\(PT\Leftrightarrow5x^2+x\left(5y-7\right)+5y^2-14y=0\)

\(\Delta=\left(5y-7\right)^2-4.5.\left(5y^2-14y\right)\)

   \(=196-3\left(5y-7\right)^2\)

Để phương trình có nghiệm thì \(\Delta\ge0\Rightarrow\left(5y-7\right)^2\le65\)

Mặt khác \(5y-7\equiv3\left(mod5\right)\)

\(\Rightarrow\left(5y-7\right)^2\equiv4\left(mod5\right)\)

do đó \(\left(5y-7\right)^2\in\left\{4,9,14,19,24,29,34,39,44,49,54,59,64\right\}\)

mà (5y-7)² là số chính phưng nên \(\left(5y-7\right)^2\in\left\{4,9,64\right\}\)

Từ đó tính ra

\(5\left(x^2+xy+y^2\right)=7\left(x+2y\right)\)

\(\Leftrightarrow5x^2+5xy+5y^2-7x-14y=0\)

\(\Leftrightarrow5x^2+x\left(5y-7\right)+5y^2-14y=0\)

\(\Rightarrow\Delta_x=\left(5y-7\right)^2-4\cdot5\cdot\left(5y^2-14y\right)\)

\(=-75y^2+210y+49\)

\(=196-3\left(25y^2-2\cdot5y\cdot7+79\right)\ge0\)

\(=196-3\left(5y-7\right)^2\ge0\)

Để phương trình có nghiệm nguyên thì \(\Delta_x\ge0\Leftrightarrow\left(5y-7\right)^2\le65\)

Nhận thấy \(5y-7\equiv3\left(mod5\right)\Rightarrow\left(5y-7\right)^2\equiv4\left(mod5\right)\)

Do đó \(\left(5y-7\right)^2\in\left\{4;9;14;19;24;29;34;39;44;49;54;59\right\}\)

Mà \(\left(5y-7\right)^2\)chinh phương nên \(\left(5y-7\right)^2\in\left\{4;9;49\right\}\)

Đến đây ta xét trường hợp là ra.

Ta có \(\left(a-1\right)^2\left(a^2+a+1\right)\ge0\)\(\Leftrightarrow\left(a^2-2a+1\right)\left(a^2+a+1\right)\ge0\)

\(\Leftrightarrow a^4-a^3-a+1\ge0\)

\(\Leftrightarrow a^4-a^3+1\ge a\)

\(\Leftrightarrow a^4-a^3+ab+2\ge a+ab+1\)

\(\Rightarrow\frac{1}{\sqrt{a^4-a^3+ab+2}}\le\frac{1}{\sqrt{ab+a+1}}\)

Tương tự \(\frac{1}{\sqrt{b^4-b^3+bc+2}}\le\frac{1}{\sqrt{bc+b+1}}\)

             \(\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\frac{1}{\sqrt{ca+c+1}}\)

Cộng từng vế các bđt trên ta được

\(VT\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\)

Áp dụng bđt Bunhiacopski ta có

\(VT\le\sqrt{3\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)}\)\(=\sqrt{3\left(\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{a^2bc+abc+ab}\right)}=\sqrt{3}\)

Dấu "=" xảy ra khi a=b=c=1

Đoán xem

Từ \(\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}+\frac{5d}{1+d}\le1\)

\(\Rightarrow1-\frac{a}{1+a}+2-\frac{2b}{1+b}+3-\frac{3c}{1+c}+5-\frac{5d}{1+d}\ge10\)

\(\Rightarrow\frac{1}{1+a}+\frac{2}{1+b}+\frac{3}{1+c}+\frac{5}{1+d}\ge10\)

Áp dụng BĐT AM-GM ta có: 

\(\frac{1}{a+1}\ge\)\(\frac{2b}{1+b}+\frac{3c}{1+c}+\frac{5d}{1+d}\ge10\sqrt[10]{\frac{b^2c^3d^5}{\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^5}}\)

