{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4"],["1"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Tr\u00ean tia Oy l\u1ea5y hai \u0111i\u1ec3m A, B sao cho OA = 2cm, OB = 6cm. G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OB. T\u00ednh AB, AM. <br\/> \u0110\u00e1p \u00e1n l\u00e0: AB = _input_ cm; AM = _input_ cm","hint":"M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a OB n\u00ean MO = MB = $\\dfrac{OB}{2}$","explain":" <span class='basic_left'><span class='basic_green'>T\u00f3m t\u1eaft:<\/span><br\/> Tr\u00ean tia Oy <br\/> OA = 2cm <br\/> OB = 6cm <br\/> M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a OB <br\/> T\u00ednh AM, AB <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i: <\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/bai6/lv3/img\/H616_K101.png' \/><\/center> <br\/> V\u00ec M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OB n\u00ean: <br\/> MB = OM = $\\dfrac{OB}{2} = \\dfrac{6}{2} = 3$cm <br\/> Tr\u00ean tia Oy c\u00f3 OA = 2cm, OB = 6cm <br\/> V\u00ec 2cm < 6cm n\u00ean OA < OB. Do \u0111\u00f3 \u0111i\u1ec3m A n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m O v\u00e0 B <br\/> $\\begin{align*} OA + AB &= AB \\\\ 2cm + AB &= 6cm \\\\ AB &= 6cm - 2cm \\\\ AB &= 4cm \\end{align*}$ <br\/> Tr\u00ean tia Oy c\u00f3 OA = 2cm, OM = 3cm <br\/> V\u00ec 2cm < 3cm n\u00ean OA < OM <br\/> Do \u0111\u00f3 \u0111i\u1ec3m A n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m O v\u00e0 M <br\/> $\\begin{align*} OA + AM &= OM \\\\ 2cm + AM &= 3cm \\\\ AM &= 3cm - 2cm \\\\ AM &= 1 cm \\end{align*}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: AB = 4cm, AM = 1cm<\/span> <br\/> <span class='basic_green'> <i> L\u01b0u \u00fd: V\u1edbi nh\u1eefng b\u00e0i t\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng n\u1ebfu b\u00e0i ch\u01b0a cho \u0111i\u1ec3m n\u00e0o n\u1eb1m gi\u1eefa th\u00ec tr\u01b0\u1edbc ti\u00ean ta ph\u1ea3i ch\u1ec9 ra \u0111i\u1ec3m n\u1eb1m gi\u1eefa r\u1ed3i m\u1edbi \u00e1p d\u1ee5ng t\u00ednh ch\u1ea5t c\u1ed9ng \u0111\u1ed9 d\u00e0i \u0111\u1ec3 t\u00ednh <\/b> <\/span> "}]}],"id_ques":2031},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Tr\u00ean tia Ot l\u1ea5y hai \u0111i\u1ec3m A, B sao cho OA = 3cm, OB = 7cm. G\u1ecdi I, K l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OA, OB. T\u00ednh IK. <br\/> \u0110\u00e1p \u00e1n l\u00e0: IK = _input_ cm","hint":"T\u00ednh OI, OK r\u1ed3i t\u00ednh IK","explain":" <span class='basic_left'><span class='basic_green'>T\u00f3m t\u1eaft:<\/span><br\/> Tr\u00ean tia Ot <br\/> OA = 3cm <br\/> OB = 7cm <br\/> I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a OA <br\/> K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a OB <br\/> T\u00ednh IK = ? <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i: <\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/bai6/lv3/img\/H616_K102.png' \/><\/center> <br\/> V\u00ec I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a OA n\u00ean OI = IA = $\\dfrac{OA}{2} = \\dfrac{3}{2} = 1,5cm$ <br\/> K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a OB n\u00ean OK = KB = $\\dfrac{OB}{2} = \\dfrac{7}{2} = 3,5cm$ <br\/> Tr\u00ean tia Ot c\u00f3: OI = 1,5cm, OK = 3,5cm <br\/> V\u00ec 1,5cm < 3,5cm n\u00ean OI < OK <br\/> Do \u0111\u00f3 \u0111i\u1ec3m I n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m O v\u00e0 K <br\/> $\\begin{align*} OI + IK &= OK \\\\ 1,5cm + IK &= 3,5cm \\\\ IK &= 3,5cm - 1,5cm \\\\ IK &= 2cm \\end{align*}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: IK = 2cm <\/span> "}]}],"id_ques":2032},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho \u0111o\u1ea1n th\u1eb3ng AB = 4cm, I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia AI l\u1ea5y \u0111i\u1ec3m C sao cho AC = 1cm. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia BI l\u1ea5y \u0111i\u1ec3m D sao cho BD = 1cm. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang.","select":["A. I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a CD","B. D l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BI","C. C l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AI","D. B l\u00e0 trung \u0111i\u1ec3m c\u1ee7a ID"],"hint":"V\u1ebd h\u00ecnh, quan s\u00e1t h\u00ecnh v\u1ebd \u0111\u1ec3 l\u1eadp lu\u1eadn ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","explain":" <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/bai6/lv3/img\/H616_K105.png' \/><\/center> <br\/> V\u00ec I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB n\u00ean $AI = IB = \\dfrac{AB}{2} = \\dfrac{4}{2} = 2cm$ <br\/> Tia IA v\u00e0 IB l\u00e0 hai tia \u0111\u1ed1i nhau (1) <br\/> C thu\u1ed9c tia \u0111\u1ed1i c\u1ee7a tia AI n\u00ean A, C n\u1eb1m c\u00f9ng ph\u00eda \u0111\u1ed1i v\u1edbi \u0111i\u1ec3m I (2) <br\/> D thu\u1ed9c tia \u0111\u1ed1i c\u1ee7a tia IB n\u00ean B, D n\u1eb1m c\u00f9ng ph\u00eda \u0111\u1ed1i v\u1edbi \u0111i\u1ec3m I (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow$ IC v\u00e0 ID l\u00e0 hai tia \u0111\u1ed1i nhau hay \u0111i\u1ec3m I n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m C v\u00e0 D (4)<br\/> V\u00ec C thu\u1ed9c tia \u0111\u1ed1i c\u1ee7a tia AI n\u00ean \u0111i\u1ec3m A n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m C v\u00e0 I <br\/> $\\begin{align*} CI &= AC + AI \\\\ CI &= 1cm + 2cm \\\\ CI &= 3cm \\end{align*}$ <br\/> D thu\u1ed9c tia \u0111\u1ed1i c\u1ee7a tia BI n\u00ean \u0111i\u1ec3m B n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m D v\u00e0 I <br\/> $\\begin{align*} DI &= BI + BD \\\\ DI &= 2cm + 1cm \\\\ DI &= 3cm \\end{align*}$ <br\/> V\u00ec CI = 3cm, DI = 3cm m\u00e0 3cm = 3cm n\u00ean CI = DI (5) <br\/> T\u1eeb (4) v\u00e0 (5) $\\Rightarrow$ \u0110i\u1ec3m I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng CD <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> ","column":2}]}],"id_ques":2033},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho \u0111o\u1ea1n th\u1eb3ng $AB = 2^{2003}cm$. G\u1ecdi $M_{1}$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB, $M_{2}$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $M_{1}B$, $M_{3}$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $M_{2}B$;.....; $M_{2003}$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $M_{2003}B$. <br\/> \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $M_{2003}B$ l\u00e0:","select":["A. $2cm$","B. $1cm$","C. $2^{2002}cm$","D. $2^{2001}cm$"],"hint":"D\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t: N\u1ebfu M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB th\u00ec $AM = MB = \\dfrac{AB}{2}$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $M_{1}B = \\dfrac{AB}{2}$ <br\/> \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $M_{2}B = \\dfrac{M_{1}B}{2} = \\dfrac{AB}{2^2}$ <br\/>..... <br\/> T\u01b0\u01a1ng t\u1ef1 nh\u01b0 v\u1eady ta t\u00ecm quy lu\u1eadt \u0111\u1ec3 t\u00ednh $M_{2003}B$ <\/span> <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/bai6/lv3/img\/H616_K106.png' \/><\/center> <br\/> V\u00ec $M_{1}$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB n\u00ean: <br\/> $M_{1}B = \\dfrac{AB}{2} = \\dfrac{2^{2003}}{2} = 2^{2002}$ <br\/> $M_{2}$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $M_{1}B$ n\u00ean: <br\/> $M_{2}B = \\dfrac{M_{1}B}{2} = \\dfrac{2^{2002}}{2} = \\dfrac{2^{2003}}{2^2} = 2^{2001}$ <br\/> $M_{3}$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $M_{2}B$ n\u00ean: <br\/> $M_{3}B = \\dfrac{M_{2}B}{2} = \\dfrac{2^{2001}}{2} = \\dfrac{2^{2003}}{2^3} = 2^{2000}$ <br\/> .......<br\/> $M_{2003}$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $M_{2002}B$ n\u00ean: <br\/> $M_{2003}B = \\dfrac{M_{2002}B}{2} = \\dfrac{2^{2003}}{2^{2003}} =1$ <br\/> V\u1eady \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $M_{2003}B$ l\u00e0 1cm <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0: B. 1cm <\/span> ","column":2}]}],"id_ques":2034},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["f","f","f","t"]],"list":[{"point":10,"image":"","col_name":["Kh\u1eb3ng \u0111\u1ecbnh","\u0110\u00fang","Sai"],"arr_ques":["N\u1ebfu 3 \u0111i\u1ec3m A, M, B th\u1eb3ng h\u00e0ng th\u00ec AM = MB = $\\dfrac{AB}{2}$","N\u1ebfu \u0111i\u1ec3m M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B th\u00ec $AM = MB = \\dfrac{AB}{2}$","N\u1ebfu MA + MB = AB th\u00ec MA = MB = $\\dfrac{AB}{2}$","N\u1ebfu \u0111i\u1ec3m M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A; B v\u00e0 M c\u00e1ch \u0111\u1ec1u hai \u0111\u1ea7u \u0111o\u1ea1n th\u1eb3ng AB th\u00ec $AM = MB = \\dfrac{AB}{2}$ "],"hint":"","explain":["SAI v\u00ec ch\u1ec9 khi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB ta m\u1edbi c\u00f3 $AM = MB = \\dfrac{AB}{2}$ ","<br\/>SAI: V\u00ec n\u1ebfu \u0111i\u1ec3m M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B ta ch\u1ec9 suy ra \u0111\u01b0\u1ee3c AM + MB = AB "," <br\/>SAI: V\u00ec n\u1ebfu MA + MB = AB ta ch\u1ec9 suy ra \u0111\u01b0\u1ee3c \u0111i\u1ec3m M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B nh\u01b0ng ch\u01b0a ch\u1eafc $MA = MB = \\dfrac{AB}{2}$ ","<br\/> \u0110\u00daNG: V\u00ec N\u1ebfu M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B v\u00e0 M c\u00e1ch \u0111\u1ec1u hai \u0111\u1ea7u \u0111o\u1ea1n th\u1eb3ng AB <br\/> $\\Rightarrow$ \u0110i\u1ec3m M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB <br\/> $\\Rightarrow$ $MA = MB = \\dfrac{AB}{2}$"]}]}],"id_ques":2035},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["f","t","t","t"]],"list":[{"point":10,"image":"","col_name":["Kh\u1eb3ng \u0111\u1ecbnh","\u0110\u00fang","Sai"],"arr_ques":["N\u1ebfu \u0111i\u1ec3m M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B ho\u1eb7c \u0111i\u1ec3m M c\u00e1ch \u0111\u1ec1u hai \u0111\u1ea7u \u0111o\u1ea1n th\u1eb3ng AB th\u00ec AM = MB = $\\dfrac{AB}{2}$","N\u1ebfu M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB th\u00ec $AM = MB = \\dfrac{AB}{2}$","N\u1ebfu