{"segment":[{"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":"X\u00e9t h\u00e0m s\u1ed1 $y=f\\left( x \\right)={{\\left( x+1 \\right)}^{2}}$ . V\u1edbi $-2\\le x\\le 2$ th\u00ec $f\\left( x \\right)$ nh\u1eadn c\u00e1c gi\u00e1 tr\u1ecb nh\u01b0 th\u1ebf n\u00e0o?","select":["A. $0\\le f\\left( x \\right)\\le 4$ ","B. $0\\le f\\left( x \\right)\\le 9$ ","C. $1\\le f\\left( x \\right)\\le 5$ ","D. $0\\le f\\left( x \\right)\\le 4$ "],"hint":"\u0110\u1eb7t $X=x+1$. D\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 $y=aX^2$ \u0111\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n.","explain":"<span class='basic_left'>\u0110\u1eb7t $X=x+1.$ <br\/> Khi \u0111\u00f3, h\u00e0m s\u1ed1 $y=f\\left( x \\right)=g\\left( X \\right)={{X}^{2}}$ c\u00f3 d\u1ea1ng $y=a{{x}^{2}}$ <br\/>V\u1edbi $-2 \\le x \\le 2$ th\u00ec $-1\\le X\\le 3$<br\/>V\u00ec $a=1>0$ n\u00ean h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn khi $X>0;$ ngh\u1ecbch bi\u1ebfn khi $X < 0 $ v\u00e0 \u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t l\u00e0 $y=0$ t\u1ea1i $X=0$<br\/>+ X\u00e9t $-1\\le X\\le 0$, h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn n\u00ean ta c\u00f3: <br\/>$g\\left( -1 \\right)\\ge g\\left( X \\right)\\ge g\\left( 0 \\right)\\Leftrightarrow 1\\ge f\\left( x \\right)\\ge 0$ <br\/>+ X\u00e9t $0\\le X\\le 3$, h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn n\u00ean ta c\u00f3:<br\/> $g\\left( 0 \\right)\\le g\\left( X \\right)\\le g\\left( 3 \\right)\\Leftrightarrow 0\\le f\\left( x \\right)\\le 9$ Suy ra v\u1edbi $-2\\le x\\le 2$ th\u00ec $0\\le f\\left( x \\right)\\le 9$.<br\/><br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":811},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy,$ cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):$ $y = -x + 2$ v\u00e0 parabol $(P):$ $ y = x^2.$<br\/>G\u1ecdi $A, B$ l\u00e0 hai giao \u0111i\u1ec3m c\u1ee7a $(d)$ v\u00e0 $(P).$ T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $OAB.$<br\/><b> \u0110\u00e1p s\u1ed1:<\/b> Di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ l\u00e0 _input_ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: V\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $(P)$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $(d).$ T\u1eeb \u0111\u00f3 x\u00e1c \u0111\u1ecbnh t\u1ecda \u0111\u1ed9 $A$ v\u00e0 $B$.<br\/>B\u01b0\u1edbc 2: H\u1ea1 $AH\\bot \\text{Ox}$ t\u1ea1i $H,$ $BK$ $\\bot$ $Ox$ t\u1ea1i $K$. T\u1eeb \u0111\u00f3 t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ qua di\u1ec7n t\u00edch $\\Delta{HAC}, \\Delta{AHO},\\Delta{OBC}$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/>Ta c\u00f3 \u0111\u1ed3 th\u1ecb c\u1ee7a $(P)$ v\u00e0 $(d)$:<br\/><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.2.png' \/><\/center><br\/><br\/>D\u1ef1a v\u00e0o h\u00ecnh v\u1ebd, ta c\u00f3 $A(-2;4)$ v\u00e0 $B(1;1)$<br\/>H\u1ea1 $AH\\bot Ox$ t\u1ea1i $H,$ $BK$ $\\bot$ $Ox$ t\u1ea1i $K$ v\u00e0 $C=\\left( d \\right)\\,\\bigcap Ox$ <br\/>Suy ra $H(-2;0),$ $K(1;0)$ v\u00e0 $C(2;0)$<br\/>Suy ra $AH=4,$ $BK=1$ v\u00e0 $HC=4.