Và \(\frac{1}{1+b}\ge\)\(\frac{a}{1+a}+\frac{b}{b+1}+\frac{3c}{c+1}+\frac{5d}{d+1}\)

\(\ge10\sqrt[10]{\frac{abc^3d^5}{\left(1+a\right)\left(1+b\right)\left(1+c\right)^3\left(1+d\right)^5}}\)

Và \(\frac{1}{1+c}\ge\frac{a}{1+a}+\frac{2b}{b+1}+\frac{2c}{c+1}+\frac{5d}{d+1}\)

\(\ge10\sqrt[10]{\frac{ab^2c^2d^5}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^2\left(1+d\right)^5}}\)

Và \(\frac{1}{1+d}\ge\frac{a}{a+1}+\frac{2b}{b+1}+\frac{3c}{c+1}+\frac{4d}{d+1}\)

\(\ge10\sqrt[10]{\frac{ab^2c^3d^4}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^4}}\)

Nhân theo vế 4 BĐT có: \(\frac{1}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^5}\)

\(\ge10^{1+2+3+5}\sqrt[10]{\frac{a^{2+3+5}b^{2+2+6+10}c^{3+6+6+15}d^{5+10+15+20}}{\left(1+a\right)^{10}\left(1+b\right)^{20}\left(1+c\right)^{30}\left(1+d\right)^{50}}}\)

Tương đương với \(ab^2c^3d^5\le\frac{1}{10^{11}}\) (ĐPCM)

kho ko

a, Ta co 2 bo de quen thuoc sau : FC la phan giac ^EFD, FB la phan giac PFD

ma QR//EP nen

\(\widehat{PFB}=\widehat{FQD}=\widehat{QFD}\Rightarrow\Delta DFQ\) can tai D => DF=DQ (1)

mat khac theo tinh chat tia phan giac ngoai ^PFD co \(\frac{FD}{FP}=\frac{CD}{CP}\) 

ma \(\frac{CD}{CP}=\frac{DT}{PF}\) (DT//PF)

suy ra \(\frac{DF}{PF}=\frac{DT}{PF}\Rightarrow DT=DF\) (2)

Tu(1)va (2) suy ra DT=DQ hay D la trung diem QT

b, Goi S la trung diem BC ta chung minh PQSR noi tiep 

Co \(\Delta PSE~\Delta ESD\left(G-G\right)\Rightarrow\frac{PS}{ES}=\frac{ES}{SD}\Leftrightarrow ES^2=PS.DS\)

lai co ES=SB=SC do S la trung diem canh huyen BC cua tam giac vuong BEC

suy ra \(BS^2=PS.SD=DS\left(PD+DS\right)=SD^2+PD.DS\)

=> \(PD.DS=BS^2-SD^2=\left(BS-DS\right)\left(BS+DS\right)=BD.DC\) (3)

Mat khac ^DQB=^PFB(cmt)

^PFB=^RCD( BFEC nt)

suy ra ^DQB=^RCD=> BQCR noi tiep

=> \(BD.DC=DQ.DR\) (4)

Tu (3),(4) suy ra DP.DS=DQ.DR => PQDR noi tiep 

=> (PQR) di qua S la trung diem BC co dinh

c,lay H' doi xung voi H qua BC, ta co H' thuoc (O) .

ta lai co bo de sau : \(BD.DC=DH.DA\) (quen thuoc)

suy ra \(DP.DS=DH.DA\left(=DB.DC\right)\)

<=> \(\frac{DH}{DP}=\frac{DS}{DA}\)

ma ^HDP=^SDA=90

suy ra \(\Delta DHP~\Delta DSA\left(c-g-c\right)\Rightarrow\widehat{DHP}=\widehat{DSA}\)

va \(\widehat{DSA}=\widehat{AHK}\left(phu\widehat{DAS}\right)\)

=>\(\widehat{DHP}=\widehat{AHK}\) => P,H,K thang hang

lai co \(\widehat{AFH}=\widehat{AKH}=\widehat{AEH}=90\)

