<br\/> V\u1edbi c\u00e1c bi\u1ec3u th\u1ee9c $A, B, C$ m\u00e0 $A\\ge 0$ v\u00e0 $A\\ne {{B}^{2}}$ ta c\u00f3:$\\dfrac{C}{\\sqrt{A}\\pm B}=\\dfrac{C\\left( \\sqrt{A}\\mp B \\right)}{A-{{B}^{2}}}$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/> $\\begin{align} & \\,\\,\\,\\,\\dfrac{1}{\\sqrt{5}-2}-\\dfrac{1}{\\sqrt{5}+2} \\\\ & =\\dfrac{\\sqrt{5}+2}{(\\sqrt{{5}})^2-{{2}^{2}}}-\\dfrac{\\sqrt{5}-2}{(\\sqrt{{5}})^2-{{2}^{2}}} \\\\ & =\\sqrt{5}+2-\\sqrt{5}+2 \\\\ & =4 \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $4$<\/span><\/span>"}]}],"id_ques":562},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{1}{3}$","B. $\\dfrac{2}{3}$","C. $\\dfrac{5}{3}$"],"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh:$5\\sqrt{12x}-4\\sqrt{3x}+2\\sqrt{48x}=14$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${?}","hint":"\u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c.<br\/><b>B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng: V\u1edbi $B\\ge 0$ ta c\u00f3 $\\sqrt{{{A}^{2}}B}=\\left| A \\right|\\sqrt{B}\\,$$=\\left\\{ \\begin{align} & \\begin{matrix} \\,\\,\\,A\\sqrt{B} & \\text{n\u1ebfu}\\,\\,\\, A\\ge0 \\\\\\end{matrix} \\\\ & \\begin{matrix} -A\\sqrt{B} & \\text{n\u1ebfu} \\,\\,\\,A < 0 \\\\\\end{matrix} \\\\ \\end{align} \\right.$ <br\/><b>B\u01b0\u1edbc 3:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ge 0$ <br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,5\\sqrt{12x}-4\\sqrt{3x}+2\\sqrt{48x}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=14 \\\\ & \\Leftrightarrow 5\\sqrt{4.3x}-4\\sqrt{3x}+2\\sqrt{16.3x}=14 \\\\ & \\Leftrightarrow 5.2\\sqrt{3x}-4\\sqrt{3x}+2.4.\\sqrt{3x}=14 \\\\ & \\Leftrightarrow 10\\sqrt{3x}-4\\sqrt{3x}+8\\sqrt{3x}\\,\\,\\,\\,\\,\\,\\,=14 \\\\ & \\Leftrightarrow 14\\sqrt{3x}\\,=14 \\\\ & \\Leftrightarrow \\sqrt{3x}=1 \\\\ & \\Leftrightarrow x=\\dfrac{1}{3} \\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ \\end{align}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ \\dfrac{1}{3} \\right\\}$"}]}],"id_ques":563},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $2\\sqrt{7}$","B. $3\\sqrt{7}$","C. $\\sqrt{7}$"],"ques":"T\u00ednh $\\dfrac{1}{\\sqrt{7}-\\sqrt{5}}+\\dfrac{1}{\\sqrt{7}+\\sqrt{5}}=$ ?","hint":"\u00c1p d\u1ee5ng quy t\u1eafc: Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>\u00c1p d\u1ee5ng quy t\u1eafc: Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu: <br\/>V\u1edbi c\u00e1c bi\u1ec3u th\u1ee9c $A, B, C$ m\u00e0 $A\\ge 0,\\,\\,B\\ge 0$ v\u00e0 $A\\ne B$ ta c\u00f3: $\\dfrac{C}{\\sqrt{A}\\pm \\sqrt{B}}=\\dfrac{C\\left( \\sqrt{A}\\mp \\sqrt{B} \\right)}{A-B}$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\dfrac{1}{\\sqrt{7}-\\sqrt{5}}+\\dfrac{1}{\\sqrt{7}+\\sqrt{5}} \\\\ & =\\dfrac{\\sqrt{7}+\\sqrt{5}}{(\\sqrt{7})^{2}-(\\sqrt{5})^{2}}+\\dfrac{\\sqrt{7}-\\sqrt{5}}{(\\sqrt{7})^{2}-(\\sqrt{5})^{2}} \\\\ & =\\dfrac{\\sqrt{7}+\\sqrt{5}+\\sqrt{7}-\\sqrt{5}}{2} \\\\ & =\\sqrt{7} \\\\ \\end{align}$<\/span> "}]}],"id_ques":564},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu c\u1ee7a ph\u00e2n th\u1ee9c $\\dfrac{2}{\\sqrt{3}+1}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:","select":["A. $2\\sqrt{3}-1$","B. $\\sqrt{3}-2$","C. $\\sqrt{3}-1$ ","D. $\\sqrt{3}+1$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>V\u1edbi c\u00e1c bi\u1ec3u th\u1ee9c $A, B, C$ m\u00e0 $A\\ge 0$ v\u00e0 $A\\ne {{B}^{2}}$ ta c\u00f3:<br\/>$\\dfrac{C}{\\sqrt{A}\\pm B}=\\dfrac{C\\left( \\sqrt{A}\\mp B \\right)}{A-{{B}^{2}}}$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/>$\\dfrac{2}{\\sqrt{3}+1}=\\dfrac{2.