{"segment":[{"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":[[["30,64"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"H\u00ecnh b\u00ecnh h\u00e0nh $ABCD$ c\u00f3 $AB= 5 cm$; $AD= 8 cm$ v\u00e0 $\\widehat{A}=130^{o}$<br\/>Di\u1ec7n t\u00edch h\u00ecnh b\u00ecnh h\u00e0nh $ABCD$ l\u00e0:_input_ ($cm^2$)<br\/>(K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai)","hint":"K\u1ebb $CH$ vu\u00f4ng g\u00f3c v\u1edbi $AD$, t\u00ednh $CH$ v\u00e0 t\u00ednh di\u1ec7n t\u00edch h\u00ecnh b\u00ecnh h\u00e0nh $ABCD$.","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai4/lv3/img\/H914_K10.png' \/><\/center><br\/>Ta c\u00f3 $ABCD$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh (gi\u1ea3 thi\u1ebft)<br\/>Suy ra: $CD= AB=5 cm$ ; $\\widehat{D}=180^o -130^o=50^o$ (t\u00ednh ch\u1ea5t h\u00ecnh b\u00ecnh h\u00e0nh)<br\/>K\u1ebb $CH\\bot AD$, ($H \\in AD$)<br\/>X\u00e9t tam gi\u00e1c $CDH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $CD=5\\,cm$<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c vu\u00f4ng $CHD$, ta c\u00f3:<br\/>$CH=CD.sin\\,{\\widehat {CDH}}=5.sin\\,50^o\\approx 3,83\\,(cm)$<br\/>V\u1eady di\u1ec7n t\u00edch h\u00ecnh b\u00ecnh h\u00e0nh $ABCD$ l\u00e0:<br\/>$CH.AD\\approx 3,83.8 \\approx 30,64\\,(cm^2)$<br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $30,64$<\/span><\/span><span class='basic_left'><span class='basic_green'>L\u01b0u \u00fd:<\/span> M\u1ed9t c\u00e1ch t\u1ed5ng qu\u00e1t ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c <br\/>Di\u1ec7n t\u00edch h\u00ecnh b\u00ecnh h\u00e0nh b\u1eb1ng t\u00edch c\u1ee7a hai c\u1ea1nh k\u1ec1 nh\u00e2n v\u1edbi sin c\u1ee7a g\u00f3c nh\u1ecdn t\u1ea1o b\u1edfi hai \u0111\u01b0\u1eddng th\u1eb3ng ch\u1ee9a hai c\u1ea1nh \u0111\u00f3<br\/><b>C\u00e1ch 2:<\/b><br\/> $\\begin{align} S_{ADC}&=\\dfrac{1}{2}AD.DC.\\sin{\\widehat {ADC}}\\\\ &=\\dfrac{1}{2}.5.8.\\sin 50^{o}\\\\& \\approx 15,32 \\,(cm^2) \\\\ \\end{align}$<br\/>Suy ra: $S_{ABCD}=2S_{ADC}$$\\approx 30,64 \\,(cm^2)$<\/span>"}]}],"id_ques":1391},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["7,5"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AB= 3, BC= 5$. C\u00e1c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong v\u00e0 ph\u00e2n gi\u00e1c ngo\u00e0i c\u1ee7a g\u00f3c $B$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $AC$ t\u1ea1i $M$ v\u00e0 $N$. \u0110\u1ed9 d\u00e0i $MN$ l\u00e0:_input_","hint":"Tam gi\u00e1c $MBN$ vu\u00f4ng t\u1ea1i $B$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn <\/span><br\/><b> B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago t\u00ednh c\u1ea1nh $AC$<br\/><b> B\u01b0\u1edbc 2: <\/b> \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c \u0111\u1ec3 t\u00ednh $MA$ <br\/><b> B\u01b0\u1edbc 3: <\/b> T\u00ednh c\u1ea1nh $AN$ d\u1ef1a v\u00e0o h\u1ec7 th\u1ee9c $h^2=b'c'$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai4/lv3/img\/H914_K5.png' \/><\/center><br\/><span class='basic_left'>X\u00e9t $\\Delta ABC$ c\u00f3: $ \\widehat{A}=90^o$. \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago ta c\u00f3:<br\/>$BC^2=AB^2+AC^2 \\\\ \\Rightarrow AC^2=5^2-3^2\\\\ \\Rightarrow AC^2 =16 \\Rightarrow AC=4$<br\/> Ta l\u1ea1i c\u00f3 $BM$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $B$ (gi\u1ea3 thi\u1ebft)<br\/> $\\Rightarrow \\dfrac{MA}{MC}=\\dfrac{BA}{BC}=\\dfrac{3}{5}$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c)<br\/>$\\Rightarrow \\dfrac{MA}{3}=\\dfrac{MC}{5}$$=\\dfrac{MA+MC}{3+5}=\\dfrac{4}{8}=\\dfrac{1}{2}$<br\/>$\\Rightarrow MA=3.