M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB th\u00ec \u0111i\u1ec3m M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A, B","N\u1ebfu \u0111i\u1ec3m M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB th\u00ec \u0111i\u1ec3m M thu\u1ed9c \u0111o\u1ea1n th\u1eb3ng AB "],"hint":"","explain":["SAI v\u00ec n\u1ebfu $AM = MB = \\dfrac{AB}{2}$ th\u00ec \u0111i\u1ec3m M ph\u1ea3i th\u1ecfa m\u00e3n c\u1ea3 hai \u0111i\u1ec1u ki\u1ec7n: <br\/> + \u0110i\u1ec3m M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B <br\/> +) \u0110i\u1ec3m M c\u00e1ch \u0111\u1ec1u hai \u0111\u1ea7u \u0111o\u1ea1n th\u1eb3ng AB ","<br\/>\u0110\u00daNG: V\u00ec n\u1ebfu \u0111i\u1ec3m M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB th\u00ec M c\u00e1ch \u0111\u1ec1u hai \u0111\u1ea7u \u0111o\u1ea1n th\u1eb3ng AB n\u00ean $AM = MB = \\dfrac{AB}{2}$ "," <br\/>\u0110\u00daNG: V\u00ec n\u1ebfu M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB th\u00ec M ph\u1ea3i th\u1ecfa m\u00e3n hai \u0111i\u1ec1u ki\u1ec7n: <br\/> +) \u0110i\u1ec3m M n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B <br\/> +) \u0110i\u1ec3m M c\u00e1ch \u0111\u1ec1u hai \u0111\u1ea7u \u0111o\u1ea1n th\u1eb3ng AB","<br\/> \u0110\u00daNG: V\u00ec n\u1ebfu M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB th\u00ec M ph\u1ea3i n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B n\u00ean M $\\in$ AB"]}]}],"id_ques":2036},{"time":24,"part":[{"title":"","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho \u0111o\u1ea1n th\u1eb3ng MN = 10cm, \u0111i\u1ec3m T n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M, N v\u00e0 MT = 2cm. \u0110i\u1ec3m R n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m T v\u00e0 N sao cho TR = 6cm. G\u1ecdi O l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng MN. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang?","select":["A. OR = OT","B. OR < OT","C. OR > OT","D. MT > OT"],"hint":"T\u00ednh OM, ON tr\u01b0\u1edbc r\u1ed3i t\u00ednh OT, OR sau \u0111\u00f3 so s\u00e1nh v\u00e0 ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> V\u00ec O l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng MN n\u00ean t\u00ednh \u0111\u01b0\u1ee3c OM, ON <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ec9 ra \u0111i\u1ec3m T n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M v\u00e0 O sau \u0111\u00f3 t\u00ednh OT <br\/> <b> B\u01b0\u1edbc 3: <\/b> Ch\u1ec9 ra \u0111i\u1ec3m O n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m T v\u00e0 R r\u1ed3i t\u00ednh OR <br\/> <b> B\u01b0\u1edbc 4: <\/b> So s\u00e1nh c\u00e1c \u0111o\u1ea1n MT, OR, OT v\u00e0 ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang <\/span> <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <span class='basic_left'> <b> T\u00f3m t\u1eaft: <\/b> <br\/> MN = 10cm <br\/> T n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M, N <br\/> MT = 2cm <br\/> R n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m T, N <br\/> TR = 6cm <br\/> O l\u00e0 trung \u0111i\u1ec3m MN <br\/> ? So s\u00e1nh OT, MT, OR <\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/bai6/lv3/img\/H616_K110.png' \/><\/center> <br\/> V\u00ec \u0111i\u1ec3m O l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng MN n\u00ean: <br\/> $OM = ON = \\dfrac{MN}{2} = \\dfrac{10cm}{2} = 