(\\sqrt{3}-1)}{(\\sqrt{3}+1)(\\sqrt{3}-1)}\\,$$=\\dfrac{2(\\sqrt{3}-1)}{2}=\\sqrt{3}-1$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span> <\/span>","column":2}]}],"id_ques":565},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu c\u1ee7a ph\u00e2n th\u1ee9c $\\dfrac{5}{3\\sqrt{2}}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:","select":["A. $\\dfrac{10\\sqrt{2}}{3}$","B. $\\dfrac{5\\sqrt{2}}{3}$","C. $\\dfrac{5\\sqrt{2}}{12}$ ","D. $\\dfrac{5\\sqrt{2}}{6}$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>V\u1edbi c\u00e1c bi\u1ec3u th\u1ee9c $A, B$ m\u00e0 $B > 0$. ta c\u00f3 $\\dfrac{A}{\\sqrt{B}}=\\dfrac{A\\sqrt{B}}{B}$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\dfrac{5}{3\\sqrt{2}}=\\dfrac{5.\\sqrt{2}}{3\\sqrt{2}.\\sqrt{2}}=\\dfrac{5\\sqrt{2}}{6}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span> <\/span>","column":2}]}],"id_ques":566},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Ph\u01b0\u01a1ng tr\u00ecnh $\\sqrt{9x} - \\sqrt{4x}=3$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0:","select":["A. $\\dfrac{3}{5}$","B. $\\dfrac{9}{5}$","C. $9$ ","D. $3$"],"hint":"\u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 bi\u1ec3u th\u1ee9c trong c\u0103n c\u00f3 ngh\u0129a. <br\/><b>B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n:<br\/> V\u1edbi $B\\ge 0$, ta c\u00f3 $\\sqrt{{{A}^{2}}B}=\\left| A \\right|\\sqrt{B}\\,$$=\\left\\{ \\begin{align} & \\begin{matrix} \\,\\,\\,A\\sqrt{B} & \\text {n\u1ebfu}\\,\\,\\, A\\ge0 \\\\\\end{matrix} \\\\ & \\begin{matrix} -A\\sqrt{B} & \\text {n\u1ebfu} \\,\\,\\,A < 0 \\\\\\end{matrix} \\\\ \\end{align} \\right.$<br\/><b>B\u01b0\u1edbc 3:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh $x\\ge 0$<br\/>$\\sqrt{9x}-\\sqrt{4x}=3$<br\/> $\\begin{align} & \\Leftrightarrow 3\\sqrt{x}-2\\sqrt{x}=3 \\\\ & \\Leftrightarrow \\sqrt{x}=3 \\\\ & \\Leftrightarrow x=9 \\,\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ \\end{align}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=9$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":567},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch $x + 3\\sqrt{x}$ th\u00e0nh nh\u00e2n t\u1eed l\u00e0:","select":["A. $\\sqrt{x}(\\sqrt{x}+3)$","B. $-\\sqrt{x}(\\sqrt{x}+3)$","C. $\\sqrt{x}(\\sqrt{x}-3)$ ","D. $-\\sqrt{x}(\\sqrt{x}-3)$"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$x + 3\\sqrt{x}=(\\sqrt{x})^2+3\\sqrt{x}\\,$$=\\sqrt{x}(\\sqrt{x}+3)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A <\/span>","column":2}]}],"id_ques":568},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai4/lv1/img\/5.jpg' \/><\/center> Cho $a=6\\sqrt{2}$ v\u00e0 $b=2\\sqrt{18}$. So s\u00e1nh $a$ v\u00e0 $b$. ","select":["A. $a > b$","B. $a < b$","C. $a = b$ "],"hint":"\u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 v\u00e0o trong d\u1ea5u c\u0103n ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b>B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng: V\u1edbi $A\\ge 0;\\,\\,B\\ge 0$ th\u00ec $A\\sqrt{B}=\\sqrt{{{A}^{2}}B}$ <br\/><b>B\u01b0\u1edbc 2:<\/b> So s\u00e1nh $a$ v\u00e0 $b$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/> $a=6\\sqrt{2}=\\sqrt{6^2.2}=\\sqrt {72}$;<br\/> $b=2\\sqrt{18}=\\sqrt{2^2.18}=\\sqrt {72}$<br\/> V\u1eady $6\\sqrt{2} =2\\sqrt{18} \\Rightarrow a = b $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":3}]}],"id_ques":569},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai4/lv1/img\/5.jpg' \/><\/center> Cho $m=3\\sqrt{10}$ v\u00e0 $n=2\\sqrt{15}$. So s\u00e1nh $m$ v\u00e0 $n$. ","select":["A. $m > n$","B. $m < n$","C. $m=n$ "],"hint":"\u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 v\u00e0o trong d\u1ea5u c\u0103n ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b>B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng: V\u1edbi $A\\ge 0;\\,\\,B\\ge 0$ th\u00ec $A\\sqrt{B}=\\sqrt{{{A}^{2}}B}$ <br\/><b>B\u01b0\u1edbc 2:<\/b> So s\u00e1nh $m$ v\u00e0 $n$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/> $m=3\\sqrt{10}=\\sqrt{3^2.10}=\\sqrt {90}$;<br\/> $n=2\\sqrt{15}=\\sqrt{2^2.15}=\\sqrt {60}$<br\/>Do $\\sqrt {90} > \\sqrt {60} \\Rightarrow 3\\sqrt{10} > 2\\sqrt{15} \\Rightarrow m > n$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A <\/span>","column":3}]}],"id_ques":570},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai4/lv1/img\/6.jpg' \/><\/center> $\\sqrt{a^2b}=-a\\sqrt{b}$ khi $a \\ge 0,\\, b \\ge 0$ ","select":["A. \u0110\u00fang","B. Sai "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $ \\sqrt{a^2b}=|a|\\sqrt{b}=a\\sqrt{b}$ (V\u00ec $a \\ge 0$ n\u00ean $|a|=a$)<br\/>Do \u0111\u00f3 kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span>","column":2}]}],"id_ques":571},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai4/lv1/img\/6.jpg' \/><\/center> $\\sqrt{a^6b}=a^3\\sqrt{b}$ khi $a \\ge 0,\\, b \\ge 0$ ","select":["A. \u0110\u00fang","B. Sai "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\sqrt{a^6b}=|a^3|\\sqrt{b}=a^3\\sqrt{b}$ (V\u00ec $a\\ge 0$ n\u00ean $a^3\\ge 0$)<br\/>Do \u0111\u00f3 kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang. <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A <\/span>","column":2}]}],"id_ques":572},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u $> , < , =$ v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["<"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai4/lv1/img\/5.jpg' \/><\/center><br\/>$4\\sqrt{7}$ _input_$3\\sqrt{13}$ ","hint":"\u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 v\u00e0o trong d\u1ea5u c\u0103n","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> V\u1edbi $A\\ge 0;\\,\\,B\\ge 0$ th\u00ec $A\\sqrt{B}=\\sqrt{{{A}^{2}}B}$ <br\/><b>B\u01b0\u1edbc 2:<\/b> So s\u00e1nh <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/>$4\\sqrt{7}=\\sqrt{4^2.7}=\\sqrt{112}$;<br\/> $3\\sqrt{13}=\\sqrt{3^2.13}=\\sqrt{117}$ <br\/>V\u00ec $\\sqrt {112} < \\sqrt {117}$ n\u00ean $4\\sqrt{7} < 3\\sqrt{13}$ <br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $<$ <\/span> <\/span><\/span>"}]}],"id_ques":573},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u $> , < , =$ v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[[">"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai4/lv1/img\/5.jpg' \/><\/center><br\/>$2\\sqrt{7}$ _input_$3\\sqrt{3}$ ","hint":"\u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 v\u00e0o trong d\u1ea5u c\u0103n","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> V\u1edbi $A\\ge 0;\\,\\,B\\ge 0$ th\u00ec $A\\sqrt{B}=\\sqrt{{{A}^{2}}B}$ <br\/><b>B\u01b0\u1edbc 2:<\/b> So s\u00e1nh <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/>$2\\sqrt{7}=\\sqrt{2^2.7}=\\sqrt{28}$; <br\/>$3\\sqrt{3}=\\sqrt{3^2.3}=\\sqrt{27}$ <br\/>V\u00ec $\\sqrt {28} > \\sqrt{27}$ n\u00ean $2\\sqrt{7} > 3\\sqrt{3}$ <br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $>$ <\/span> <\/span>"}]}],"id_ques":574},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $-\\sqrt{3}$","B. $-2\\sqrt{3}$","C. $-5\\sqrt{3}$"],"ques":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c:$\\sqrt{48}-3\\sqrt{12}-\\sqrt{27}=$?","hint":"\u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\sqrt{48}-3\\sqrt{12}-\\sqrt{27} \\\\ & =\\sqrt{16.3}-3\\sqrt{4.3}-\\sqrt{9.3} \\\\ & =4\\sqrt{3}-6\\sqrt{3}-3\\sqrt{3} \\\\ & =-5\\sqrt{3} \\\\ \\end{align}$"}]}],"id_ques":575},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $2\\sqrt{5}$","B. $4\\sqrt{5}$","C. $5\\sqrt{5}$"],"ques":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c:$\\sqrt{45}-2\\sqrt{20}+\\sqrt{125}=$?","hint":"\u00c1p d\u1ee5ng quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n. ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\sqrt{45}-2\\sqrt{20}+\\sqrt{125} \\\\ & =\\sqrt{9.5}-2\\sqrt{4.5}+\\sqrt{25.5} \\\\ & =3\\sqrt{5}-4\\sqrt{5}+5\\sqrt{5} \\\\ & =4\\sqrt{5} \\\\ \\end{align}$"}]}],"id_ques":576},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{20}$","B. $\\sqrt{10}$","C. $2\\sqrt{10}$"],"ques":"\u0110\u01b0a th\u1eeba s\u1ed1 v\u00e0o b\u00ean trong d\u1ea5u c\u0103n<br\/>$2\\sqrt{5}=$?","hint":"\u00c1p d\u1ee5ng: V\u1edbi $A\\ge 0;\\,\\,B\\ge 0$ th\u00ec $A\\sqrt{B}=\\sqrt{{{A}^{2}}B}$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $2\\sqrt{5}=\\sqrt{2^2.5}=\\sqrt{20}$"}]}],"id_ques":577},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\sqrt{9}$","B. $\\sqrt{27}$","C. $\\sqrt{36}$"],"ques":"\u0110\u01b0a th\u1eeba s\u1ed1 v\u00e0o b\u00ean trong d\u1ea5u c\u0103n <br\/>$3\\sqrt{3}=$?","hint":"\u00c1p d\u1ee5ng: V\u1edbi $A\\ge 0;\\,\\,B\\ge 0$ th\u00ec $A\\sqrt{B}=\\sqrt{{{A}^{2}}B}$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $3\\sqrt{3}=\\sqrt{3^2.3}=\\sqrt{27}$"}]}],"id_ques":578},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $4\\sqrt{3}$","B. $3\\sqrt{3}$","C. $\\sqrt{3}$"],"ques":"\u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n <br\/>$\\sqrt{48}=$?","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> V\u1edbi $B\\ge 0$ ta c\u00f3 $\\sqrt{{{A}^{2}}B}=\\left| A \\right|\\sqrt{B}\\,$$=\\left\\{ \\begin{align} & \\begin{matrix} \\,\\,\\,A\\sqrt{B} & \\text{n\u1ebfu}\\,\\,\\, A\\ge0 \\\\\\end{matrix} \\\\ & \\begin{matrix} -A\\sqrt{B} & \\text{n\u1ebfu} \\,\\,\\,A < 0 \\\\\\end{matrix} \\\\ \\end{align} \\right.$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3 $\\sqrt{48}=\\sqrt{16.3}=4\\sqrt{3}$<\/span> "}]}],"id_ques":579},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $6\\sqrt{2}$","B. $5\\sqrt{2}$","C. $7\\sqrt{2}$"],"ques":"\u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n <br\/>$\\sqrt{72}=$?","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span> <br\/>V\u1edbi $B\\ge 0$ ta c\u00f3 $\\sqrt{{{A}^{2}}B}=\\left| A \\right|\\sqrt{B}\\,$$=\\left\\{ \\begin{align} & \\begin{matrix} \\,\\,\\,A\\sqrt{B} & \\text{n\u1ebfu}\\,\\,\\, A\\ge0 \\\\\\end{matrix} \\\\ & \\begin{matrix} -A\\sqrt{B} & \\text{n\u1ebfu} \\,\\,\\,A < 0 \\\\\\end{matrix} \\\\ \\end{align} \\right.$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $\\sqrt{72}=\\sqrt{36.2}=6\\sqrt{2}$<\/span>"}]}],"id_ques":580}],"lesson":{"save":0,"level":1}}