\\dfrac{1}{2}=1,5$<br\/>V\u00ec $BM$ v\u00e0 $BN$ l\u00e0 hai tia ph\u00e2n gi\u00e1c c\u1ee7a hai g\u00f3c k\u1ec1 b\u00f9 n\u00ean $BM \\bot BN$<br\/>X\u00e9t $\\Delta BMN$ c\u00f3: $ \\,\\widehat{MBN}=90^o,$ $BA\\bot MN$ t\u1ea1i $A$. <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng.<br\/>$\\Rightarrow BA^2=AM.AN$$\\Rightarrow AN=\\dfrac{BA^2}{AM}=\\dfrac{3^2}{1,5}=6$<br\/>Do \u0111\u00f3, $MN= AN + AM $$=6 + 1,5=7,5$<br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $7,5$<\/span><\/span><\/span>"}]}],"id_ques":1392},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'>H\u00ecnh thang $ABCD$ c\u00f3 $AB \/\/ CD$; $\\,\\widehat{C}+\\widehat{D}=90^o$. Bi\u1ebft $AD= 4; BC= 7$ <br\/><b> C\u00e2u 1: <\/b> T\u00ednh s\u1ed1 \u0111o c\u1ee7a g\u00f3c $C$ v\u00e0 g\u00f3c $D$<\/span>","select":["A. $\\widehat{C}\\approx 29^o44'$; $\\,\\widehat{D}\\approx 60^o16'$","B. $\\widehat{C}\\approx 60^o16'$; $\\,\\widehat{D}\\approx 29^o44'$","C. $\\widehat{C}\\approx 28^o35'$; $\\,\\widehat{D}\\approx 61^o15'$","D. $\\widehat{C}\\approx 61^o15'$; $\\,\\widehat{D}\\approx 28^o35'$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> K\u1ebb th\u00eam $BE \/\/CD$<br\/><b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $\\Delta BCE$ vu\u00f4ng<br\/><b> B\u01b0\u1edbc 3: <\/b> T\u00ednh g\u00f3c $C$v\u00e0 g\u00f3c $D$<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai4/lv3/img\/H914_K4b.png' \/><\/center> K\u1ebb $BE \/\/AD$ ($E \\in\\, CD$) $\\Rightarrow \\widehat{E_{1}} =\\widehat{D}$ (hai g\u00f3c \u0111\u1ed3ng v\u1ecb) <br\/>Do \u0111\u00f3, $\\widehat{C}+\\widehat{E_{1}}=$$\\widehat{C}+\\widehat{D}=90^o$<br\/>$\\Rightarrow \\widehat{CBE}=180^o-(\\widehat C +\\widehat {E_1})=180^o-90^o=90^o$<br\/>V\u1eady $\\Delta BCE $ vu\u00f4ng t\u1ea1i $B$<br\/>Ta l\u1ea1i c\u00f3 $AB\/\/DE$(gi\u1ea3 thi\u1ebft) v\u00e0 $\\,AD\/\/BE$ (theo c\u00e1ch d\u1ef1ng).<br\/> Suy ra $ABED$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh. (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)<br\/>$\\Rightarrow AD=BE=4$ (t\u00ednh ch\u1ea5t h\u00ecnh b\u00ecnh h\u00e0nh)<br\/>Tam gi\u00e1c $EBC$ vu\u00f4ng t\u1ea1i $B$, ta c\u00f3: $tg\\, C=\\dfrac{BE}{BC}=\\dfrac {4}{7} $<br\/>$\\Rightarrow \\widehat{C}\\approx 29^o44'$; $\\,\\widehat{D}\\approx 60^o16'$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":1393},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["31,3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>H\u00ecnh thang $ABCD$ c\u00f3 $AB \/\/ CD$; $\\,\\widehat{C}+\\widehat{D}=90^o$. Bi\u1ebft $AD= 4; BC= 7$<br\/><b> C\u00e2u 2: <\/b> Bi\u1ebft $CD=13$. Di\u1ec7n