$<br\/>Ta c\u00f3:<br\/>$\\begin{align} & {{S}_{\\vartriangle HAC}}={{S}_{\\vartriangle AHO}}+{{S}_{\\vartriangle OAB}}+{{S}_{\\vartriangle OBC}} \\\\ & \\Rightarrow {{S}_{\\vartriangle OAB}}={{S}_{\\vartriangle HAC}}-{{S}_{\\vartriangle AHO}}-{{S}_{\\vartriangle OBC}} \\\\ \\end{align}$ <br\/>+ ${{S}_{\\Delta HAC}}=\\dfrac{1}{2}AH.HC=\\dfrac{1}{2}.4.4=8$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>+ ${{S}_{\\Delta AHO}}=\\dfrac{1}{2}AH.HO=\\dfrac{1}{2}.4.2=4$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>+ ${{S}_{\\Delta OBC}}=\\dfrac{1}{2}BK.OC=\\dfrac{1}{2}.1.2=1$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>Suy ra ${{S}_{\\Delta OAB}}=8-4-1=3$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3$<br\/><\/span><\/span>"}]}],"id_ques":812},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho parabol $y=2{{x}^{2}}$ (P). G\u1ecdi $M$ l\u00e0 \u0111i\u1ec3m t\u00f9y \u00fd n\u1eb1m tr\u00ean parabol $(P).$ G\u1ecdi $N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OM.$ Khi $M$ di chuy\u1ec3n tr\u00ean $(P)$ th\u00ec $N$ di chuy\u1ec3n tr\u00ean \u0111\u01b0\u1eddng n\u00e0o?. <\/span>","select":["A. $y={{x}^{2}}$ ","B. $y=\\dfrac{1}{2}{{x}^{2}}$ ","C. $y=3{{x}^{2}}$ ","D. $y=4{{x}^{2}}$ "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $N$ qua t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $M.$<br\/>B\u01b0\u1edbc 2: T\u00ecm m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa ho\u00e0nh \u0111\u1ed9 v\u00e0 tung \u0111\u1ed9 c\u1ee7a $N$ t\u1eeb \u0111i\u1ec1u ki\u1ec7n $M$ thu\u1ed9c $(P)$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.9.png' \/><\/center>Gi\u1ea3 s\u1eed $M(m;n)$ l\u00e0 \u0111i\u1ec3m b\u1ea5t k\u00ec thu\u1ed9c (P). Suy ra $n=2{{m}^{2}}$ (1)<br\/>$N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $MO$ n\u00ean $N\\left( x;y \\right)=N\\left( \\dfrac{m}{2};\\dfrac{n}{2} \\right)$ .<br\/>Suy ra $\\left\\{ \\begin{align} & m=2x \\\\ & n=2y \\\\ \\end{align} \\right.$ <br\/>Thay v\u00e0o (1), ta c\u00f3 $2y=2{{\\left( 2x \\right)}^{2}}\\Leftrightarrow y=4{{x}^{2}}$<br\/> V\u1eady khi $M$ ch\u1ea1y tr\u00ean $(P)$ th\u00ec $N$ ch\u1ea1y tr\u00ean parabol $y=4{{x}^{2}}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":813},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u $(>,=,<)$ th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[[">"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=f\\left( x \\right)=a{{x}^{2}}$ . Bi\u1ebft $x=\\dfrac{1}{4}$ th\u00ec $y=\\dfrac{3}{16}$ <br\/>V\u1edbi ${{x}_{1}}<0<{{x}_{2}}$ v\u00e0 $f\\left( {{x}_{1}} \\right)>f\\left( {{x}_{2}} \\right)$, ta c\u00f3 $\\left| {{x}_{1}} \\right|$ _input_ $\\left| {{x}_{2}} \\right|$ <\/span>","hint":"Ch\u1ee9ng minh $f\\left( \\left| {{x}_{1}} \\right| \\right)>f\\left( \\left| {{x}_{2}} \\right| \\right)$. T\u1eeb \u0111\u00f3 d\u1ef1a v\u00e0o t\u00ednh \u0111\u1ed3ng bi\u1ebfn khi $x>0$ \u0111\u1ec3 so s\u00e1nh $\\left| {{x}_{1}} \\right|$ v\u00e0 $\\left| {{x}_{2}} \\right|$","explain":"<span class='basic_left'>Ta c\u00f3 $x=\\dfrac{1}{4}$ th\u00ec $y=\\dfrac{3}{16}$ n\u00ean<br\/> $\\dfrac{3}{16}=a.