=> A,F,H,K,E cung thuoc 1 duong tron =. FHKE noi tiep

=>\(PF.PE=PH.PK\) (5)

ma BFEC noi tiep => \(PF.PE=PB.PC\) (6)

(5)+(6)Suy ra \(PH.PK=PB.PC\) => BHKC noi tiep

Vi H' ,I doi xung voi H,K qua BC ma BHKC noi tiep => BH'IC noi tiep

do vay \(I\in\left(BH'C\right)=\left(ABH'C\right)=\left(O\right)\)

e,Goi tam (CJL) la U, (U) cat (O) tai V, BC giao OG tai X

=> \(\widehat{VBG}=\widehat{VJG}\left(=\widehat{VCB}\right)\) =>BJVG noi tiep

=> B,J,X,V,G cung thuoc 1 duong tron => ^BVG=^BXG=90

lai co ^XVG +^XBG=180 hay ^XVG+^BAC=180

va ^BVC+^BAC=180

suy ra ^XVG=^BVC

hay 90 +^XVB=^XVB+^XVC

=> ^XVC=90

=> V thuoc duong tron dk XC

mat khac V cung thuoc (O)

suy ra V co dinh ,C co dinh 

suy ra tam U di chuyen tren trung truc VC co dinh (dpcm)

em mới lớp 5 lên ko bít bài này

tu gia thiet => \(4x+4y+4z+4\sqrt{xyz}=16\)

Xet \(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left(16-4y-4z+yz\right)}\)

= \(\sqrt{x\left(4x+4y+4y+4\sqrt{xyx}-4y-4z+yz\right)}\)

=\(\sqrt{x\left(4x+4\sqrt{xyz}+yz\right)}\)

=\(\sqrt{4x^2+4x\sqrt{xyx}+xyz}=\sqrt{\left(2x+\sqrt{xyz}\right)^2}\)

= \(2x+\sqrt{xyz}\)

tuong tu va suy ra \(\sqrt{x\left(4-y\right)\left(4-z\right)}+\sqrt{y\left(4-z\right)\left(4-x\right)}+\sqrt{z\left(4-x\right)\left(4-y\right)}\)

= \(2\left(x+y+z\right)+3\sqrt{xyz}\)

hinh nhu de bai bn viet thieu \(-\sqrt{xyz}\)

neu dung de thi goi bieu thuc can tinh la A

ta co \(A=2\left(x+y+z\right)+2\sqrt{xyz}=2\left(x+y+z+\sqrt{xyz}\right)=2.4=8\)

Chuc ban hoc tot 

jhfghjgjtfyt tjgfyjjtf

trước tiên ta phải cm: \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\left(#\right)\left(\forall a,b,c\in R;x,y,z>0\right)\)

dấu = xảy ra khi zà chỉ khi\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

thật zậy , zới \(a,b\in R;x,y>0\)ta có \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\left(##\right)\left(a,b\in R;x,y>0\right)\)

\(\Leftrightarrow\left(a^2y+b^2x\right)\left(x+y\right)\ge xy\left(a+b\right)^2\Leftrightarrow\left(bx-ay\right)^2\ge0\)( luôn đúng )

 dấu = xảy ra khi zà chỉ khi\(\frac{a}{x}=\frac{b}{y}\)

* áp dụng bất đẳng thức (##) ta được 

\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b\right)^2}{x+y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)

dấu = xảy ra khi zà chỉ khi \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)\

* áp dụng bất đẳng thức (#) ta có

vt = \(\frac{x^2}{x\left(x^2-yz+2010\right)}+\frac{y^2}{y\left(y^2-xz+2010\right)}+\frac{z^2}{z\left(z^2-xy+2010\right)}\)

   =\(\frac{x^2}{x\left(x^2-yz+2010\right)}+\frac{y^2}{y\left(y^2-xz+2010\right)}+\frac{z^2}{z\left(z^2-xy+2010\right)}\)

