{"segment":[{"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":"S\u1eafp x\u1ebfp c\u00e1c s\u1ed1 sau theo th\u1ee9 t\u1ef1 t\u0103ng d\u1ea7n: $4\\sqrt{2},\\,\\,\\,\\sqrt{37},\\,\\,\\,3\\sqrt{7},\\,\\,\\,\\,2\\sqrt{15}$ ","select":["A. $\\sqrt{37};4\\sqrt{2};3\\sqrt{7};2\\sqrt{15}$","B. $4\\sqrt{2};\\sqrt{37};3\\sqrt{7};2\\sqrt{15}$","C. $4\\sqrt{2};\\sqrt{37};2\\sqrt{15};3\\sqrt{7}$","D. $\\sqrt{37};4\\sqrt{2};2\\sqrt{15};3\\sqrt{7}$ "],"hint":"\u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\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: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 v\u00e0o trong d\u1ea5u c\u0103n<br\/>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 c\u00e1c c\u0103n th\u1ee9c t\u00ecm \u0111\u01b0\u1ee3c. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $4\\sqrt{2}=\\sqrt{32};$<br\/>$3\\sqrt{7}=\\sqrt{63};$<br\/>$2\\sqrt{15}=\\sqrt{60}$ <br\/>V\u00ec $\\sqrt{32}<\\sqrt{37}<\\sqrt{60}<\\sqrt{63}$ n\u00ean $4\\sqrt{2}<\\sqrt{37}<2\\sqrt{15}<3\\sqrt{7}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":581},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["6"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c: $C=\\dfrac{{{a}^{2}}+\\sqrt{a}}{a-\\sqrt{a}+1}$ t\u1ea1i $a =4$<br\/>\u0110\u00e1p s\u1ed1: $C=$_input_ ","hint":"Ph\u00e2n t\u00edch t\u1eed th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed: S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t nh\u00e2n t\u1eed chung v\u00e0 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^3+b^3$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Ph\u00e2n t\u00edch t\u1eed s\u1ed1 th\u00e0nh $a^2+\\sqrt{a}=\\sqrt{a}(\\sqrt{a^3}+1)$<br\/><b>B\u01b0\u1edbc 2:<\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/><b>B\u01b0\u1edbc 3:<\/b> Thay $a = 4$ v\u00e0o bi\u1ec3u th\u1ee9c $C$ \u0111\u00e3 r\u00fat g\u1ecdn. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $a\\ge 0$ <br\/> Ta c\u00f3:<br\/>$\\begin{align} C &=\\dfrac{{{a}^{2}}+\\sqrt{a}}{a-\\sqrt{a}+1} \\\\ & =\\dfrac{(\\sqrt{a})^{4}+\\sqrt{a}}{a-\\sqrt{a}+1} \\\\ & =\\dfrac{\\sqrt{a}\\left[(\\sqrt{a})^{3}+1 \\right]}{a-\\sqrt{a}+1} \\\\ & =\\dfrac{\\sqrt{a}\\left( \\sqrt{a}+1 \\right)\\left( a-\\sqrt{a}+1 \\right)}{a-\\sqrt{a}+1} \\\\ & =\\sqrt{a}\\left( \\sqrt{a}+1 \\right) \\\\ \\end{align}$ <br\/>Thay $a= 4$ v\u00e0o bi\u1ec3u th\u1ee9c $C$ \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $C=\\sqrt{4}\\left( \\sqrt{4}+1 \\right)=2\\left( 2+1 \\right)=6$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $6$<\/span><br\/><b>Nh\u1eadn x\u00e9t:<\/b> <br\/>Ngo\u00e0i c\u00e1ch gi\u1ea3i tr\u00ean, ta c\u00f3 th\u1ec3 th\u1ef1c hi\u1ec7n t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a $C$ t\u1ea1i $a=4$ b\u1eb1ng c\u00e1ch thay tr\u1ef1c ti\u1ebfp $a=4$ v\u00e0o bi\u1ec3u th\u1ee9c $C$ ban \u0111\u1ea7u.