$\dpi{150} \small Ta co:x^6+y^4\geq 2\sqrt{x^6y^4}=2x^3y^2\rightarrow \frac{2x}{x^6+y^4}\leq \frac{1}{x^2y^2}\rightarrow \sum \frac{1}{x^2y^2}\geq \sum \frac{2x^6}{x^6+y^4}.Lai co \sum \frac{1}{x^2y^2}\leq \sum \frac{1}{x^4}(BDT Co si)\rightarrow dpcm$

http://diendantoanho...-p8k1/?p=374335

cach lam o day ne

1. x^3-x^2-x=1/3tuong duong voi 3x^3-3x^2-3x=1  tuong duong voi 4x^3=(x+1)^3ta tim ra dc x ok nhe ban
2.                                

$\dpi{150} \small \:Dat \:b _{n}=\frac{1}{u_{n}}\rightarrow \:b_{1}=2 \:va \: b_{n+1}=b_{n}^2-b_{n}+1\rightarrow \frac{1}{b_{n+1}-1}=\frac{1}{b_{n}(b_{n}-1)}=\frac{1}{b_{n}}-\frac{1}{b_{n}-1}\rightarrow \frac{1}{b_{n}}=\frac{1}{b_{n-1}}-\frac{1}{b_{n+1}-1}\ \:Ta \:có \::\sum_{i=1}^{n} u_{i}=u1+u2+...+un=\frac{1}{b_{1}}+\frac{1}{b_{2}}+.....+\frac{1}{b_{n}}= \frac{1}{b_{1}-1}-\frac{1}{b_{2}-1}+\frac{1}{b_{2}-1}-\frac{1}{b_{3}-1}+.....+\frac{1}{b_{n}-1}-\frac{1}{b_{n+1}-1}=\frac{1}{b_{1}-1}-\frac{1}{b_{n+1-1}}=\frac{1}{2-1}-\frac{1}{b_{n+1}-1}< 1\rightarrow dpcm\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

bai cuoi phai co them dkien m,n nguyen duong moi lam dc

Cho tam giác $ABC$. Dựng về phía ngoài tam giác hình vuông $ABDE$ và $ACIJ$ sao cho $C$ và $D$ nằm khác phía so với $A$ và $B$. Chứng minh giao của $BI$ và $CD$ thuôc đường cao $AH$ của tam giác $ABC$.

minh làm roi nhé ban tu ve hinh

$\;BDT \;can \;cm \;\Leftrightarrow \;(a+b+c)(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{a+c})\leq 3(a^2+b^2+c^2)\Leftrightarrow \left [ (a^2+b^2)+\frac{c(a^2+b^2)}{a+b} \right ] +\left [(b^2+c^2) +\frac{a(b^2+c^2)}{b+c} \right ]+\left [(c^2+a^2)+\frac{b(c^2+a^2)}{a+c} \right ]\leq 3(a^2+b^2+c^2)\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2\Leftrightarrow \left [c^2-\frac{c(a^2+b^2)}{a+b} \right ]+\left [a^2-\frac{a(b^2+c^2)}{b+c} \right ]+\left [ b^2-\frac{b(c^2+a^2)}{a+c} \right ]\geq 0\;\Leftrightarrow \frac{bc(b-c)+ba(b-a)}{c+a}+\frac{ca(c-a)+cb(c-b)}{a+b}+\frac{ab(a-b)+ac(a-c)}{b+c}\geq 0\Leftrightarrow \frac{ca(c-a)^2}{(a+b)(b+c)}+\frac{bc(b-c)^2}{(c+a)(a+b)}+\frac{ab(a-b)^2}{(b+c)(c+a)}\geq 0\rightarrow dpcm\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;$

$\dpi{150} \:Gia \:su \:(a,b)=d\rightarrow a=k1d,b=k2d\rightarrow 2^a-1=2^{k1d}-1=(2^d-1).A,2^b-1=2^{k2d} -1=(2^d-1).B\rightarrow dpcm\: \: \: \: \: \: \: \: \: \: \: \: \:$