t\u00edch h\u00ecnh thang $ABCD\\approx$ _input_(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>(K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t)<\/span>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai4/lv3/img\/H914_K4.png' \/><\/center><br\/>Theo c\u00e2u 1, ta c\u00f3: Tam gi\u00e1c $BCE$ vu\u00f4ng t\u1ea1i $B$ <br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago cho tam gi\u00e1c vu\u00f4ng $BEC$, ta c\u00f3: <br\/>$CE^2=BC^2+BE^2\\\\ \\Rightarrow CE^2=7^2+4^2=65\\\\ \\Rightarrow CE=\\sqrt{65}\\approx 8,1$<br\/> M\u1eb7t kh\u00e1c: $AB\/\/DE; AD\/\/ BE$, suy ra: $ABED$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)<br\/>Suy ra $AB=DE=DC -EC =13 - 8,1 =4,9$<br\/>K\u1ebb $BH \\bot BC$ t\u1ea1i $H$. X\u00e9t tam gi\u00e1c $BCE$ vu\u00f4ng t\u1ea1i $B$, ta c\u00f3: <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng $EBC$, ta c\u00f3:<br\/>$BC.BE=EC.BH\\Rightarrow BH=\\dfrac{BC.BE}{EC}$$=\\dfrac{4.7}{8,1}\\approx 3,5$<br\/>Di\u1ec7n t\u00edch h\u00ecnh thang $ABCD$ l\u00e0 : <br\/>$S=\\dfrac{(AB+CD).BH}{2}$$\\approx\\dfrac{(4,9+13).3,5}{2}\\approx 31, 3 \\,$(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $31,3$<\/span><\/span>"}]}],"id_ques":1394},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["40"],["18"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho h\u00ecnh v\u1ebd, t\u00ecm $x, y$. <br\/><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai4/lv3/img\/H914_K2.png' \/><br\/>\u0110\u00e1p s\u1ed1: $\\,x=$_input_; $\\,\\,\\,\\,$$\\,y=$_input_<\/span>","hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng $c^2=c'a$ trong tam gi\u00e1c vu\u00f4ng $ABC$, ta c\u00f3:<br\/>$ \\begin{aligned} & {{30}^{2}}=y\\left( y+32 \\right) \\\\ & \\Leftrightarrow {{y}^{2}}+32y-900\\,\\,\\,\\,=0 \\\\ & \\Leftrightarrow {{y}^{2}}-18y+50y-900\\,\\,\\,\\,=0 \\\\ & \\Leftrightarrow y(y-18) + 50(y -18) = 0 \\\\ & \\Leftrightarrow \\left( y-18 \\right)\\left( y+50 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & y- 18 = 0 \\\\ & y + 50 = 0 \\\\ \\end{aligned} \\right. \\Leftrightarrow \\left[ \\begin{aligned} & y=18 \\,\\,\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ & y=-50\\,\\,\\,\\,\\text{(lo\u1ea1i)} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng $b^2=b'.a$ trong tam gi\u00e1c vu\u00f4ng $ABC$, ta c\u00f3: <br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\,\\,{{x}^{2}}=32\\left( 32+18 \\right) \\\\ & \\Leftrightarrow {{x}^{2}}={{40}^{2}} \\\\ & \\Leftrightarrow x\\,\\,\\,=40 \\\\ \\end{aligned}$ <br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $40$ v\u00e0 $18$<\/span><\/span>"}]}],"id_ques":1395},{"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":"Gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c: <br\/>$\\sin^6x+\\cos^6x$$+3\\sin^2x\\cos^2x=$_input_","hint":"\u00c1p d\u1ee5ng: $\\sin^2x+\\cos^2x=1$ v\u00e0 