<\/span><\/span>"}]}],"id_ques":582},{"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":"T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c: $A=\\dfrac{1}{\\sqrt{x}-1}-\\dfrac{1}{\\sqrt{x}+1}$ t\u1ea1i $x= 2$<br\/>\u0110\u00e1p s\u1ed1: $A=$_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\/><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: 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\/><b>B\u01b0\u1edbc 3:<\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/><b>B\u01b0\u1edbc 4:<\/b> Thay $x = 2$ v\u00e0o bi\u1ec3u th\u1ee9c $A$ \u0111\u00e3 r\u00fat g\u1ecdn r\u1ed3i t\u00ednh. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ge 0; x\\ne 1$ <br\/> Ta c\u00f3:<br\/>$\\begin{align} A& =\\dfrac{1}{\\sqrt{x}-1}-\\dfrac{1}{\\sqrt{x}+1} \\\\ & =\\dfrac{\\sqrt{x}+1-\\left( \\sqrt{x}-1 \\right)}{\\left( \\sqrt{x}-1 \\right)\\left( \\sqrt{x}+1 \\right)} \\\\ & =\\dfrac{\\sqrt{x}+1-\\sqrt{x}+1}{x-1} \\\\ & =\\dfrac{2}{x-1} \\\\ \\end{align}$ <br\/>Thay $x= 2$ v\u00e0o bi\u1ec3u th\u1ee9c $A$ \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $A=\\dfrac{2}{2-1}=2$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$<\/span><\/span><\/span>"}]}],"id_ques":583},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u $> , < , =$ v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","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/lv2/img\/5.jpg' \/><\/center><br\/>$\\dfrac{1}{2}\\sqrt{\\dfrac{5}{7}}$ _input_ $\\dfrac{1}{3}\\sqrt{\\dfrac{3}{5}}$ ","hint":"\u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 v\u00e0o trong d\u1ea5u c\u0103n r\u1ed3i so s\u00e1nh","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b>\u0110\u01b0a c\u00e1c th\u1eeba s\u1ed1 v\u00e0o trong c\u0103n th\u1ee9c b\u1eb1ng c\u00e1c \u00e1p d\u1ee5ng c\u00f4ng th\u1ee9c: V\u1edbi $A\\ge 0,\\,\\,B\\ge 0$, ta c\u00f3: $A\\sqrt B = \\sqrt {A^2B}$<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\/>$\\dfrac{1}{2}\\sqrt{\\dfrac{5}{7}}=\\sqrt{{{\\left( \\dfrac{1}{2} \\right)}^{2}}.\\dfrac{5}{7}}=\\sqrt{\\dfrac{5}{28}}$<br\/>$\\dfrac{1}{3}\\sqrt{\\dfrac{3}{5}}=\\sqrt{{{\\left( \\dfrac{1}{3} \\right)}^{2}}.\\dfrac{3}{5}}=\\sqrt{\\dfrac{1}{15}}$<br\/>V\u00ec $\\dfrac{5}{28}>\\dfrac{5}{75} = \\dfrac{1}{15} \\Rightarrow \\sqrt {\\dfrac{5}{28}}>\\sqrt {\\dfrac{1}{15}} \\Rightarrow \\dfrac{1}{2}\\sqrt{\\dfrac{5}{7}}>\\dfrac{1}{3}\\sqrt{\\dfrac{3}{5}}$ <br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $>$ <\/span><\/span><\/span>"}]}],"id_ques":584},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["12"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"T\u00ednh. $\\dfrac{\\sqrt{7}+\\sqrt{5}}{\\sqrt{7}-\\sqrt{5}}+\\dfrac{\\sqrt{7}-\\sqrt{5}}{\\sqrt{7}+\\sqrt{5}} =$_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\/><b>B\u01b0\u1edbc 1:<\/b> Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu <br\/><b>B\u01b0\u1edbc 