\(\ge\frac{\left(x+y+z\right)^3}{x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)}\left(1\right)\)

Lưu ý nhé : \(x\left(x^2-yz+2010\right)=x\left(x^2+xy+zx+1340\right)>0\)

                  \(y\left(y^2-xz+2010\right)>0\)

                  \(z\left(z^2-xy+2010\right)>0\)

Ta có \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

                                                      \(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(xy+yz+xz\right)\right]\)

                                       do dó       \(x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)\)        \(\)

                                                     =\(\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(xy+yz+zx\right)+2010\right]\)

                                                     =\(\left(x+y+z\right)^3\left(2\right)\)

Từ (1) zà (2) suy ra

vt \(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}\)

dấu = xảy ra khi zà chỉ khi \(x=y=z=\frac{\sqrt{2010}}{3}\)

thí chủ có link koooooo

Bài hay quá!

Đặt \(a=\frac{3x}{x+y+z};b=\frac{3y}{x+y+z};c=\frac{3z}{x+y+z}\left(x;y;z>0\right)\)

Sau khi quy đồng cần chứng minh:

\(2\, \left( x+y+z \right) \left( {x}^{4}y+{x}^{4}z+3\,{x}^{3}{y}^{2}- 11\,{x}^{3}yz+3\,{x}^{3}{z}^{2}+3\,{x}^{2}{y}^{3}+3\,{x}^{2}{y}^{2}z+3 \,{x}^{2}y{z}^{2}+3\,{x}^{2}{z}^{3}+x{y}^{4}-11\,x{y}^{3}z+3\,x{y}^{2} {z}^{2}-11\,xy{z}^{3}+x{z}^{4}+{y}^{4}z+3\,{y}^{3}{z}^{2}+3\,{y}^{2}{z }^{3}+y{z}^{4} \right) \geq 0 \)(gõ Latex, không biết ad đã fix lỗi chưa, nếu nó không hiện thì hỏi ad, đừng hỏi em!)

Hay là: \( \left( {x}^{4}y+{x}^{4}z+3\,{x}^{3}{y}^{2}- 11\,{x}^{3}yz+3\,{x}^{3}{z}^{2}+3\,{x}^{2}{y}^{3}+3\,{x}^{2}{y}^{2}z+3 \,{x}^{2}y{z}^{2}+3\,{x}^{2}{z}^{3}+x{y}^{4}-11\,x{y}^{3}z+3\,x{y}^{2} {z}^{2}-11\,xy{z}^{3}+x{z}^{4}+{y}^{4}z+3\,{y}^{3}{z}^{2}+3\,{y}^{2}{z }^{3}+y{z}^{4} \right) \geq 0 \)

Or:

\(9\, \left( 1/4\, \left( x-2\,z+y \right) ^{2}+3/4\, \left( -y+x \right) ^{2} \right) {z}^{3}+3\, \left( x-2\,z+y \right) ^{3}{z}^{2}+ \left( \left( 3/4\, \left( x-2\,z+y \right) ^{2}+1/4\, \left( -y+x \right) ^{2} \right) \left( -y+x \right) ^{2}+ \left( x-z \right) ^{ 4}+ \left( y-z \right) ^{4} \right) z+ \left( x-z \right) \left( y-z \right) \left( \left( x-z \right) ^{3}+3\, \left( x-z \right) ^{2} \left( y-z \right) +3\, \left( x-z \right) \left( y-z \right) ^{2}+ 21\, \left( x-z \right) \left( y-z \right) z+ \left( y-z \right) ^{3} \right) \geq 0 \)

Cách xử trí: Nếu nó không hiện: Sau khi quy đồng, ta biến đối nó về như trong link sau: https://imgur.com/D8ScX4k

Cách khác:

\(\Leftrightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge a^2+b^2+c^2+3\)

Or \(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+3\)

Or \(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2\left(ab+bc+ca\right)\ge12\)

