{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{3}{5}$","B. $\\dfrac{2}{5}$","C. $\\dfrac{1}{5}$"],"ques":"<span class='basic_left'>Cho g\u00f3c nh\u1ecdn $\\alpha$. Bi\u1ebft r\u1eb1ng $\\cos \\,\\alpha-sin\\,\\alpha=\\dfrac{1}{5}$. T\u00ednh $\\sin\\,\\alpha$<br\/>","hint":"T\u1eeb \u0111\u1eb3ng th\u1ee9c \u0111\u1ec1 cho, r\u00fat ra $cos\\,\\alpha$ r\u1ed3i thay v\u00e0o h\u1ec7 th\u1ee9c $\\,sin^2x+cos^2x=1$ \u0111\u1ec3 t\u00ednh $sin\\,\\alpha$","explain":"<span class='basic_left'> Ta c\u00f3: $\\cos \\,\\alpha - $$sin\\,\\alpha=\\dfrac{1}{5}$ <br\/> $\\,\\Rightarrow cos\\,\\alpha$$ = sin\\,\\alpha+\\dfrac{1}{5}$ <br\/> M\u00e0 $sin^2\\alpha+$$cos^2\\alpha=1$ <br\/> $\\Rightarrow sin^2\\alpha $$+\\left(sin\\,\\alpha+\\dfrac{1}{5}\\right)^2\\,=1$ <br\/> $\\Leftrightarrow sin^2\\alpha $$+sin^2\\,\\alpha+\\dfrac{2}{5}sin\\,\\alpha+\\dfrac{1}{25}\\,=1$ <br\/>$\\Leftrightarrow 25sin^2\\alpha +25sin^2\\,\\alpha+10sin\\,\\alpha +1=25$ <br\/>$\\Leftrightarrow 50sin^2\\alpha +10sin\\,\\alpha -24=0$ <br\/> $ \\Leftrightarrow 25sin^2\\alpha+$$5sin\\,\\alpha-12\\,\\,=0$ <br\/> $ \\Leftrightarrow (5sin\\,\\alpha-3)$$(5sin\\,\\alpha+4) = 0$ <br\/> $ \\Leftrightarrow$ $\\left[ \\begin{align}& sin\\, \\alpha =\\dfrac{3}{5} \\\\ & sin\\, \\alpha =-\\dfrac{4}{5} \\, \\text{(lo\u1ea1i v\u00ec}\\,\\, sin\\alpha >0) \\\\ \\end{align} \\right.$<\/span>"}]}],"id_ques":1331},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $\\dfrac{6}{17}$","B. $\\dfrac{7}{17}$","C. $\\dfrac{8}{17}$"],"ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng \u1edf $A$. N\u1ebfu $\\dfrac{AB}{AC}=\\dfrac{15}{8}$ th\u00ec $sin\\,B=$ ?","hint":"\u0110\u1eb7t $AB=15k; AC=8k$, t\u00ednh $AC$ ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai2/lv3/img\/H912_K7.png' \/><\/center><span class='basic_left'> X\u00e9t tam gi\u00e1c $ABC$c\u00f3 $\\dfrac{AB}{AC}=\\dfrac{15}{8}\\,$$\\Rightarrow \\dfrac{AB}{15}=\\dfrac{AC}{8}=k$ <br\/> Do \u0111\u00f3 $AB=15k, AC=8k$ <br\/>X\u00e9t $\\Delta ABC$ c\u00f3: $\\widehat{A}={{90}^{o}}.$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago ta c\u00f3: <br\/> $\\begin{align} & \\,\\,\\,A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}} \\\\ & \\Rightarrow B{{C}^{2}}={{\\left( 15k \\right)}^{2}}+{{\\left( 8k \\right)}^{2}} \\\\ & \\Leftrightarrow B{{C}^{2}}=289{{k}^{2}} \\\\ & \\Rightarrow BC\\,\\,=17k \\\\ \\end{align}$ <br\/> Suy ra $sinB=\\dfrac{AC}{BC}=\\dfrac{8k}{17k}=\\dfrac{8}{17}$<\/span>"}]}],"id_ques":1332},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $AB= 3\\,cm,$$ AC= 4\\,cm,$ $\\widehat{A}=30^{o}.