2:<\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\dfrac{\\sqrt{7}+\\sqrt{5}}{\\sqrt{7}-\\sqrt{5}}+\\dfrac{\\sqrt{7}-\\sqrt{5}}{\\sqrt{7}+\\sqrt{5}} \\\\ & =\\dfrac{{{\\left( \\sqrt{7}+\\sqrt{5} \\right)}^{2}}+{{\\left( \\sqrt{7}-\\sqrt{5} \\right)}^{2}}}{(\\sqrt{7})^{2}-(\\sqrt{5})^{2}} \\\\ & =\\dfrac{7+2\\sqrt{35}+5+7-2\\sqrt{35}+5}{7-5} \\\\ & =\\dfrac{24}{2}\\\\ &=12 \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $12$ <\/span> <\/span>"}]}],"id_ques":585},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{22}{5}$","B. $\\dfrac{18}{5}$","C. $\\dfrac{16}{5}$"],"ques":"T\u00ednh $\\dfrac{\\sqrt{8}+\\sqrt{3}}{\\sqrt{8}-\\sqrt{3}}+\\dfrac{\\sqrt{8}-\\sqrt{3}}{\\sqrt{8}+\\sqrt{3}} =$?","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\/><b>B\u01b0\u1edbc 1:<\/b> Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu<br\/><b>B\u01b0\u1edbc 2:<\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\dfrac{\\sqrt{8}+\\sqrt{3}}{\\sqrt{8}-\\sqrt{3}}+\\dfrac{\\sqrt{8}-\\sqrt{3}}{\\sqrt{8}+\\sqrt{3}} \\\\ & =\\dfrac{{{\\left( \\sqrt{8}+\\sqrt{3} \\right)}^{2}}+{{\\left( \\sqrt{8}-\\sqrt{3} \\right)}^{2}}}{(\\sqrt{8})^{2}-(\\sqrt{3})^{2}} \\\\ & =\\dfrac{8+2\\sqrt{24}+3+8-2\\sqrt{24}+3}{8-3} \\\\ & =\\dfrac{22}{5} \\\\ \\end{align}$<\/span> "}]}],"id_ques":586},{"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":"K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch bi\u1ec3u th\u1ee9c $x+\\sqrt{x}-12$ th\u00e0nh nh\u00e2n t\u1eed l\u00e0: ","select":["A. $\\left( 4-\\sqrt{x} \\right)\\left( \\sqrt{x}+3 \\right)$","B. $\\left( \\sqrt{x}-4 \\right)\\left( \\sqrt{x}+3 \\right)$","C. $\\left( \\sqrt{x}+4 \\right)\\left( 3-\\sqrt{x} \\right)$","D. $\\left( \\sqrt{x}+4 \\right)\\left( \\sqrt{x} -3\\right)$ "],"hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch. <br\/>Ghi nh\u1edb: V\u1edbi $x\\ge0$, $x=(\\sqrt x)^2=\\sqrt x.\\sqrt x$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & x+\\sqrt{x}-12 \\\\ & =x-3\\sqrt{x}+4\\sqrt{x}-12 \\\\ & =\\sqrt{x}\\left( \\sqrt{x}-3 \\right)+4\\left( \\sqrt{x}-3 \\right) \\\\ & =\\left( \\sqrt{x}+4 \\right)\\left( \\sqrt{x}-3 \\right) \\\\ \\end{align}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span>","column":2}]}],"id_ques":587},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch bi\u1ec3u th\u1ee9c $x+\\sqrt{x}-2$ th\u00e0nh nh\u00e2n t\u1eed l\u00e0: ","select":["A. $\\left( 1-\\sqrt{x} \\right)\\left( \\sqrt{x}+2 \\right)$","B. $\\left( \\sqrt{x}-1 \\right)\\left( \\sqrt{x}+2 \\right)$","C. $\\left( \\sqrt{x}+1 \\right)\\left( \\sqrt{x}-2 \\right)$","D. $\\left( \\sqrt{x}+1 \\right)\\left(2- \\sqrt{x} \\right)$ "],"hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch h\u1ea1ng t\u1eed.<br\/>Ghi nh\u1edb: V\u1edbi $x\\ge0$, $x=(\\sqrt x)^2=\\sqrt x.