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\sin^6x+\\cos^6x$$+3\\sin^2x\\cos^2x$<br\/>$=\\sin^6x+\\cos^6x$$+3\\sin^2x\\cos^2x(\\sin^2x+\\cos^2x)$<br\/>$= \\sin^6x+3\\sin^4x\\cos^2x\\,$$+3\\sin^2x\\cos^4x+\\cos^6x$<br\/>$=(\\sin^2x+\\cos^2x)^3$<br\/>$=1$<br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$<\/span><\/span>"}]}],"id_ques":1396},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{2}{3}$","B. $\\dfrac{2}{5}$","C. $\\dfrac{1}{3}$"],"ques":"Cho $cotg\\,\\alpha=\\dfrac{2}{3}$. Gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c: <br\/>$M=\\dfrac{2sin\\,\\alpha }{sin\\,\\alpha+3cos\\,\\alpha}=$?","hint":"\u0110\u1eb7t $cotg\\,\\alpha=t$. S\u1eed d\u1ee5ng d\u1eef ki\u1ec7n \u0111\u1ec1 b\u00e0i v\u00e0 \u00e1p d\u1ee5ng: $cotg\\,\\alpha=\\dfrac{cos\\,\\alpha}{sin\\,\\alpha}$ t\u00ednh $cos\\,\\alpha$ v\u00e0 $sin\\,\\alpha$ theo $t$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $sin\\, \\alpha$ v\u00e0 $cos\\,\\alpha$ theo $t$<br\/><b> B\u01b0\u1edbc 2: <\/b> Thay v\u00e0o $M$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3:<br\/> $cotg\\,\\alpha=\\dfrac{cos\\,\\alpha}{sin\\,\\alpha}=\\dfrac{2}{3}\\\\ \\Rightarrow \\dfrac{cos\\,\\alpha}{2}=\\dfrac{sin\\,\\alpha}{3}=t\\\\ \\Rightarrow sin\\, \\alpha= 3t ; cos\\, \\alpha=2t$<br\/>$\\begin{align} M&=\\dfrac{2sin\\,\\alpha }{sin\\,\\alpha+3cos\\,\\alpha}\\\\ &=\\dfrac{2.3t}{3t+3.2t}\\\\ &=\\dfrac{2}{3}\\\\ \\end{align}$<br\/><b>C\u00e1ch 2:<\/b> Nh\u1eadn x\u00e9t: V\u00ec $cotg\\alpha=\\dfrac{2}{3}\\Rightarrow sin\\,\\alpha \\ne 0$. <br\/>Ta c\u00f3 th\u1ec3 th\u1ef1c hi\u1ec7n chia c\u1ea3 t\u1eed v\u00e0 m\u1eabu c\u1ee7a $M$ cho $sin\\,\\alpha$. \u0110\u01b0a v\u1ec1 bi\u1ec3u th\u1ee9c ch\u1ec9 ch\u1ee9a $cotg\\,\\alpha$<\/span>"}]}],"id_ques":1397},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai ","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho \u0111o\u1ea1n th\u1eb3ng $AB=2a$ v\u00e0 trung \u0111i\u1ec3m $O$ c\u1ee7a n\u00f3. Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AB$ v\u1ebd c\u00e1c tia $Ax, By$ vu\u00f4ng g\u00f3c v\u1edbi $AB$. Qua $O$ v\u1ebd m\u1ed9t tia c\u1eaft tia $Ax$ t\u1ea1i $M$ sao cho $\\widehat{AOM}=\\alpha < 90^o$. Qua $O$ v\u1ebd tia th\u1ee9 hai c\u1eaft tia $By$ t\u1ea1i $N$ sao cho $\\widehat{MON}=90^o$. Khi \u0111\u00f3,<br\/><b> C\u00e2u 1: <\/b> Di\u1ec7n t\u00edch tam gi\u00e1c $MON$ b\u1eb1ng $ \\dfrac{a}{2\\sin\\alpha.\\cos\\alpha}$<\/span>","select":["A. \u0110\u00fang","B. Sai"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai4/lv3/img\/H914_K1a.png' \/><\/center> Theo \u0111\u1ec1 b\u00e0i ta c\u00f3: $AB=2a \\Rightarrow OA=OB=a$ <br\/>Ta c\u00f3: $\\widehat{ONB}=\\widehat{AOM}=\\alpha$ $\\,$(c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat{BON}$)<br\/>X\u00e9t $\\Delta AOM$ c\u00f3:$ \\widehat{A}=90^o$<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c, ta c\u00f3:<br\/>$ OA=OM.\\cos \\alpha \\Rightarrow OM=\\dfrac{a}{\\cos\\alpha}$<br\/>X\u00e9t $ \\Delta BON$ c\u00f3: $ \\widehat{B}=90^o$<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c, ta c\u00f3:<br\/> $OB=ON.\\sin\\alpha\\Rightarrow ON=\\dfrac{a}{\\sin\\alpha}$<br\/>V\u1eady di\u1ec7n t\u00edch tam gi\u00e1c ${MON}$ l\u00e0:<br\/>$\\dfrac{1}{2}.OM.ON$$=\\dfrac{a^2}{2\\sin \\alpha.