$<br\/> Khi \u0111\u00f3 di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ l\u00e0 _input_$(cm^2)$","hint":"T\u00ednh chi\u1ec1u cao h\u1ea1 t\u1eeb \u0111\u1ec9nh $C$ xu\u1ed1ng c\u1ea1nh $AB$","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai2/lv3/img\/H912_K4.png' \/><\/center> <br\/> H\u1ea1 $CH\\bot AB$ t\u1ea1i $H$ <br\/>X\u00e9t $ \\Delta ACH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: $sin\\,A=\\dfrac{HC}{AC}$ <br\/> $ \\Rightarrow HC=4.sin\\,30^{o}=2$ <br\/> Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0 <br\/>$\\dfrac{HC.AB}{2}\\,$ $=\\dfrac{1}{2}.2.3=3\\,(cm^2)$<br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $3$<\/span>"}]}],"id_ques":1333},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng \u1edf $A$, ta c\u00f3: $\\dfrac{AC}{AB}=\\dfrac{\\sin B}{\\sin C}$ ","select":["A. \u0110\u00fang","B. Sai"],"explain":" <span class='basic_left'> <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai2/lv3/img\/H912_K3.png' \/><\/center>Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ n\u00ean <br\/> $\\begin{align} & \\sin B=\\dfrac{AC}{BC};sinC=\\dfrac{AB}{BC} \\\\ & \\Rightarrow \\dfrac{\\sin B}{\\sin C}=\\dfrac{AC}{BC}.\\dfrac{BC}{AB}=\\dfrac{AC}{AB} \\\\ & \\Rightarrow \\dfrac{AC}{AB}=\\dfrac{\\sin B}{\\sin C} \\\\ \\end{align}$ <br\/> Kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang. <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 A<\/span>","column":2}]}],"id_ques":1334},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c <br\/> $N=tg\\,35^o.tg\\,40^o$$.tg\\,45^o.tg\\,50^o$$.tg\\,55^o $ <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $N = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span>","hint":"S\u1eed d\u1ee5ng h\u1ec7 th\u1ee9c c\u01a1 b\u1ea3n: $tg\\,\\alpha.cotg\\,\\alpha=1$ v\u00e0 \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd v\u1ec1 gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a hai g\u00f3c ph\u1ee5 nhau","explain":" <span class='basic_left'> D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c $tg\\,\\alpha.cotg\\,\\alpha=1$ <br\/>V\u00ec $35^o$ v\u00e0 $55^o$ l\u00e0 hai g\u00f3c ph\u1ee5 nhau n\u00ean $tg\\, 35^o =cotg\\, 55^o$<br\/>V\u00ec $40^o$ v\u00e0 $50^o$ l\u00e0 hai g\u00f3c ph\u1ee5 nhau n\u00ean $tg\\, 40^o =cotg\\,50^o$<br\/> Suy ra<br\/>$N=tg\\,35^o.tg\\,40^o.$$tg\\,45^o.tg\\,50^o.$$tg\\,55^o$ <br\/> $=(tg\\,35^o.tg\\,55^o).$$(tg\\,40^o.tg\\,50^o).$$tg\\,45^o$ <br\/> $=(tg\\,35^o.cotg\\,35^o).$$(tg\\,40^o.cotg\\,40^o).$$tg\\,45^o$ <br\/> $=1.1.1=1 $ <br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$<\/span><\/span>"}]}],"id_ques":1335},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c <br\/> $\\begin{align} A&=(cos\\,\\alpha - sin\\,\\alpha)^2+(cos\\,\\alpha+sin\\alpha)^2 \\\\ &=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\end{align}$","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $A$ s\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(a\\pm b)^2$ v\u00e0 h\u1ec7 th\u1ee9c c\u01a1 b\u1ea3n $\\,sin^2x+cos^2x=1$ ","explain":" <span class='basic_left'> Ta c\u00f3: <br\/>$A=(cos\\,\\alpha - sin\\,\\alpha)^2$$+(cos\\,\\alpha+sin\\alpha)^2$ <br\/> $\\,\\,\\,\\,\\,=cos^2\\alpha-2cos$$\\,\\alpha.sin\\,\\alpha+sin^2\\alpha$$+cos^2\\alpha+2cos\\,\\alpha.sin$$\\,\\alpha+sin^2\\alpha$ <br\/> $\\,\\,\\,\\,\\,$$=2(sin^2\\alpha+cos^2\\alpha)$ <br\/> $\\,\\,\\,\\,\\,=2$<br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2$<\/span><\/span>"}]}],"id_ques":1336},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho tam gi\u00e1c nh\u1ecdn $ABC$, $\\widehat{A}={{30}^{o}}$. Hai \u0111\u01b0\u1eddng cao $BH$ v\u00e0 $CK$. T\u00ednh t\u1ec9 s\u1ed1 gi\u1eefa di\u1ec7n t\u00edch tam gi\u00e1c $AHK$ v\u00e0 t\u1ee9 gi\u00e1c $BCHK$. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $S_{AHK}=$_input_$S_{BCHK}$ ","hint":"T\u00ednh $S_{AHK}$ v\u00e0 $S_{BCHK}$ theo $S_{ABC}$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh $\\Delta AHK \\sim \\Delta ABC$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<br\/><\/span><b> B\u01b0\u1edbc 1: <\/b>Ch\u1ee9ng minh $\\Delta ABH \\sim \\Delta ACK$<br\/><b> B\u01b0\u1edbc 2:<\/b> Ch\u1ee9ng minh $\\Delta AHK \\sim \\Delta ABC$ v\u00e0 t\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{S_{AHK}}{S_{ABC}}.$ <br\/><b> B\u01b0\u1edbc 3 <\/b> T\u00ednh $S_{AHK}$ v\u00e0 $S_{BCHK}$ theo $S_{ABC}$<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai2/lv3/img\/H912_K1.png' \/><\/center><br\/> $\\Delta ABH\\sim \\Delta ACK$ (do $\\widehat{A}$ chung; $\\widehat{AHB}=\\widehat{AKC}={{90}^{o}}$ )<br\/> $\\Rightarrow \\dfrac{AH}{AK}=\\dfrac{AB}{AC}$$\\Rightarrow \\dfrac{AH}{AB}=\\dfrac{AK}{AC}$ <br\/> X\u00e9t $\\Delta AHK$ v\u00e0 $\\Delta ABC$ c\u00f3: <br\/> + $\\widehat{A}$ chung; <br\/> + $\\dfrac{AH}{AB}=\\dfrac{AK}{AC}$ <br\/> $\\Rightarrow \\Delta AHK\\sim \\Delta ABC\\left( c.g.c \\right)$ <br\/> $\\Rightarrow \\dfrac{{{S}_{AHK}}}{{{S}_{ABC}}}$$={{\\left( \\dfrac{AH}{AB} \\right)}^{2}}$$=co{{s}^{2}}A$ (t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch b\u1eb1ng b\u00ecnh ph\u01b0\u01a1ng t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng) <br\/> $\\Rightarrow {{S}_{AHK}}={{S}_{ABC}}co{{s}^{2}}A$ <br\/>Ta l\u1ea1i c\u00f3 <br\/>${{S}_{BCHK}}={{S}_{ABC}}-{{S}_{AHK}}\\,$$={{S}_{ABC}}\\left( 1-co{{s}^{2}}A \\right)\\,$$={{S}_{ABC}}.sin^2A$ <br\/> V\u00ec $\\widehat{A}={{30}^{o}}$ n\u00ean ${{S}_{AHK}}={{S}_{ABC}}.\\left( \\dfrac{\\sqrt{3}}{2} \\right)^2$$=\\dfrac{3}{4}{{S}_{ABC}}$ (1) <br\/> ${{S}_{BCHK}}={{S}_{ABC}}{{\\left( \\dfrac{1}{2} \\right)}^{2}}$$=\\dfrac{1}{4}{{S}_{ABC}}$ (2) <br\/>T\u1eeb (1) v\u00e0 (2) suy ra: ${{S}_{AHK}}=3{{S}_{BHCK}}$ <br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $3$<\/span><\/span>"}]}],"id_ques":1337},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{7}{2}$","B. $\\dfrac{5}{2}$","C. $\\dfrac{3}{2}$"],"ques":"<span class='basic_left'>T\u00ednh: $M=cos^{2}15^{o}$$+\\,cos^{2}25^{o}$$+\\,cos^{2}35^{o}$$+\\,cos^{2}45^{o}$$+\\,cos^{2}55^{o}$$+\\,cos^{2}65^{o}$$+\\,cos^{2}75^{o}$ <br\/> <b>\u0110\u00e1p s\u1ed1: <\/b> $M=$ ?