5cm$ <br\/> V\u00ec MO = 5cm, MT = 2cm m\u00e0 2cm < 5cm n\u00ean MT < MO (1) <br\/> MT v\u00e0 MO l\u00e0 hai tia tr\u00f9ng nhau (2) <br\/> $\\Rightarrow$ \u0110i\u1ec3m T n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M v\u00e0 O <br\/> $\\begin{align*} MT + OT &= OM \\\\ OT &= OM - MT \\\\ OT &= 5cm - 2cm \\\\ OT &= 3cm \\end{align*}$ <br\/> V\u00ec 3 \u0111i\u1ec3m O, T, R c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng m\u00e0 TO = 3cm, TR = 6cm n\u00ean TO < TR (3) <br\/> Hai tia TO, TR l\u00e0 hai tia tr\u00f9ng nhau (4) <br\/> T\u1eeb (3) v\u00e0 (4) $\\Rightarrow$ \u0110i\u1ec3m O n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m T v\u00e0 R <br\/> $\\begin{align*} OT + OR &= TR \\\\ OR &= TR - OT \\\\ OR &= 6cm - 3cm \\\\ OR &= 3cm \\end{align*}$ <br\/> V\u00ec OT = 3cm, OR = 3cm n\u00ean OT = OR <br\/> MT = 2cm, OT = 3cm n\u00ean MT < OT <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A. OR = OT <\/span> ","column":2}]}],"id_ques":2037},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho \u0111o\u1ea1n th\u1eb3ng MN = 14cm, \u0111i\u1ec3m P n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M v\u00e0 N v\u00e0 MP = 4cm. \u0110i\u1ec3m Q n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m P, N sao cho MP = QN. G\u1ecdi R, S theo th\u1ee9 t\u1ef1 l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng MP, NQ. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. RQ > PS","B. RQ < PS","C. RQ = PS"],"hint":"T\u00ednh MR v\u00e0 SN tr\u01b0\u1edbc r\u1ed3i sau \u0111\u00f3 t\u00ednh PS v\u00e0 QR","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> V\u00ec R l\u00e0 trung \u0111i\u1ec3m c\u1ee7a MP, S l\u00e0 trung \u0111i\u1ec3m c\u1ee7a QN n\u00ean ta t\u00ednh \u0111\u01b0\u1ee3c MR v\u00e0 SN <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ec9 ra c\u00e1c \u0111i\u1ec3m P, Q, R, S n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M, N <br\/> <b> B\u01b0\u1edbc 3: <\/b> \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t c\u1ed9ng \u0111o\u1ea1n th\u1eb3ng t\u00ednh PS v\u00e0 QR <br\/> <b> B\u01b0\u1edbc 4: <\/b> So s\u00e1nh PS; QR v\u00e0 ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang <\/span> <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <span class='basic_left'> <b> T\u00f3m t\u1eaft: <\/b> <br\/> MN = 14cm <br\/> P n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M, N <br\/> MP = 4cm <br\/> Q n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m P, N <br\/> QN = MP = 4cm <br\/> R l\u00e0 trung \u0111i\u1ec3m MP <br\/> S l\u00e0 trung \u0111i\u1ec3m QN <br\/> ? So s\u00e1nh RQ, PS <\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/bai6/lv3/img\/H616_K111.png' \/><\/center> <br\/> V\u00ec \u0111i\u1ec3m R l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng MP n\u00ean: <br\/> $MR = RP = \\dfrac{MP}{2} = \\dfrac{4cm}{2} = 2cm$ <br\/> V\u00ec N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng QN m\u00e0 QN = MP = 2cm n\u00ean: <br\/> $QS = SN = \\dfrac{QN}{2} = \\dfrac{4}{2} = 2cm$ <br\/> V\u00ec P, Q n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M, N (theo \u0111\u1ec1 b\u00e0i) <br\/> M\u00e0 R l\u00e0 trung \u0111i\u1ec3m c\u1ee7a MP, S l\u00e0 trung \u0111i\u1ec3m c\u1ee7a QN n\u00ean c\u00e1c \u0111i\u1ec3m P, Q, R, S