$\dpi{120} \small Ton \: tai\: day\left \{ x1,x2,.... xn\right \}=\left \{ 1,2,.....n \right \}\: thoa\: man\: x1+x2+.....+xk\vdots k \forall k=1,2.....n$

$\dpi{200} \small CMR:a\sqrt{\frac{b}{a+c}}+b\sqrt{\frac{c}{a+b}}+c\sqrt{\frac{a}{b+c}}\leq \frac{3\sqrt{3}}{4}\sqrt{\frac{(a+b)(b+c)(c+a)}{a+b+c}}$

$\bigtriangleup ABC$ nt(O,r).các đường cao $AD,BE,CF$.$H$ la trực tâm.CM

a)$(AH+BH+CH)^2\leq a^2+b^2+c^2$ ($a,b,c$ la 3 canh tam giac)

b) $a^2+b^2+c^2\geq 8\sqrt{3}p{R}'$ ($p$ là nửa chu vi tam giac $DEF$,${R}')$ la bán kính $(DEF)$

$\bigtriangleup ABC nt\left ( 0 \right ).P nằm trong \bigtriangleup ABC.Lấy Q sao cho BQ song song voi AC,PQ song song voi AB.Lấy R thuộc BC sao cho R,Q khac phia A,P và \angle PRQ=\angle ACB.CMR(PQR)tx (O)$

Cho các số thực dương $a,b,c$. Chứng minh:

$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3(\frac{a^3+b^3+c^3}{3abc})^\frac{1}{4} $

$\:Tim \:cac \:so \:tu \:nhien \:n/d(n)^{3}=4n \: \: \: \:$

$\dpi{150} \:Neu \: p=2\:thi \: moi\:n=2k \:deu \:thoa \:man \:. \:Gia \:su \p> 2,theo \:dly \:nho \:Fermat, \:ta \: co\: 2^{m(p-1)}\equiv 1(modp)\:.Lay \:n=m(p-1) \:voi \: m\equiv-1(mod p) \:\rightarrow n=m(p-1)\equiv 1(modp) \:va2^n-n\equiv 2^n-1\equiv 0(modp). \: \:Do \:co \:vo \:so \:so \: nguyen\: duong\:m \: sao \:cho \:m\equiv-1(mod p)\rightarrow ton \:tai \: vo\: so\: so\: nguyen\:duong \: n\: TMDK\:da \:cho\rightarrow dpcm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\dpi{150} \:Gia \: su\: so\:nguyen \:to \ \:TMDK \:da\:cho\rightarrow 5^{2p^2} \equiv 1(mod p)\:.Mat\neq p^2-1\vdots p-1 ,\:theo \: dly\:nho \: Fermat\: ta\:co \: 5^{2(p^2-1)}\equiv 1(modp)\rightarrow 5^2\equiv 1(mod p)\:\rightarrow p=2hoac3 \: .Thu\:lai \:ta \:tim \:dc \=3 \: \: \: \: \: \: \:$

$\dpi{150} \;Su \;dung \;BDT \;AM-GM, \;ta \;co: \;ab+2abc=2ab(c+\frac{1}{2})\leq 2a(\frac{b+c+\frac{1}{2}}{2})^2=2a(\frac{7-2a}{4})^2. \;Nhu \;vay \;can \;CM: \; a+2a(\frac{7-2a}{4})^2\leq \frac{9}{2}(du doan min la \frac{9}{2})\Leftrightarrow (4-a)(2a-3)^2\geq 0(luon dung)\; .Dau\;= \;xay \;ra \; \Leftrightarrow \; (a,b,c)=(\frac{3}{2},1,\frac{1}{2})\; \; \; \; \; \; \; \; \; \; \;$

$$$\;Goi ; Ia\;,Ib, \;Ic \; theo\; thu\; tu\; la \; tam\;duong \;tron \;bang \;tiep \;cac \;goc \;A ,B,C cua tam giac ABC; .De\, dang \, CM \, duoc\, IbIc\, song\, song \, EF,IcIa song song FD,IaIb song song DE .Do do ton tai phep vi tu f ma bien Ia,Ib,Ic lan luot thanh D,E,F \rightarrow f \, bien \, duong \, thang\, Euler\, cua\, tam \, giac\, IaIbIc \, thanh \, duong \, thang\, Euler\, cua\, tam \, giac \, DEF \; Mat\, khac\, I \, nam \, tren\, duong\, thang\, Euler \, cua\, tam\, giac\, IaIbIc\, suy \, ra\, I\, nam\, tren\, duogn\, thang\, Euler \, cua \, tam\, giac\, DEF\, \rightarrow duong\, thang\, Euler\, cua \, tam \, giac \, DEF \, di\, qua\, I\, co \, dinh;$$$

$\;Bạn\;xem \; tài\;liệu \; chuyên\;toán \;10 \;để \;rõ \; thêm\, chi\, tiết$

@ supermember: Lần sau những bài viết nghèo nàn về nội dung & không nghiêm túc về hình thức như thế này sẽ bị xoá . Bạn rút kinh nhiệm nhé

$\; \;Goi \; J=\;AE\cap DF \;. \;Ta\, co:\Delta BIT\sim \Delta JHT \;\rightarrow \overline{IT}.\overline{TJ}=\overline{BT}.\overline{TH }=\overline{BI}.\overline{ID}=AI^2\rightarrow (AETJ)=-1\rightarrow C(AETJ)=-1\rightarrow (DFHJ)=-1= (AETJ)\rightarrow AF,DE,BC \, dong\, qui\rightarrow dpcm \;. \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;$