{{\\left( \\dfrac{1}{4} \\right)}^{2}}\\Leftrightarrow a=3$ <br\/>V\u1eady h\u00e0m s\u1ed1 t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 $y=3{{x}^{2}}$ <br\/>Do $a=3>0$ n\u00ean h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn khi $x>0$ <br\/>V\u1edbi ${{x}_{1}}<0<{{x}_{2}}$ th\u00ec <br\/>$f\\left( \\left| {{x}_{1}} \\right| \\right)=f\\left( -{{x}_{1}} \\right)=a{{\\left( -{{x}_{1}} \\right)}^{2}}=a{{\\left( {{x}_{1}} \\right)}^{2}}=f\\left( {{x}_{1}} \\right)$ <br\/>$f\\left( \\left| {{x}_{2}} \\right| \\right)=f\\left( {{x}_{2}} \\right)$<br\/>M\u00e0 $f\\left( {{x}_{1}} \\right)>f\\left( {{x}_{2}} \\right)$ n\u00ean $f\\left( \\left| {{x}_{1}} \\right| \\right)>f\\left( \\left| {{x}_{2}} \\right| \\right)$ <br\/>Do $\\left| {{x}_{1}} \\right|,\\left| {{x}_{2}} \\right|$ c\u00f9ng d\u01b0\u01a1ng v\u00e0 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn khi $x>0$ n\u00ean $\\left| {{x}_{1}} \\right|>\\left| {{x}_{2}} \\right|$ <br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $>$<\/span><\/span>"}]}],"id_ques":814},{"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":"\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=x\\left| x \\right|$ l\u00e0 <\/span>","select":["A. <img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.1c.png' \/>","B. <img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.1.png' \/>","C.<img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.1b.png' \/>"],"hint":"\u00c1p d\u1ee5ng \u0111\u1ecbnh ngh\u0129a gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i \u0111\u1ec3 vi\u1ebft l\u1ea1i h\u00e0m s\u1ed1, sau \u0111\u00f3 v\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 tr\u00ean t\u1eebng kho\u1ea3ng x\u00e1c \u0111\u1ecbnh.","explain":"<span class='basic_left'> Ta c\u00f3 $\\left| x \\right|=\\left\\{ \\begin{align} & x \\, \\text{n\u1ebfu}\\,x\\ge 0 \\\\ & -x \\,\\text{n\u1ebfu}\\,x<0 \\\\ \\end{align} \\right.$ n\u00ean $y=x\\left| x \\right|=\\left\\{ \\begin{align} & {{x}^{2}}\\, \\text{n\u1ebfu}\\,x\\ge 0 \\\\ & -{{x}^{2}}\\, \\text{n\u1ebfu}\\, x<0 \\\\ \\end{align} \\right.$ <br\/>Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9, <br\/>+ V\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=x^2,$ x\u00f3a b\u1ecf nh\u00e1nh b\u00ean tr\u00e1i tr\u1ee5c $Oy$ v\u00e0 gi\u1eef nguy\u00ean nh\u00e1nh b\u00ean ph\u1ea3i tr\u1ee5c $Oy$<br\/>+ V\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=-x^2,$ x\u00f3a b\u1ecf nh\u00e1nh b\u00ean ph\u1ea3i tr\u1ee5c $Oy$ v\u00e0 gi\u1eef nguy\u00ean nh\u00e1nh b\u00ean tr\u00e1i tr\u1ee5c $Oy$<br\/>Ph\u1ea7n \u0111\u1ed3 th\u1ecb c\u00f2n l\u1ea1i ch\u00ednh l\u00e0 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=x\\left| x \\right|$<br\/><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.1.png' \/><\/center><br\/><br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<br\/><\/span><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>+ \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 tr\u00ean lu\u00f4n \u0111\u1ed3ng bi\u1ebfn v\u1edbi m\u1ecdi $x$<br\/>+ B\u00e0i to\u00e1n li\u00ean quan: Bi\u1ec7n lu\u1eadn s\u1ed1 nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh: $x\\left| x \\right|=m$<br\/>H\u01b0\u1edbng d\u1eabn: D\u1ef1a v\u00e0o \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1, ta c\u00f3 v\u1edbi m\u1ecdi $m$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 nghi\u1ec7m duy nh\u1ea5t<\/span>","column":3}]}],"id_ques":815},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho \u0111i\u1ec3m $A\\left( 0;\\dfrac{1}{2} \\right)$ v\u00e0 h\u00e0m s\u1ed1 $y=a{{x}^{2}}$ c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $B(-2;2).$ G\u1ecdi $M(x;y)$ l\u00e0 \u0111i\u1ec3m di chuy\u1ec3n tr\u00ean parabol n\u00f3i tr\u00ean. Ta c\u00f3 $MA=\\left| y-\\dfrac{1}{2} \\right|,$ <b> \u0111\u00fang<\/b> hay <b>sai<\/b>? <\/span>","select":["A. \u0110\u00fang ","B. Sai "],"hint":"V\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. K\u1ebb $MH\\bot Oy$. T\u00ednh $MA$","explain":"<span class='basic_left'>Parabol $y=a{{x}^{2}}$ c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $B(-2;2)$ n\u00ean ta c\u00f3 $2=a{{\\left( -2 \\right)}^{2}}\\Leftrightarrow a=\\dfrac{1}{2}$ <br\/>V\u1eady parabol t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 $y=\\dfrac{1}{2}{{x}^{2}}$ <br\/>Ta c\u00f3 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=\\dfrac{1}{2}{{x}^{2}}$:<br\/><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.4.png' \/><\/center><br\/><br\/>L\u1ea5y $M(x;y)$ b\u1ea5t k\u00ec tr\u00ean (P). K\u1ebb $MH\\bot Oy$. <br\/>Ta lu\u00f4n c\u00f3 $MH=\\left| x \\right|;AH=\\left| OH-OA \\right|=\\left| y-\\dfrac{1}{2} \\right|$ <br\/>X\u00e9t $\\Delta MHA$ vu\u00f4ng t\u1ea1i $H.$ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3: <br\/>$\\begin{aligned} & M{{A}^{2}}=M{{H}^{2}}+A{{H}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{x}^{2}}+{{\\left| y-\\dfrac{1}{2} \\right|}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{x}^{2}}+{{y}^{2}}-y+\\dfrac{1}{4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ \\end{aligned}$ <br\/>Do $y=\\dfrac{1}{2}{{x}^{2}}$ n\u00ean thay $x^2=2y$ v\u00e0o (1), ta c\u00f3:<br\/>$\\begin{aligned} & M{{A}^{2}}=2y+{{y}^{2}}-y+\\dfrac{1}{4} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{y}^{2}}+y+\\dfrac{1}{4} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{\\left( y+\\dfrac{1}{2} \\right)}^{2}} \\\\ \\end{aligned}$ <br\/>Do $y\\ge 0$ n\u00ean $MA=y+\\dfrac{1}{2}$<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<br\/><\/span><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>a) C\u00f3 th\u1ec3 t\u00ednh kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111i\u1ec3m $\\left( {{x}_{1}};{{y}_{1}} \\right)$ v\u00e0 $\\left( {{x}_{2}};{{y}_{2}} \\right)$ theo c\u00f4ng th\u1ee9c:<br\/> $d=\\sqrt{{{\\left( {{x}_{2}}-{{x}_{1}} \\right)}^{2}}+{{\\left( {{y}_{2}}-{{y}_{1}} \\right)}^{2}}}$ <br\/>b) Do $MA=y+\\dfrac{1}{2}$ n\u00ean m\u1ecdi \u0111i\u1ec3m $M$ thu\u1ed9c parabol $y=\\dfrac{1}{2}{{x}^{2}}$ \u0111\u1ec1u c\u00f3 kho\u1ea3ng c\u00e1ch $MA$ b\u1eb1ng kho\u1ea3ng c\u00e1ch t\u1eeb $M$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $y=-\\dfrac{1}{2}$.