n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m M, N <br\/> $\\begin{align*} \\Rightarrow MP + PS + SN &= MN \\\\ PS &= MN - MP - SN \\\\ PS &= 14cm - 4cm - 2cm \\\\ PS &= 8cm \\end{align*}$ <br\/> $\\begin{align*} MR + QR + QN &= MN \\\\ QR &= MN - MR - QN \\\\ QR &= 14cm - 2cm - 4cm \\\\ QR &= 8cm \\end{align*}$ <br\/> V\u00ec QR = 8cm, RS = 8cm n\u00ean QR = PS <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C. QR = PS <\/span> ","column":3}]}],"id_ques":2038},{"time":14,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","4"]],"list":[{"point":10,"img":"","ques":"Tr\u00ean tia Oy cho 3 \u0111i\u1ec3m A, B, C sao cho OA = 1,5cm, OB = 3cm, OC = 4,5cm. H\u00e3y ch\u1ecdn c\u00e1c kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau:","hint":"V\u1ebd h\u00ecnh r\u1ed3i ki\u1ec3m tra xem c\u00e1c \u0111i\u1ec3m n\u1eb1m gi\u1eefa xem c\u00f3 l\u00e0 trung \u0111i\u1ec3m kh\u00f4ng","column":2,"number_true":2,"select":["A. \u0110i\u1ec3m B l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AC","B. \u0110i\u1ec3m B l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OC","C. \u0110i\u1ec3m C l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OB","D. \u0110i\u1ec3m A l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OB"],"explain":"<span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span><br\/> <span class='basic_left'> <b> T\u00f3m t\u1eaft:<\/b> <br\/> Tr\u00ean tia Oy <br\/> OA = 1,5cm <br\/> OB = 3cm <br\/> OC = 4,5cm <br\/> H\u1ecfi \u0111i\u1ec3m n\u00e0o l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng n\u00e0o? <\/span> <br\/>Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/bai6/lv3/img\/H616_K114.png' \/><\/center> <br\/> Tr\u00ean tia Oy c\u00f3: OA = 1,5cm; OB = 3cm <br\/> V\u00ec 1,5cm < 3cm do \u0111\u00f3 OA < OB n\u00ean \u0111i\u1ec3m A n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m O v\u00e0 B <br\/> $\\begin{align*} \\Rightarrow OA + AB &= OB (1) \\\\ 1,5cm + AB &= 3cm \\\\ AB &= 3cm - 1,5cm \\\\ AB &= 1,5cm \\end{align*}$ <br\/> M\u00e0 OA = 1,5cm n\u00ean OA = AB (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ <b> \u0110i\u1ec3m A l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OB <\/b> <br\/> Tr\u00ean tia Oy c\u00f3: OB = 3cm, OC = 4,5cm <br\/> V\u00ec 1,5cm < 4,5cm n\u00ean OA < OC do \u0111\u00f3 \u0111i\u1ec3m A n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m O v\u00e0 C <br\/> $\\begin{align*} \\Rightarrow OA + AC &= OC \\\\ 1,5cm + AC &= 4,5cm \\\\ AC &= 4,5cm - 1,5cm \\\\ AC &= 3cm \\end{align*}$ <br\/> Tr\u00ean tia Oy c\u00f3: OB = 3cm; OC = 4,5cm <br\/> V\u00ec 3cm < 4,5cm n\u00ean OB < OC do \u0111\u00f3 \u0111i\u1ec3m B n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m B n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m O v\u00e0 C <br\/> $\\begin{align*} \\Rightarrow OB + BC&= OC \\\\ 3cm + BC &= 4,5cm \\\\ BC &= 4,5cm - 3cm \\\\ BC &= 1,5cm \\end{align*}$ <br\/> Ta c\u00f3: AB = 1,5cm; BC = 1,5cm n\u00ean AB = BC (3) <br\/> V\u00e0 AB + Bc = 1,5cm + 1,5cm = 3cm <br\/> M\u00e0 AC = 3cm $\\Rightarrow$ AB + BC = AC (4) <br\/> T\u1eeb (3) v\u00e0 (4) suy ra <b> \u0111i\u1ec3m B l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AC <\/b> <br\/> <span class='basic_pink'>V\u1eady \u0111i\u1ec3m A l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OB <br\/> \u0110i\u1ec3m B l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AC <\/span> <br\/><span class='basic_green'><i> L\u01b0u \u00fd: \u0110\u1ec3 ch\u1ee9ng t\u1ecf m\u1ed9t \u0111i\u1ec3m l\u00e0 trung \u0111i\u1ec3m c\u1ee7a m\u1ed9t \u0111o\u1ea1n th\u1eb3ng \u0111i\u1ec3m \u0111\u00f3 c\u1ea7n th\u1ecfa m\u00e3n hai \u0111i\u1ec1u ki\u1ec7n sau: <br\/> +) \u0110i\u1ec3m \u0111oa n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m c\u00f2n l\u1ea1i <br\/> +) \u0110i\u1ec3m \u0111\u00f3 c\u00e1ch \u0111\u1ec1u hai \u0111\u1ea7u m\u00fat c\u1ee7a \u0111o\u1ea1n th\u1eb3ng <\/i><\/span> "}]}],"id_ques":2039},{"time":14,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","3"]],"list":[{"point":10,"img":"","ques":"Cho \u0111i\u1ec3m O n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng mn. Tr\u00ean tia Om l\u1ea5y \u0111i\u1ec3m A, B sao cho OA = 3cm, OB = 6cm. Tr\u00ean tia On l\u1ea5y \u0111i\u1ec3m C sao cho OC = 3cm. H\u00e3y ch\u1ecdn c\u00e1c kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau:","hint":"V\u1ebd h\u00ecnh r\u1ed3i ki\u1ec3m tra xem c\u00e1c \u0111i\u1ec3m n\u1eb1m gi\u1eefa xem c\u00f3 l\u00e0 trung \u0111i\u1ec3m kh\u00f4ng","column":2,"number_true":2,"select":["A. \u0110i\u1ec3m O l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AC","B. \u0110i\u1ec3m C l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OA","C. \u0110i\u1ec3m A l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OB","D. \u0110i\u1ec3m B l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AC"],"explain":" <span class='basic_left'> <b> T\u00f3m t\u1eaft:<\/b> <br\/> O $\\in$ \u0111\u01b0\u1eddng th\u1eb3ng mn <br\/> A, B $\\in$ tia Om <br\/> OA = 3cm <br\/> OB = 6cm <br\/> OC = 3cm <br\/> H\u1ecfi \u0111i\u1ec3m n\u00e0o l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng n\u00e0o? <\/span> <br\/> <span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span><br\/> Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/hinhhoc/bai6/lv3/img\/H616_K115.png' \/><\/center> <br\/> \u0110i\u1ec3m O thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng mn $\\Rightarrow$ Om v\u00e0 On l\u00e0 hai tia \u0111\u1ed1i nhau <br\/> V\u00ec \u0111i\u1ec3m A thu\u1ed9c tia Om, \u0111i\u1ec3m C thu\u1ed9c tia On <br\/> $\\Rightarrow$ \u0110i\u1ec3m O n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 C <br\/> Theo b\u00e0i ra ta c\u00f3: OA = OC = 3cm (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ <b> \u0110i\u1ec3m O l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AC <\/b> <br\/> Tr\u00ean tia Om c\u00f3: OA = 3cm; OB = 6cm <br\/> V\u00ec 3cm < 6cm n\u00ean OA < OB do \u0111\u00f3 \u0111i\u1ec3m A n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m O v\u00e0 B <br\/> $\\begin{align*} \\Rightarrow OA + AB &= OB (3) \\\\ 3cm + AB &= 6cm \\\\ AB &= 6cm - 3cm \\\\ AB &= 3cm \\end{align*}$ <br\/> $\\Rightarrow$ OA = AB (4) <br\/> T\u1eeb (3) v\u00e0 (4) $\\Rightarrow$ <b> \u0110i\u1ec3m A l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OB <\/b> <br\/> <span class='basic_pink'>V\u1eady \u0111i\u1ec3m O l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AC <br\/> \u0110i\u1ec3m A l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OB <\/span> "}]}],"id_ques":2040}],"lesson":{"save":0,"level":3}}