\\sqrt x$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & x+\\sqrt{x}-2 \\\\ & =x+2\\sqrt{x}-\\sqrt{x}-2 \\\\ & =\\sqrt{x}\\left( \\sqrt{x}+2 \\right)-\\left( \\sqrt{x}+2 \\right) \\\\ & =\\left( \\sqrt{x}-1 \\right)\\left( \\sqrt{x}+2 \\right) \\\\ \\end{align}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span>","column":2}]}],"id_ques":588},{"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":"Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu c\u1ee7a ph\u00e2n th\u1ee9c $\\dfrac{12}{\\sqrt{3}+\\sqrt{5}}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: ","select":["A. $6\\left( \\sqrt{5}-\\sqrt{3} \\right)$","B. $6\\left( \\sqrt{5}+\\sqrt{3} \\right)$","C. $6\\sqrt{5}+\\sqrt{3}$","D. $6\\sqrt{3}-6\\sqrt{5}$ "],"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 $A\\ge 0,\\,\\,B\\ge 0$ 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\/>$\\dfrac{12}{\\sqrt{3}+\\sqrt{5}}=\\dfrac{12\\left( \\sqrt{5}-\\sqrt{3} \\right)}{(\\sqrt{5})^{2}-(\\sqrt{3})^{2}}\\,$$=\\dfrac{12\\left( \\sqrt{5}-\\sqrt{3} \\right)}{2}=6\\left( \\sqrt{5}-\\sqrt{3} \\right)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A <\/span><\/span>","column":2}]}],"id_ques":589},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu c\u1ee7a ph\u00e2n th\u1ee9c $\\sqrt{\\dfrac{{{\\left( \\sqrt{3}-\\sqrt{10} \\right)}^{2}}}{3}}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: ","select":["A. $\\dfrac{\\sqrt{3}-\\sqrt{10}}{3}$","B. $\\dfrac{\\sqrt{10}-\\sqrt{3}}{3}$","C. $\\dfrac{\\sqrt{30}-3}{3}$","D. $\\dfrac{3-\\sqrt{30}}{3}$ "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b>B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng: Quy t\u1eafc tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu: <br\/>V\u1edbi c\u00e1c bi\u1ec3u th\u1ee9c $A, B$ m\u00e0 $A.B \\ge 0$ v\u00e0 $B\\ne 0$, ta c\u00f3: $\\sqrt{\\dfrac{A}{B}}=\\dfrac{\\sqrt{AB}}{\\left| B \\right|}$<br\/><b>B\u01b0\u1edbc 2:<\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\sqrt{\\dfrac{{{\\left( \\sqrt{3}-\\sqrt{10} \\right)}^{2}}}{3}}=\\dfrac{\\left| \\sqrt{3}-\\sqrt{10} \\right|}{\\sqrt{3}}\\,$$=\\dfrac{\\sqrt{10}-\\sqrt{3}}{\\sqrt{3}}=\\dfrac{\\sqrt{30}-3}{3}$ (V\u00ec $\\sqrt{10}>\\sqrt{3}$)<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span> <\/span>","column":2}]}],"id_ques":590},{"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/lv2/img\/4.jpg' \/><\/center>V\u1edbi $x < 0$ th\u00ec $x\\sqrt{\\dfrac{-7}{2x}}=\\dfrac{-\\sqrt{-14x}}{2}$ ","select":["A. \u0110\u00fang","B. Sai "],"hint":"\u00c1p d\u1ee5ng: Quy t\u1eafc tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c v\u1ebf tr\u00e1i. ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$x\\sqrt{\\dfrac{-7}{2x}}=x.\\sqrt{\\dfrac{-14x}{4{{x}^{2}}}}\\,$$=\\dfrac{x\\sqrt{-14x}}{\\left| 2x \\right|}=\\dfrac{x\\sqrt{-14x}}{-2x}\\,$$=-\\dfrac{\\sqrt{-14x}}{2}$ (v\u00ec $x < 0$ n\u00ean $|x|=-x$) <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":591},{"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/lv2/img\/2.jpg' \/><\/center>V\u1edbi $x > 0$ th\u00ec $x\\sqrt{\\dfrac{5}{x}}=-\\sqrt{5x}$ ","select":["A. \u0110\u00fang","B. Sai "],"hint":"\u00c1p d\u1ee5ng: Quy t\u1eafc tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c v\u1ebf tr\u00e1i. ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$x\\sqrt{\\dfrac{5}{x}}=x.\\sqrt{\\dfrac{5x}{{{x}^{2}}}}=\\dfrac{x\\sqrt{5x}}{\\left| x \\right|}\\,$$=\\dfrac{x\\sqrt{5x}}{x}=\\sqrt{5x}$ (v\u00ec $x > 0$ n\u00ean $|x|=x$) <br\/>Do \u0111\u00f3 kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span>","column":2}]}],"id_ques":592},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["65"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\sqrt{36x-36}-\\sqrt{9x-9}-\\sqrt{4x-4}=16-\\sqrt{x-1}$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $S=${_input_}","hint":"\u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 bi\u1ec3u th\u1ee9c trong c\u0103n c\u00f3 ngh\u0129a<br\/><b>B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n<br\/> V\u1edbi $B\\ge 0$ ta c\u00f3 $\\sqrt{{{A}^{2}}B}=\\left| A \\right|\\sqrt{B}\\,$$=\\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: $x-1\\ge 0\\Leftrightarrow x\\ge 1$<br\/>Ta c\u00f3:<br\/> $ \\sqrt{36x-36}-\\sqrt{9x-9}-\\sqrt{4x-4}$$\\,=16-\\sqrt{x-1}$<br\/>$ \\Leftrightarrow\\sqrt{36(x-1)}-\\sqrt{9(x-1)}-\\sqrt{4(x-1)}$$\\,=16-\\sqrt{x-1}$<br\/>$ \\Leftrightarrow 6\\sqrt{x-1}-3\\sqrt{x-1}-2\\sqrt{x-1}\\,$$=16-\\sqrt{x-1} $<br\/>$ \\Leftrightarrow 2\\sqrt{x-1}=16 $<br\/>$ \\Leftrightarrow \\sqrt{x-1}=8$<br\/>$ \\Leftrightarrow x-1=64 $<br\/>$\\Leftrightarrow x=65 $ (th\u1ecfa m\u00e3n)<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{65\\}$<br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $65$ <\/span><\/span>"}]}],"id_ques":593},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. {$1;\\dfrac{15}{9}$}","B. {$1;\\dfrac{16}{9}$}","C. {$2;\\dfrac{17}{9}$}"],"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $3x-7\\sqrt{x}+4=0$ <br\/> T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $S=${?;?}","hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><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> Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch t\u00e1ch $-7\\sqrt{x} = -3\\sqrt{x} -4\\sqrt{x}$<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{aligned} & \\,\\,\\,\\,\\,\\,\\,\\,\\,3x-7\\sqrt{x}+4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=0 \\\\ & \\Leftrightarrow 3x-3\\sqrt{x}-4\\sqrt{x}+4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=0 \\\\ & \\Leftrightarrow 3\\sqrt{x}\\left( \\sqrt{x}-1 \\right)-4\\left( \\sqrt{x}-1 \\right)=0 \\\\ & \\Leftrightarrow \\left( \\sqrt{x}-1 \\right)\\left( 3\\sqrt{x}-4 \\right)=0 \\\\ \\end{aligned}$<br\/>$ \\Leftrightarrow \\left[ \\begin{aligned} & \\sqrt{x}-1=0 \\\\ & 3\\sqrt{x}-4=0 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left[ \\begin{aligned} & \\sqrt{x}=1 \\\\ & \\sqrt{x}=\\dfrac{4}{3} \\\\ \\end{aligned} \\right.