\\cos\\alpha}$<br\/>V\u1eady 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><\/span>","column":2}]}],"id_ques":1398},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["45"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho \u0111o\u1ea1n th\u1eb3ng $AB=2a$ v\u00e0 trung \u0111i\u1ec3m $O$ c\u1ee7a n\u00f3. Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AB$ v\u1ebd c\u00e1c tia $Ax, By$ vu\u00f4ng g\u00f3c v\u1edbi $AB$. Qua $O$ v\u1ebd m\u1ed9t tia c\u1eaft tia $Ax$ t\u1ea1i $M$ sao cho $\\widehat{AOM}=\\alpha < 90^o$. Qua $O$ v\u1ebd tia th\u1ee9 hai c\u1eaft tia $By$ t\u1ea1i $N$ sao cho $\\widehat{MON}=90^o$. <br\/><b> C\u00e2u 2: <\/b> X\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb c\u1ee7a $\\alpha$ \u0111\u1ec3 tam gi\u00e1c $MON$ c\u00f3 di\u1ec7n t\u00edch nh\u1ecf nh\u1ea5t<br\/><b> \u0110\u00e1p s\u1ed1:<\/b> $\\alpha=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$<\/span>","hint":"S\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy v\u00e0 $\\sin^2\\alpha+\\cos^2\\alpha=1$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai4/lv3/img\/H914_K1a.png' \/><\/center>G\u1ecdi di\u1ec7n t\u00edch tam gi\u00e1c $MON$ l\u00e0 $S_{MON}$.<br\/>Theo c\u00e2u 1, ta c\u00f3: $S_{MON}=\\dfrac{a^2}{2\\sin \\alpha.\\cos\\alpha}\\,\\,(1)$<br\/>\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy ta c\u00f3: <br\/>$\\sin^2\\alpha+\\cos^2\\alpha \\ge 2\\sin\\alpha.\\cos\\alpha\\\\ \\Rightarrow 1\\ge 2\\sin\\alpha.cos\\alpha$<br\/>T\u1eeb (1) ta c\u00f3: $S_{MON}=\\dfrac{a^2}{2\\sin\\alpha.\\cos\\alpha}\\ge \\dfrac{a^2}{1}=a^2$<br\/>D\u1ea5u b\u1eb1ng x\u1ea3y ra $\\Leftrightarrow \\sin \\alpha=\\cos\\alpha\\\\ \\Leftrightarrow \\sin\\alpha=\\sin(90^o-\\alpha)\\\\ \\Leftrightarrow\\alpha=90^o-\\alpha \\\\ \\Leftrightarrow \\alpha=45^o$<br\/>V\u1eady min $S_{MON}=a^2$ khi $\\alpha=45^o$ <br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $45$<\/span><br\/><span class='basic_green'>L\u01b0u \u00fd<\/span><br\/>N\u1ebfu $O$ kh\u00f4ng ph\u1ea3i l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB, O$ ch\u1ec9 n\u1eb1m gi\u1eefa $A$ v\u00e0 $B, OA= a$; $OB=b$ th\u00ec c\u0169ng ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 nh\u01b0 tr\u00ean ta \u0111\u01b0\u1ee3c min $S=ab \\Leftrightarrow \\alpha=45^o$<br\/><\/span>"}]}],"id_ques":1399},{"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":[[["30"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"T\u00ednh s\u1ed1 \u0111o g\u00f3c nh\u1ecdn $x$, bi\u1ebft: $cos^2x-sin^2x=\\dfrac{1}{2}$<br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $x=$_input_ $^o$","hint":"$\\sin^2x+\\cos^2x=1$","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$\\,\\,\\,\\,\\,\\,cos^2x-sin^2x=\\dfrac{1}{2}\\\\ \\Leftrightarrow cos^2x-(1-cos^2x)=\\dfrac{1}{2}\\\\ \\Leftrightarrow 2cos^2x=\\dfrac{3}{2}\\\\ \\Leftrightarrow cos x=\\dfrac{\\sqrt{3}}{2} \\,\\,(\\text{V\u00ec}\\, 0^o\\le x \\le 90^o\\Rightarrow cos\\, x >0)\\\\ \\Leftrightarrow x= 30^o$<br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $30$<\/span><\/span><\/span>"}]}],"id_ques":1400}],"lesson":{"save":0,"level":3}}