BẠN oi C o dau the

tuong ve phai la -1 chu

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

$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$

$\inline Ta có:(8x+1)^{2}<100x^{2}+39x+\sqrt{3}<(10x+2)^{2}(dung voi moi x)\Rightarrow 8x+1<\sqrt{100x^{2}+39x+\sqrt{3}}<10x+2 \Rightarrow (4x+1)^{2}<16x^{2}+\sqrt{100x^{2}+39x+\sqrt{3}}<16x^{2}+10x+2<(4x+2)^{2} \Rightarrow 4x+1<\sqrt{16x^{2}+\sqrt{100x^{2}+39x+\sqrt{3}}}<4x+2 \Rightarrow (2x+1)^{2}<4x^{2}+\sqrt{16x^{2}+\sqrt{100x^{2}+39x+\sqrt{3}}}<4x^{2}+4x+2<(2x+2)^{2} \Rightarrow 2x+1<\sqrt{4x^{2}+\sqrt{16x^{2}+\sqrt{100x^{2}+39x+\sqrt{3}}}}<2x+2 \Rightarrow (x+1)^{2}

to chua quen go

$\Leftrightarrow \overline{ab}^{3}=\overline{abcde}$$\Leftrightarrow \overline{ab}^{3}=\overline{abcde}$$\Leftrightarrow $Đặt\overline{ab}=x,\overline{cde}=y (10\leq x\leq 99),100\leq y\leq 999$$ta co x^{3}=1000x+y,y\geq 0\rightarrow x^{3}\geq 1000x\rightarrow x(x^{2})-1000)>0\rightarrow x^{2}> 1000\rightarrow x> 31$$.Tip theo ta co y< 1000\rightarrow x^{3}< 1000x+1000\rightarrow x^{3}-1000x< 1000\rightarrow x(x^{2}-1000)< 1000.Neu x\geq 33\rightarrow x(x^{2}-1000)\geq 1000\rightarrow x< 33\rightarrow x=32$$vay ta dc so can tim la 32768 va ab=32$

$\dpi{150} \small Dat a=\overline{xy},b=\overline{ztu}.Tu gt suy ra:(a+b)^2=1000a+b\Leftrightarrow (a+b)(a+b-1)=999a(1).Lai co (a+b)^2\leq 9999\rightarrow a+b\leq 316.Ket hop voi (1)\rightarrow (a+b)\vdots 27 va (a+b-1)\vdots 37 hoac (a+b)\vdots 37 va (a+b-1)\vdots 27. TH1:a+b=27m,a+b-1=37n\rightarrow 27m=37n+1.GIai ta dc m=37k+11,n=27k+8(k\geq 0).THAy vao (1) ta dc a=mn=(37k+11)(27k+8)(k\geq 0).Vi 10\leq a\leq 99 nen k=0.khi do a=88,b=209 va\overrightarrow{xyztu}=88209.TH2: ta co 37m=27n+1 giai ta dc m=27k-8 va n=37k-11 THAy vao (1) ta dc a=mn=(27k-8)(37k-11)vi k\geq 1\rightarrow a\geq 19.26> 100(loai).Vay x=y=8,z=2,t=0,u=9$

$\;Xet \;dk \;thu \; 2:\;Voi \;n=1,2,3 \;thi \;m< \frac{1}{4} (thay\: vao\: BDT\:la \:dc )\;.Voi \;n=4 \; thi\; m=\frac{1}{4}.\; Nhu\; vay\; ta\;se \;chon \;m=\frac{1}{4} \;va \;diem \;xuat \; \;phat \;qui \; nap\;la \;n=4 \;Voi \;n=4 \;thay \;vao \;BDt \;ta \; dc\; 1066<1071(thoa man).\;Gia \;su \;BDT \;dg \;voi \;n=k\rightarrow S_{k}=\frac{1}{k+1}+\frac{1}{k+2}+.....+\frac{1}{2k} <\frac{7}{10}-\frac{1}{4k}\;,ta \; phai\;Cm \;BDT \;dg \;voi \; n=k+1\; \;hay S_{k+1} =\frac{1}{k+2}+\frac{1}{k+3}+....+\frac{1}{2k+2}< \frac{7}{10}-\frac{1}{4(k+1)}.Theo\;gt \;qui \;nap \;ta \;co \;S_{k+1}=S_{k}-\frac{1}{k+1} +\frac{1}{2k+1}+\frac{1}{2k+2}=Sk+\frac{1}{2(k+1)(2k+1)}< \frac{7}{10}-\frac{1}{4k}+\frac{1}{2(k+1)(2k+1)}.\;Nhu \; vay\;chi \;can \;CM \;\frac{-1}{4k}+\frac{1}{2(k+1)(2k+1)}< \frac{-1}{4(k+1)}\Leftrightarrow \frac{2}{(k+1)(2k+1)}< \frac{1}{k(k+1)}\Leftrightarrow 2k<2k+1\Leftrightarrow 0<1\rightarrow Bai \;toan \;dc \;CM \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;$