<\/span>","hint":"S\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a hai g\u00f3c ph\u1ee5 nhau v\u00e0 h\u1ec7 th\u1ee9c $\\,sin^2x + cos^2x=1$ \u0111\u1ec3 nh\u00f3m c\u00e1c h\u1ea1ng t\u1eed.","explain":"<span class='basic_left'> V\u00ec $15^o$ v\u00e0 $75^o$ l\u00e0 hai g\u00f3c ph\u1ee5 nhau n\u00ean $sin\\,75^o=cos\\,15^o$.<br\/>T\u01b0\u01a1ng t\u1ef1: $sin\\,25^o=cos\\,65^o$; $sin\\,35^o=cos\\,55^o$<br\/> Suy ra:<br\/>$M = cos^{2}15^{o}$$+\\,cos^{2}25^{o}$$+\\,cos^{2}35^{o}$$+cos^{2}45^{o}$$+\\,cos^{2}55^{o}$$+cos^{2}65^{o}$$+\\,cos^{2}75^{o}$<br\/> $ M=(cos^215^{o}+cos^275^{o})+\\,$$(cos^225^{o}+cos^265^{o})+\\,$$(cos^235^{o}+cos^255^{o})$$+\\,cos^245^{o}$ <br\/> $M= (cos^215^{o}+sin^215^{o})+\\,$$(cos^225^{o}+sin^225^{o})+\\,$$(cos^235^{o}+sin^235^{o})$$+\\,cos^245^{o}$ <br\/> $M=1+1+1+$${{\\left( \\dfrac{\\sqrt{2}}{2} \\right)}^{2}}$$=\\dfrac{7}{2}$<\/span>"}]}],"id_ques":1338},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"V\u1edbi g\u00f3c nh\u1ecdn $\\alpha$ t\u00f9y \u00fd, ta c\u00f3: $1+tg^2\\alpha =\\dfrac{1}{sin^2\\alpha}$","select":["A. \u0110\u00fang ","B. Sai "],"hint":"S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a c\u1ee7a c\u00e1c t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c v\u00e0 h\u1ec7 th\u1ee9c c\u01a1 b\u1ea3n: $\\,sin^2x+cos^2x=1$","explain":" <span class='basic_left'> X\u00e9t v\u1ebf tr\u00e1i ta \u0111\u01b0\u1ee3c: <br\/>$\\begin{align}1+tg^2\\alpha &=1+\\left(\\dfrac{sin \\,\\alpha}{cos\\,\\alpha}\\right)^2\\\\&=\\dfrac{cos^2\\alpha+sin^2 \\alpha}{cos^2\\alpha}\\\\&=\\dfrac{1}{cos^2\\alpha}\\end{align}$<br\/>Suy ra v\u1ebf tr\u00e1i kh\u00e1c v\u1ebf ph\u1ea3i. <br\/>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 B<\/span>","column":2}]}],"id_ques":1339},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["45"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"T\u00ecm $x$ bi\u1ebft: $sin x + cos x =$ $\\sqrt{2}$<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $x=$ _input_$^{o}$","hint":"B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf \u0111\u1eb3ng th\u1ee9c \u0111\u00e3 cho v\u00e0 s\u1eed d\u1ee5ng h\u1ec7 th\u1ee9c $sin^2x+cos^2x=1$ \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i.","explain":"<span class='basic_left'> D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c $sin^2x+cos^2x=1$<br\/> Ta c\u00f3: $ sin x + cos x = $$\\sqrt{2}$ <br\/> $\\Rightarrow (sin x + cos x)^2 = $$(\\sqrt{2})^2$ <br\/> $ \\Leftrightarrow {{\\sin }^{2}}x+2\\sin x.cos\\,x$$+{{\\cos }^{2}}x=2 $ <br\/> $ \\Leftrightarrow 1+$$2\\sin \\,x.cos\\,x=2 $ <br\/> $\\Leftrightarrow 1-$$2\\sin \\,x.cos\\,x=0 $ <br\/> $ \\Leftrightarrow {{\\sin }^{2}}x-$$2\\sin x.cos\\,x+{{\\cos }^{2}}x$$=0 $ <br\/> $\\Leftrightarrow {{\\left( \\sin \\,x-c\\text{os}\\, x \\right)}^{2}}$$=0 $ <br\/> $ \\Leftrightarrow \\sin \\,x=cos\\,x $ <br\/> Do \u0111\u00f3 $tg \\,x$$=\\dfrac{sin \\,x}{cos\\,x}=1\\,$ $\\Rightarrow x={{45}^{o}}$<br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $45$<\/span>"}]}],"id_ques":1340}],"lesson":{"save":0,"level":3}}