Or: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\left(ab+bc+ca\right)\ge6\)

Giả sử \(\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab\ge a+b-1\)(*)

Do đó: \(VT=\frac{ab+bc+ca}{abc}+ab+bc+ca\)

\(\ge\frac{a+b+c\left(a+b\right)-1}{\frac{c\left(a+b\right)^2}{4}}+a+b+c\left(a+b\right)-1\)

\(=\frac{4\left(c+1\right)\left(a+b\right)-4}{c\left(a+b\right)^2}+\left(c+1\right)\left(a+b\right)-1\)

\(=\frac{4\left(c+1\right)\left(3-c\right)-4}{c\left(3-c\right)^2}+\left(c+1\right)\left(3-c\right)-1\ge6\)

Last inequality\(\Leftrightarrow\frac{\left(2-c\right)^3\left(c-1\right)^2}{c\left(c-3\right)^2}\ge0\). Nếu c < 2 thì ta có đpcm.

Nếu \(c\ge2\)

\(VT=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\left(ab+bc+ca\right)\)

\(>\frac{4}{a+b}+ab+c\left(a+b\right)\ge\frac{4}{a+b}+2\left(a+b\right)\ge2\sqrt{8}>3\)

\(\sqrt{a^2+\left(2^{a-3}+2^{-a-1}\right)^2}+\sqrt{a^4+a^2+2}=\sqrt{\left(a^2+a+1\right)^2+\left(1+2^{a-3}+2^{-a-1}\right)^2}\)

đề thế cơ mà , làm t nghĩ mà đell nghĩ đc j .

làm này . 

Không mất tính tổng quát 

đặt \(x=a>0,y=2^{a-3}+2^{-a-1}>0,z=a^2+1>0,t=1>0\)

khi đó phương trình trở thành

\(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}=\sqrt{\left(x+z\right)^2+\left(y+t\right)^2}\left(1\right)\)

Mặt khác ta cũng có :\(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\ge\sqrt{\left(x+z\right)^2+\left(y+t\right)^2}\)(2) zới mọi \(x,y,z,t>0\)

\(\Leftrightarrow x^2+y^2+z^2+t^2+2\sqrt{x^2z^2+x^2t^2+y^2z^2+y^2t^2}\ge x^2+y^2+z^2+t^2+2\left(xz+yt\right)\)( biến đổi từ cái trên nhá )

\(\Leftrightarrow x^2z^2+x^2t^2+y^2z^2+y^2t^2\ge x^2+y^2+z^2+t^2+2\left(xz+yt\right)\)

\(\Leftrightarrow x^2z^2+x^2t^2+y^2z^2+y^2t^2\ge x^2z^2+y^2t^2+2xyzt\Leftrightarrow\left(yz-xt\right)^2\ge0\)(luôn đúng zới mọi x,y,z,t > 0)

zậy từ (1) zà (2) xảy ra khi zà chỉ khi yz=xt

=>\(\left(2^{a-3}+2^{-a-1}\right)\left(a^2+1\right)=a\Leftrightarrow\left(2^{a-3}+2^{-a-1}\right)=\frac{a}{a^2+1}\left(3\right)\)(zì \(a^2+1>0\)

mà lại có \(\frac{a}{a^2+1}\le\frac{1}{2}\)(zì \(\left(a-1\right)^2\ge0\), dấu "=" xảy ra khi a=1 (4)

zà \(\left(2^{a-3}+2^{-a-1}\right)=\frac{2^a}{8}+\frac{1}{2.2^a}\ge\frac{1}{2}\)(theo cô-si nha) ,dấu "=" xảy ra khi a=1 (5)

zậy từ (3) , (4) , (5) \(=>a=1\)là giá trị nguyên dương duy nhất cần tìm

à thì ra ghi dài quá nó cho xuống dòng

làm t cứ tưởng

hì hì

đánh có dấu hộ mình với bạn ơi

may mik hong Vietkey mong ban thong cam