Đay chinh la BDT Holder cho 3 so ma ban

$\;Dat \;S_{k}=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n} \;. \; Neu\;giai \;bai \; toan\; bang\;p^2 \; qui\; nap\; thong\;thuong \;thi \;kho \;ma \;giai \;dc \;. \; Ta\;se \; tim\;1 \;so \;thuc \; m/BDT\;sau \; dung:\;\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}<\frac{7}{10}-\frac{m}{n} \; .So\; m\;phai \;thoa \; man\; 2\;dk: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (+)\;Buoc \;chuyen \;qui \;nap \;tu \;k \;sang \; k+1\;phai \; lam\; dc\; \; \; \; \; \; \; \; \(+); \;BDT \; tren\; phai\;dung \; voi\; gia\; tri\; dau\;cua \;n(co\:the \:\neq gia \:tri \:dau\:cua \:BDT \\:de \: bai ) \;.Xet \;dk \;1,ta \; co:S_{k+1}=Sk+\frac{1}{2(k+1)(2k+1)}< \frac{7}{10}-\frac{m}{k}+\frac{1}{2(k+1)(2k+1)}\Leftrightarrow \frac{1}{2(k+1)(2k+1)}+\frac{-m}{k}< \frac{-m}{k+1}\Leftrightarrow \frac{1}{2(k+1)(2k+1)}< \frac{m}{k(k+1)} \Leftrightarrow \;2m(2k+1)>k \Leftrightarrow (4m-1)k+2m>0\\;BDT \;cuoi \;nay \; dung\; voi\;moi \;k\Leftrightarrow m\geq \frac{1}{4} \; \; \; \; \; \; \; \; \; \; \; \;$

$\dpi{150} \small $\dpi{150} \small \:Theo \:Bdt \: Holder\: ta\: co\: \sum \sqrt{\frac{b+c}{a}})^2\left [ \sum \frac{1}{a^2(b+c)} \right ]\geq(\sum \frac{1}{}a)^3 \:Do \:đo \:ta \:chi \:can \:CM \ \sum \frac{1}{a})^3\geq \frac{6(a+b+c)}{\sqrt[3]{abc}}\sum \frac{1}{a^2(b+c)}.\:Đặt \:a=\frac{1}{x} ,b= \frac{1}{y},c= \frac{1}{z}\:Bdt\Leftrightarrow \:(x+y+z)^3\geq \6(xy+yz+zx)(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y})\:hay \:\frac{(x+y+z)^3}{\sqrt[3]{xyz}} \geq 6(x^2+y^2+z^2)+6xyz(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y})\:Áp \dụng \:Bdt Cauchy-Schwarz \: ,\:ta \:có \:6xyz(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y})\leq 6xyz(\frac{1}{4x}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4x})= 3(xy+yz+zx)\:\frac{(x+y+z)^4}{\sqrt[3]{xyz}}\geq \3(x+y+z)^3. \:Như \:vậy \:ta \:chỉ \:cần \:CM \::3(x+y+z)^3\geq 6(x^2+y^2+z^2)+3(xy+yz+zx).Sau khi thu gọn ta dc dpcm.Đay \:là \:cách \: làm\:của \:em \moi : \:nguoi\:xem \:ho \:em \:nhe \: \: \: \: \: \: \: \: \: \:$$

D la diem sao cho HD song song voi AB

$\dpi{200} \small Ta có x=y=z=0 ko phai la1 no cua he phuong trinh.Xét xyz\neq 0 Ta có:Đây là hệ hoán vị vòng quanh,dễ cm dc (x,y,z)la no thi x=y=z. Vậy x=y=z=\frac{-8}{7}$

$\dpi{150} \small \:hinh \:nhu \:la \:nam \:1766 \:dung \:ko \: a\: \: \:$

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