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

$\bigtriangleup ABC$ 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 $(DEF)$

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

$\dpi{150} \small Ta có:\angle DIB=\frac{\angle C}{2}+\angle IBC(1),\angle DIB=\frac{180-\angle A}{2}=\frac{\angle B+\angle C}{2}(2).Tu (1),(2)\rightarrow BI la phan giac \angle B\rightarrow cung AK=cung KC\rightarrow \Delta KAC cân$

$\dpi{150} \small b) Theo cau a ta co:BI la phan giac \angle B\rightarrow I la tam duong tron noi tiep \Delta ABC\rightarrow AI di qua diem chinh gia cung BC cố dinh\rightarrow dpcm$

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

cau a bai 4 de lam ban oi

 dat canh hv abcd=x roi ta co stam giac apq=s hinh vuong-s3hinh con lai roi la ra

$\dpi{120} \small Pt\Leftrightarrow \sqrt{\frac{x+7}{x+1}}-\sqrt{2x-1}=2x^2-8\Leftrightarrow \frac{(x+7)-(2x-1)(x+1)}{\sqrt{2x-1}+\sqrt{\frac{x+7}{x+1}}}= 2x^2-8\Leftrightarrow \frac{8-2x^2}{\sqrt{2x-1}+\sqrt{\frac{x+7}{x+1}}}= 2x^2-8\Leftrightarrow (2x^2-8)(1+\frac{1}{\sqrt{2x-1}+\sqrt{\frac{x+7}{x+1}}})=0\Leftrightarrow 2x^2-8=0\Leftrightarrow x=2.Vậy x=2 la nghiem duy nhat cua phuong trinh$

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

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

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

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

=7$

BAI 1 phai la n khac 0 moi dc

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

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

toi tuong=1

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

roi danh gia trong doan 0 ,1/2 la ra

$Đặt P la biểu thức ở vế trái cua BDT cần cm Áp dụng BDT \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z},với mọi x,y,z>0 ta có: P=\frac{1}{x(1-x)}+\frac{1}{y(1-y)}+\frac{1}{z(1-z)}=(\frac{1}{x}+\frac{1}{1-x})+(\frac{1}{y}+\frac{1}{1-y})+(\frac{1}{z}+\frac{1}{1-z})=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+(\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z})\geq \frac{9}{x+y+z}+\frac{9}{3-(x+y+z)} Chú y rang $0\leq x,y,z\leq \frac{1}{2}.Đăt S=x+y+z ta co P\geq \frac{9}{S}+\frac{9}{3-S}=\frac{27}{S(3-S)}= \frac{27}{\frac{9}{4}-(\frac{3}{2}-S)^2}.

$\,Ap \, dung\,BDT \, Bunhia,ta\,co \, (\sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab})^2\leq 3(a^2+b^2+c^2+2(ab+bc+ca)).Ta\, co\,a^2+b^2+c^2\geq ab+bc+ca\rightarrow \frac{1}{2} (a^2+b^2+c^2)\geq \frac{1}{2}(ab+bc+ca)\rightarrow \frac{3}{2}(a^2+b^2+c^2)-(a^2+b^2+c^2)\geq \,2(ab+bc+ca)-\frac{3}{2}(ab+bc+ca) \rightarrow \frac{3}{2}(a^2+b^2+c^2+ab+bc+ca)\geq a^2+b^2+c^2+2(ab+bc+ca)\rightarrow \frac{3}{4}(a+b+c)^2\geq a^2+b^2+c^2+2(ab+bc+ca)\rightarrow dpcm\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,$

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

$\small Kẻ HE song song voi AC\rightarrow HE vuong goc voi BH\rightarrow HB\leq EB,HC\leq CD.Do tu giac HEAD la hbh\rightarrow AH=ED\leq AE+AD\rightarrow HA+HB+HC\leq AB+AC(1) .CMTT\rightarrow HA+HB+HC\leq AB+BC(2),HA+HB+HC\leq AC+BC.Tu (1) (2) (3\rightarrow )HA+HB+HC\leq \frac{2}{3}(AB+BC+CA) \rightarrow dpcm$

D la diem sao cho HD song song voi AB

$cho a,b,c la 3 số thực ko am thoa man a^2+4b^2+9c^2=14.CMR:3b+8c+abc\leq 12$$cho a,b,c la 3 số thực ko am thoa man a^2+4b^2+9c^2=14.CMR:3b+8c+abc\leq 12$

$Vơi x1,x2,....,xn;a1,a2,.....an\geq 0.Ta co:x1a1+x2a2+....xnan\geq (x1+x2+....+xn)(a1^x1.a2^x2.....an^xn)^{\frac{1}{a1+a2+....an}}$