\\,$$\\,\\,\\,\\Leftrightarrow \\left[ \\begin{aligned} & x=1\\,\\,\\,\\text{(th\u1ecfa m\u00e3n)} \\\\ & x=\\dfrac{16}{9} \\,\\,\\,\\text{(th\u1ecfa m\u00e3n)} \\\\ \\end{aligned} \\right.$ <br\/> T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $S=\\left\\{ 1;\\dfrac{16}{9} \\right\\}$<\/span> "}]}],"id_ques":594},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. {$0;\\dfrac{9}{4}$}","B. {$1;\\dfrac{9}{4}$}","C. {$2;\\dfrac{9}{4}$}"],"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $2x-3\\sqrt{x}=0$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $S=${?;?}","hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t nh\u00e2n t\u1eed chung. <br\/> Ghi nh\u1edb: V\u1edbi $x>0$, $x=(\\sqrt x)^2=\\sqrt x \\sqrt x$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><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> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t nh\u00e2n t\u1eed chung<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\/>$2x-3\\sqrt{x}\\,\\,=0\\,$<br\/>$\\Leftrightarrow \\sqrt{x}\\left( 2\\sqrt{x}-3 \\right)=0 $<br\/>$\\Leftrightarrow \\left[ \\begin{aligned} & \\sqrt{x}=0 \\\\ & 2\\sqrt{x}-3=0 \\\\ \\end{aligned} \\right.$<br\/>$\\,\\Leftrightarrow \\left[ \\begin{aligned} & \\sqrt{x}=0 \\\\ & \\sqrt{x}=\\dfrac{3}{2} \\\\ \\end{aligned} \\right.$<br\/>$\\,\\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\,\\,\\,\\text {(th\u1ecfa m\u00e3n)}\\\\ & x=\\dfrac{9}{4} \\,\\,\\,\\text {(th\u1ecfa m\u00e3n)}\\\\ \\end{aligned} \\right. $<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 0;\\dfrac{9}{4} \\right\\}$<\/span> "}]}],"id_ques":595},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{a}$","B. $\\sqrt{b}$","C. $\\sqrt{ab}$"],"ques":"V\u1edbi $a > 0,\\,\\,b> 0$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c:<br\/> $\\dfrac{a+\\sqrt{ab}}{\\sqrt{a}+\\sqrt{b}}=$?","hint":"Ph\u00e2n t\u00edch t\u1eed th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'>V\u1edbi $a > 0,\\,\\,b> 0$. Ta c\u00f3:<br\/>$\\dfrac{a+\\sqrt{ab}}{\\sqrt{a}+\\sqrt{b}} =\\dfrac{\\sqrt{a}\\left( \\sqrt{a}+\\sqrt{b} \\right)} {\\sqrt{a}+\\sqrt{b}}=\\sqrt{a}$<br\/> <i>Ghi nh\u1edb:<\/i> V\u1edbi $a\\ge 0$, $a=(\\sqrt a)^2=\\sqrt a \\sqrt a$<\/span>"}]}],"id_ques":596},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $-\\sqrt{a}$","B. $-2\\sqrt{a}$","C. $-3\\sqrt{a}$"],"ques":"V\u1edbi $0\\le a < 5$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau<br\/>$\\left( a-5 \\right).\\sqrt{\\dfrac{4a}{{{a}^{2}}-10a+25}} =$?","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'>V\u1edbi $0\\le a < 5$. Ta c\u00f3: <br\/>$\\begin{align} & \\left( a-5 \\right).\\sqrt{\\dfrac{4a}{{{a}^{2}}-10a+25}} \\\\ & =\\left( a-5 \\right).\\sqrt{\\dfrac{4a}{(a-5)^2}} \\\\ & =\\left( a-5 \\right).\\dfrac{\\sqrt{4a}}{\\sqrt{{{\\left( a-5 \\right)}^{2}}}} \\,\\,(\\text {V\u00ec } a\\ge 0)\\\\ & =\\left( a-5 \\right).\\dfrac{2\\sqrt{a}}{\\left| a-5 \\right|} \\\\ & =\\left( a-5 \\right).\\dfrac{2\\sqrt{a}}{5-a}\\,\\,\\,\\,(V\u00ec\\,\\,0\\le a < 5) \\\\ &=-2\\sqrt{a} \\\\ \\end{align}$<br\/><i>Ghi nh\u1edb<\/i> <br\/>- V\u1edbi $A$ l\u00e0 m\u1ed9t bi\u1ec3u th\u1ee9c, ta c\u00f3 $\\sqrt{A^2}=|A|$$\\,=\\left\\{ \\begin{align} & A\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,{A}\\ge {0} \\\\ & -A\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,{A}< {0} \\\\\\end{align} \\right.$<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.$<\/span>"}]}],"id_ques":597},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $-\\sqrt{2a}$","B. $-2\\sqrt{a}$","C. $-\\sqrt{a}$"],"ques":"V\u1edbi $a > 2$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau:<br\/>$\\left( 2-a \\right)\\sqrt{\\dfrac{2a}{{{\\left( a-2 \\right)}^{2}}}}=$?","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'>V\u1edbi $a > 2$. Ta c\u00f3:<br\/>$\\left( 2-a \\right)\\sqrt{\\dfrac{2a}{{{\\left( a-2 \\right)}^{2}}}} \\,$<br\/>$=\\left( 2-a \\right).\\dfrac{\\sqrt{2a}}{\\left| a-2 \\right|}\\,$ (V\u00ec $a>2$ n\u00ean $a>0$)<br\/>$=\\left( 2-a \\right).\\dfrac{\\sqrt{2a}}{a-2}=-\\sqrt{2a}$ (V\u00ec $a> 2$ n\u00ean $|a-2|=a-2$)"}]}],"id_ques":598},{"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{6}$","B. $\\sqrt{6}$","C. $-\\sqrt{6}$"],"ques":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau<br\/>$\\sqrt{\\dfrac{2}{3}}-\\sqrt{24}+2\\sqrt{\\dfrac{3}{8}}+\\sqrt{\\dfrac{1}{6}} =$?","hint":"\u00c1p d\u1ee5ng: Quy t\u1eafc tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu v\u00e0 \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{\\dfrac{2}{3}}-\\sqrt{24}+2\\sqrt{\\dfrac{3}{8}}+\\sqrt{\\dfrac{1}{6}} \\\\ & =\\sqrt{\\dfrac{6}{{{3}^{2}}}}-2\\sqrt{6}+2\\sqrt{\\dfrac{6}{{{4}^{2}}}}+\\sqrt{\\dfrac{6}{{{6}^{2}}}} \\\\ & =\\dfrac{1}{3}\\sqrt{6}-2\\sqrt{6}+\\dfrac{1}{2}\\sqrt{6}+\\dfrac{1}{6}\\sqrt{6} \\\\ & =-\\sqrt{6} \\\\ \\end{align}$<br\/><i>Ghi nh\u1edb:<\/i> <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\/> - V\u1edbi c\u00e1c bi\u1ec3u th\u1ee9c $A, B$ m\u00e0 $B > 0$ ta c\u00f3 $\\dfrac{A}{\\sqrt{B}}=\\dfrac{A\\sqrt{B}}{B}$<\/span>"}]}],"id_ques":599},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $\\sqrt{5}$","B. $5\\sqrt{5}$","C. $7\\sqrt{5}$"],"ques":"T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c <br\/>$\\sqrt{45}-2\\sqrt{80}+\\sqrt{14,4.50} =$?","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\/>$\\begin{align} & \\,\\,\\,\\sqrt{45}-2\\sqrt{80}+\\sqrt{14,4.50} \\\\ & =\\sqrt{9.5}-2\\sqrt{16.5}+\\sqrt{144.5} \\\\ & =3\\sqrt{5}-8\\sqrt{5}+12\\sqrt{5} \\\\ & =7\\sqrt{5} \\\\ \\end{align}$"}]}],"id_ques":600}],"lesson":{"save":0,"level":2}}