$\;Ap \; dung\;dly \; ham\; so\; sin\;trong \; \Delta BOC,\;ta \; co\;:R_{1}=\frac{BC}{2sin\angle BOC} =\frac{a}{2sin\left [ 180-\frac{\angle B+\angle C}{2} \right ]}= \frac{2RsinA}{2sin\frac{\angle B+\angle C}{2}}=\frac{2Rsin\frac{A}{2}cos\frac{A}{2}}{2cos\frac{A}{2}}=2Rsin\frac{A}{2},\;tuong \; tu\;cho \;R_{2} ,R_{3}\rightarrow R_{1}.R_{2}.R_{3}=8R^3sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\;.Lai \;co \; r=4Rsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}(\;bo \; de\;nay \;de \; ban\;tu \;CM\: nhe )\rightarrow \;R_{1}.R_{2}.R_{3}=2R^2r\rightarrow dpcm \; \;$

$\dpi{150} \small \:Ban \:tu \: ve\: hinh\:nha \:. \: \: \: \: \: \: \: \: \: \: \: \:Bai \:lam \:cua \: minh:\:Goi \:tam \:duong \:tron \:ngoai \: tiep\:\Delta ABC\rightarrow OA=OB=OC=R\:.Ta \:de \:dang \:cm\:dc\::OA\ vuong\:goc \: voi\: KI,\:OB \: vuong\:goc \: voi\: KH\:, \:OC \:vuong \:goc \: voi\: HI\:(cai\:nay \:ban \:tu \:CM \:nhe \:(de lam )\: ) \: \rightarrow \: S\:tu \:giac \:OKAI,OKBH,OICH \:lan \: luot\:=R.KI,R.KH,R.IH \: \rightarrow S\Delta ABC=R(KI+KH+IH)=R.p\Delta HIK\: \:Goi \:tam \:duong \: tron\: noi \:tiep \: \Delta HIK\:la \: r\rightarrow \SDelta HIK=r.p.\Delta HIK\rightarrow \frac{S\Delta HIK}{S\Delta ABC}=\frac{r.p\Delta HIK}{R.p\Delta HIK}=\frac{r}{R}$

$\dpi{150} \small \: Ap\:dung \: BDT\: C-S,\:ta \: co\:: \:\frac{MB^2}{AB^2}+\frac{MC^2}{AC^2} \geq \frac{(MB+MC)^2}{AB^2+AC^2}\geq \frac{BC^2}{AB^2+AC^2}=1\rightarrow dpcm\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$x^{2}+y^{3}=7$

$x^{2}+y^{3}

=7$

$\sqrt{\frac{b+c}{a}} +\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\geq \sqrt{\frac{16(a+b+c)^{3}}{3(a+b)(b+c)(a+c)}}$

D lý wilson a anh

$\small Đat:b+c-a=x,a+c-b=y,a+b-c=z\rightarrow x+y+z=a+b+c\rightarrow a=\frac{y+z}{2},b=\frac{x+z}{2},c=\frac{x+y}{2}.Sau đó the vao bieu thuc dung bdt Cauchy la ra$

$\dpi{150} \small \:AB=2a\:\rightarrow AH=a. \: Ta\:có \:\OH^2=R^2-a^2\rightarrow \:OH=\sqrt{R^2-a^2}\rightarrow HE=OE-OH=R-\sqrt{R^2-a^2} \:Ta \: có\:AE^2=AH^2+HE^2=a^2+(R-\sqrt{R^2-a^2})^2=2R^2-2R\sqrt{R^2- a^2}+a^2=R^2-2R\sqrt{R^2-a^2}+R^2-a^2+2a^2=(R-\sqrt{R^2-a^2})^2+2a^2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Cho $a,b,c$ là các số thực dương. Chứng minh rằng:

$\frac{\sqrt{a+b}}{c}+\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$

$\dpi{150} \small Dat \sqrt[3]{12-x}=a,\sqrt[3]{14+x}=b.Ta co:\left\{\begin{matrix} & a+b=2(1) & \\ & a^3+b^3=26(2) & \end{matrix}\right.(1)\rightarrow a=2-b,thay vao (2),ta co:(2-b)^3+b^3=26\Leftrightarrow 8-12b+6b^2=26\Leftrightarrow 6b^2-12b-18=0\Leftrightarrow 6(b-3)(b+1)=0\rightarrow b=3, a=-1 hoac b=-1,a=3\rightarrow x=13 hoac x=-15$

Ta \; co\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;(m-n)^3+(m+n)^3=2m^3+6mn^2 \; \; \; \; \; \; \; \; \; \; \;Ta \;tim \;nghiem \;\; \;co \;dang \; (x,y,z,t)=(a-b,a+b,\frac{c}{2}-\frac{d}{2},\frac{c}{2}+\frac{d}{2})\; voi\;a,b,c,d \;la \;cac \;so \;nguyen \;.De \;thay \;(x,y,z,t)=(10,10,-1,0) \; la\;1 \;nghiem \;.Ta \; se\; tim\; nghiem\;voia=10 \;va \;c=1 co \; dinh\;. \;De \; thay\;(x,y,z,t)=(10-b,10+b,\frac{-1}{2}-\frac{d}{2},\frac{-1}{2}+\frac{d}{2}) \; la\;nghiem \;cua \;phuong \;trinh \;da \;cho\Leftrightarrow (2000+60b^2)-\frac{1+3d^2}{4} =1999\;hay \; \;d^2-80b^2=1 \;.De \;thay \; phuong\; trinh\:Pell \: d^2-80b^2=1\: co\:nghiem \:nguyen \:duong \:nho \:nhat \:la \:(d1,b1)=(9,1) \: .Do\:Pt \:Pell \:co \:vo \:so \:nghiem \: nen\:Pt \: da\:cho \:cung \:\:co \: vo\: so\:nghiem\\rightarrow dpcm \: \: