Bài tập nâng cao bài Ôn tập chương I

Home Lớp 9 Toán lớp 9 Ôn tập chương I - Hình học 9 Bài tập nâng cao

{"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":"<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\/>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $30,64$<\/span><\/span>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\/>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}$$\\approx 30,64 \\,(cm^2)$<\/span>"}]}],"id_ques":1391},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["7,5"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AB= 3, BC= 5$. C\u00e1c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong v\u00e0 ph\u00e2n gi\u00e1c ngo\u00e0i c\u1ee7a g\u00f3c $B$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $AC$ t\u1ea1i $M$ v\u00e0 $N$. \u0110\u1ed9 d\u00e0i $MN$ l\u00e0:_input_","hint":"Tam gi\u00e1c $MBN$ vu\u00f4ng t\u1ea1i $B$","explain":"H\u01b0\u1edbng d\u1eabn <\/span><br\/> B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago t\u00ednh c\u1ea1nh $AC$<br\/> B\u01b0\u1edbc 2: <\/b> \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c \u0111\u1ec3 t\u00ednh $MA$ <br\/> B\u01b0\u1edbc 3: <\/b> T\u00ednh c\u1ea1nh $AN$ d\u1ef1a v\u00e0o h\u1ec7 th\u1ee9c $h^2=b'c'$ <br\/>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai4/lv3/img\/H914_K5.png' \/><\/center><br\/>X\u00e9t $\\Delta ABC$ c\u00f3: $ \\widehat{A}=90^o$. \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago ta c\u00f3:<br\/>$BC^2=AB^2+AC^2 \\\\ \\Rightarrow AC^2=5^2-3^2\\\\ \\Rightarrow AC^2 =16 \\Rightarrow AC=4$<br\/> Ta l\u1ea1i c\u00f3 $BM$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $B$ (gi\u1ea3 thi\u1ebft)<br\/> $\\Rightarrow \\dfrac{MA}{MC}=\\dfrac{BA}{BC}=\\dfrac{3}{5}$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c)<br\/>$\\Rightarrow \\dfrac{MA}{3}=\\dfrac{MC}{5}$$=\\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\/>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":"H\u00ecnh thang $ABCD$ c\u00f3 $AB \/\/ CD$; $\\,\\widehat{C}+\\widehat{D}=90^o$. Bi\u1ebft $AD= 4; BC= 7$ <br\/> 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":"H\u01b0\u1edbng d\u1eabn<\/span><br\/> B\u01b0\u1edbc 1: <\/b> K\u1ebb th\u00eam $BE \/\/CD$<br\/> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $\\Delta BCE$ vu\u00f4ng<br\/> B\u01b0\u1edbc 3: <\/b> T\u00ednh g\u00f3c $C$v\u00e0 g\u00f3c $D$<br\/> 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\/>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":"H\u00ecnh thang $ABCD$ c\u00f3 $AB \/\/ CD$; $\\,\\widehat{C}+\\widehat{D}=90^o$. Bi\u1ebft $AD= 4; BC= 7$<br\/> 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":"<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}$$\\approx\\dfrac{(4,9+13).3,5}{2}\\approx 31, 3 \\,$(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $31,3$<\/span><\/span>"}]}],"id_ques":1394},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["40"],["18"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho h\u00ecnh v\u1ebd, t\u00ecm $x, y$. <br\/><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai4/lv3/img\/H914_K2.png' \/><br\/>\u0110\u00e1p s\u1ed1: $\\,x=$_input_; $\\,\\,\\,\\,$$\\,y=$_input_<\/span>","hint":"","explain":"\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\/>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$$+3\\sin^2x\\cos^2x=$_input_","hint":"\u00c1p d\u1ee5ng: $\\sin^2x+\\cos^2x=1$ v\u00e0 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c.","explain":"Ta c\u00f3:<br\/>$\\sin^6x+\\cos^6x$$+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\/>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":"H\u01b0\u1edbng d\u1eabn<\/span><br\/> B\u01b0\u1edbc 1: <\/b> T\u00ednh $sin\\, \\alpha$ v\u00e0 $cos\\,\\alpha$ theo $t$<br\/> B\u01b0\u1edbc 2: <\/b> Thay v\u00e0o $M$ <br\/>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\/>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":"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\/> 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":"<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\/>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":"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\/> 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\/> \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":"<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\/>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $45$<\/span><br\/>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\/> \u0110\u00e1p \u00e1n: <\/b> $x=$_input_ $^o$","hint":"$\\sin^2x+\\cos^2x=1$","explain":"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\/>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $30$<\/span><\/span><\/span>"}]}],"id_ques":1400}],"lesson":{"save":0,"level":3}}

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Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên	+ 5 điểm
Trong khoảng 1 phút -> 2 phút	+ 4 điểm
Trong khoảng 2 phút -> 3 phút	+ 3 điểm
Trong khoảng 3 phút -> 4 phút	+ 2 điểm
Trong khoảng 4 phút -> 5 phút	+ 1 điểm

Quá 5 phút: không được cộng điểm

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