<br\/> Nh\u01b0 v\u1eady n\u1ebfu cho tr\u01b0\u1edbc \u0111i\u1ec3m $A\\left( 0;\\dfrac{1}{2} \\right)$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $d$ c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $y=-\\dfrac{1}{2}$ th\u00ec t\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m c\u00f3 kho\u1ea3ng c\u00e1ch \u0111\u1ebfn \u0111i\u1ec3m $A$ b\u1eb1ng kho\u1ea3ng c\u00e1ch \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $d$ l\u00e0 parabol $y=\\dfrac{1}{2}{{x}^{2}}$<br\/><b>T\u1ed5ng qu\u00e1t: <\/b> Cho tr\u01b0\u1edbc \u0111i\u1ec3m $A(0;a),$ $a \\ne 0$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $d$ c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $y=-a.$ T\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m c\u00f3 kho\u1ea3ng c\u00e1ch \u0111\u1ebfn $A$ b\u1eb1ng kho\u1ea3ng c\u00e1ch \u0111\u1ebfn d l\u00e0 parabol $y=\\dfrac{1}{4a}{{x}^{2}}$ <\/span>","column":2}]}],"id_ques":816},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'>Cho \u0111i\u1ec3m $A(0;a),$ g\u1ecdi d l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $y=-a.$ Ch\u1ee9ng minh qu\u1ef9 t\u00edch \u0111i\u1ec3m $M(x;y)$ sao cho kho\u1ea3ng c\u00e1ch $MH$ t\u1eeb $M$ \u0111\u1ebfn $d$ b\u1eb1ng $MA$ l\u00e0 m\u1ed9t parabol.<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.6.png' \/><\/center><\/span>","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c l\u1eddi gi\u1ea3i \u0111\u00fang","temp":"sequence","correct":[[[5],[1],[4],[2],[3]]],"list":[{"point":10,"image":"","left":["Do \u0111\u00f3 qu\u1ef9 t\u00edch c\u1ee7a $M$ l\u00e0 parabol $y=\\dfrac{1}{4{{a}^{2}}}{{x}^{2}}$","K\u1ebb $MH\\bot Ox$. Suy ra $MH=\\left| y+a \\right|$ n\u00ean $M{{H}^{2}}={{\\left( y+a \\right)}^{2}}={{y}^{2}}+2ay+{{a}^{2}}$ ","<span class='basic_left'>$\\begin{align} & \\Leftrightarrow {{x}^{2}}+{{y}^{2}}-2ay+{{a}^{2}}={{y}^{2}}+2ay+{{a}^{2}} \\\\ & \\Leftrightarrow {{x}^{2}}=4ay \\\\ & \\Leftrightarrow y=\\dfrac{1}{4a}{{x}^{2}} \\\\ \\end{align}$ <\/span>","<span class='basic_left'>L\u1ea1i c\u00f3:<br\/> $\\begin{align} & M{{A}^{2}}={{\\left( x-0 \\right)}^{2}}+{{\\left( y-a \\right)}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{x}^{2}}+{{y}^{2}}-2ay+{{a}^{2}} \\\\ \\end{align}$ <\/span>","<span class='basic_left'>Theo \u0111\u1ec1 ra: $M{{A}^{2}}=M{{H}^{2}}$<\/span>"],"top":110,"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.5.png' \/><\/center><br\/><br\/>K\u1ebb $MH\\bot Ox$. Suy ra $MH=\\left| y+a \\right|$ n\u00ean $M{{H}^{2}}={{\\left( y+a \\right)}^{2}}={{y}^{2}}+2ay+{{a}^{2}}$<br\/>L\u1ea1i c\u00f3:<br\/>$\\begin{align} & M{{A}^{2}}={{\\left( x-0 \\right)}^{2}}+{{\\left( y-a \\right)}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{x}^{2}}+{{y}^{2}}-2ay+{{a}^{2}} \\\\ \\end{align}$ <br\/>Theo \u0111\u1ec1 ra: $M{{A}^{2}}=M{{H}^{2}}$ <br\/>$\\begin{align} & \\Leftrightarrow {{x}^{2}}+{{y}^{2}}-2ay+{{a}^{2}}={{y}^{2}}+2ay+{{a}^{2}} \\\\ & \\Leftrightarrow {{x}^{2}}=4ay \\\\ & \\Leftrightarrow y=\\dfrac{1}{4a}{{x}^{2}} \\\\ \\end{align}$ <br\/>Do \u0111\u00f3 qu\u1ef9 t\u00edch c\u1ee7a $M$ l\u00e0 parabol $y=\\dfrac{1}{4{{a}^{2}}}{{x}^{2}}$ <\/span>"}]}],"id_ques":817},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$ cho \u0111i\u1ec3m $A(0;2)$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $\\Delta $ : $y=-2.$ T\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m $M(x;y)$ c\u00e1ch \u0111\u1ec1u \u0111i\u1ec3m $A$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $\\Delta $ l\u00e0 parabol c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh ","select":["A. $y=\\dfrac{1}{4}{{x}^{2}}$ ","B. $y=\\dfrac{1}{3}{{x}^{2}}$","C. $y=-\\dfrac{1}{8}{{x}^{2}}$","D. $y=\\dfrac{1}{8}{{x}^{2}}$"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.7.png' \/><\/center><br\/><br\/>K\u1ebb $MH\\bot \\Delta $ t\u1ea1i $H$. Suy ra $H\\left( x;-2 \\right)$<br\/>Theo gi\u1ea3 thi\u1ebft, ta c\u00f3 $MH=MA$ <br\/>$\\begin{align} & \\Leftrightarrow \\sqrt{{{\\left( y+2 \\right)}^{2}}}=\\sqrt{{{x}^{2}}+{{\\left( y-2 \\right)}^{2}}} \\\\ & \\Leftrightarrow {{y}^{2}}+4y+4={{x}^{2}}+{{y}^{2}}-4y+4 \\\\ & \\Leftrightarrow 8y={{x}^{2}} \\\\ & \\Leftrightarrow y=\\dfrac{1}{8}{{x}^{2}} \\\\ \\end{align}$ <br\/>V\u1eady qu\u1ef9 t\u00edch \u0111i\u1ec3m $M$ l\u00e0 parabol $y=\\dfrac{1}{8}{{x}^{2}}$<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span><\/span>","column":2}]}],"id_ques":818},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho parabol $(P):$ $y=a{{x}^{2}}.$ Bi\u1ebft $(P)$ \u0111i qua $A\\left( \\sqrt{3};-3 \\right).$ G\u1ecdi $d$ l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua $A$ v\u00e0 song song v\u1edbi tr\u1ee5c ho\u00e0nh, c\u1eaft $(P)$ t\u1ea1i $A$ v\u00e0 $B.$ Tam gi\u00e1c $OAB$ l\u00e0 tam gi\u00e1c g\u00ec?<\/span>","select":["A. Tam gi\u00e1c c\u00e2n ","B. Tam gi\u00e1c vu\u00f4ng c\u00e2n ","C. Tam gi\u00e1c \u0111\u1ec1u"],"hint":"T\u00ednh \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c.","explain":"<span class='basic_left'>V\u00ec $(P)$ \u0111i qua $A\\left( \\sqrt{3};-3 \\right)$ n\u00ean ta c\u00f3 $-3=a.{{\\left( \\sqrt{3} \\right)}^{2}}\\Leftrightarrow a=-1$ . <br\/>Suy ra $(P):$ $y=-{{x}^{2}}$ <br\/>\u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $(P)$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $d:$<br\/><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.3.png' \/><\/center><br\/><br\/>Ta c\u00f3 $A\\left( \\sqrt{3};-3 \\right)$n\u00ean $B\\left( -\\sqrt{3};-3 \\right)$ <br\/>Suy ra $AB=\\sqrt{{{\\left( {{x}_{A}}-{{x}_{B}} \\right)}^{2}}+{{\\left( {{y}_{A}}-{{y}_{B}} \\right)}^{2}}}=\\sqrt{{{\\left( 2\\sqrt{3} \\right)}^{2}}+{{0}^{2}}}=2\\sqrt{3}$<br\/>T\u01b0\u01a1ng t\u1ef1 ta t\u00ednh \u0111\u01b0\u1ee3c $OA=2\\sqrt{3};OB=2\\sqrt{3}$ <br\/>V\u1eady $OA=OB=AB$<br\/>Suy ra tam gi\u00e1c $OAB$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<br\/><\/span><\/span>","column":3}]}],"id_ques":819},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'>Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh sau b\u1eb1ng \u0111\u1ed3 th\u1ecb: $ x^2 < x+2 $ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 <\/span>","select":["A. $x<-1$ ho\u1eb7c $x>2$ ","B. $x<-1$ ","C. $x>2$ ","D. $ -1 < x < 2 $ "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: V\u1ebd \u0111\u1ed3 th\u1ecb hai h\u00e0m s\u1ed1 $y=x^2$ (P)v\u00e0 h\u00e0m s\u1ed1 $y=x+2$ (d)<br\/>B\u01b0\u1edbc 2: X\u00e1c \u0111\u1ecbnh giao \u0111i\u1ec3m $A,B$ c\u1ee7a hai \u0111\u1ed3 th\u1ecb. <br\/>B\u01b0\u1edbc 3: Ph\u1ea7n parabol n\u1eb1m ph\u00eda d\u01b0\u1edbi $(d)$ l\u00e0 cung $AOB$ c\u1ee7a $(P)$. Chi\u1ebfu vu\u00f4ng g\u00f3c cung n\u00e0y l\u00ean $Ox$ ta \u0111\u01b0\u1ee3c t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>V\u1ebd \u0111\u1ed3 th\u1ecb parabol $y={{x}^{2}}$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $\\left( d \\right):y=x+2$ tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. <br\/><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai19/lv3/img\/K6.8.png' \/><\/center><br\/><br\/>Ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a $(P)$ v\u00e0 $d$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh:<br\/> $\\begin{aligned} & {{x}^{2}}=x+2\\Leftrightarrow {{x}^{2}}-x-2=0\\Leftrightarrow \\left( x+1 \\right)\\left( x-2 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-1 \\\\ & x=2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>+ V\u1edbi $x=-1$ th\u00ec $y=1$<br\/>+ V\u1edbi $x=2$ th\u00ec $y=4$<br\/>V\u1eady $(P)$ c\u1eaft $(d)$ c\u1eaft nhau t\u1ea1i $A(-1;1)$ v\u00e0 $B(2;4).$ <br\/>Ph\u1ea7n parabol n\u1eb1m ph\u00eda d\u01b0\u1edbi $(d)$ l\u00e0 cung $AOB$ c\u1ee7a $(P)$. Chi\u1ebfu vu\u00f4ng g\u00f3c cung n\u00e0y l\u00ean $Ox$ ta \u0111\u01b0\u1ee3c t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $ -1 < x < 2 $ <br\/><br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/><b>(a) C\u00e1ch 2: <\/b> \u0110\u01b0a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh t\u00edch: $f\\left( x \\right).g\\left( x \\right)<0$ <br\/>Khi \u0111\u00f3 ta c\u00f3 hai tr\u01b0\u1eddng h\u1ee3p:<br\/>+ $\\left\\{ \\begin{aligned} & f\\left( x \\right)<0 \\\\ & g\\left( x \\right)>0 \\\\ \\end{aligned} \\right.$ <br\/>+ $\\left\\{ \\begin{aligned} & f\\left( x \\right)>0 \\\\ & g\\left( x \\right)<0 \\\\ \\end{aligned} \\right.$<br\/><b>(b) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $f(x) \\ge g(x)$ <\/b> ta c\u0169ng c\u00f3 c\u00e1ch gi\u1ea3i b\u1eb1ng \u0111\u1ed3 th\u1ecb t\u01b0\u01a1ng t\u1ef1, tuy nhi\u00ean ta ch\u1ec9 x\u00e9t ph\u1ea7n parabol n\u1eb1m ph\u00eda tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng (d). Chi\u1ebfu vu\u00f4ng g\u00f3c ph\u1ea7n parabol \u0111\u00f3 xu\u1ed1ng tr\u1ee5c ho\u00e0nh ta \u0111\u01b0\u1ee3c t\u1eadp nghi\u1ec7m. <\/span>","column":2}]}],"id_ques":820}],"lesson":{"save":0,"level":3}}