gọi các điểm như trên hình

I là giao 2 đường tiếp tuyến HI và AC=>OI là phân giác góc EOK (1) và IE=IK

C là giao 2 tiếp tuyến AC và BC => OC là phân giác góc KOD (2) và KC=DC

(1) và (2) => tam giác IOC vuông tại O, có đường cao OK =>OK²=IK.KC <=> OK²=IE.DC

CM tương tự ta được OJ² = EH.BD

mà \(\text{OK=OJ=r}\) 

=>\(\text{IE.DC=EH.BD}\)

=>\(\frac{EH}{EI}=\frac{CD}{BD}\)

Ta có : \(\text{HI // BC}\)

=>\(\frac{EI}{MC}=\frac{AI}{AC}=\frac{AH}{AB}=\frac{EH}{BM}\)

=> \(\frac{BM}{MC}=\frac{EH}{EI}\)

=>\(\frac{BM}{CM}=\frac{EH}{EI}=\frac{CD}{BD}\)

=> \(1+\frac{BM}{CM}=1+\frac{CD}{BD}\)\(\Leftrightarrow\frac{BC}{CM}=\frac{BC}{BD}\Rightarrow CM=BD\)

83110=Hello

\(5\le xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\)\(\Leftrightarrow\)\(x+y+z\ge\sqrt{15}\)

\(\frac{x^2}{\sqrt{8x^2+3y^2+14xy}}=\frac{x^2}{\sqrt{8x^2+2xy+3y^2+12xy}}\ge\frac{x^2}{\sqrt{9x^2+12xy+4y^2}}=\frac{x^2}{3x+2y}\)

\(A\ge sigma\frac{x^2}{3x+2y}\ge\frac{\left(x+y+z\right)^2}{5\left(x+y+z\right)}=\frac{x+y+z}{5}\ge\sqrt{\frac{3}{5}}\)

Dấu "=" xảy ra khi \(x=y=z=\sqrt{\frac{5}{3}}\)

h2r r1000

ta có

\(0\le\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\left(\forall x,y,z>0\right)\)

\(\Leftrightarrow2xy+2yz+2zx\le2\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\)(1)

dấu  = xảy ra khi

\(x=y=z=0\)

theo giả thiết ta có

\(x\left(x+1\right)+y\left(y+1\right)+z\left(z+1\right)\le18\)

\(\Leftrightarrow x^2+y^2+z^2\le18-\left(x+y+z\right)\left(2\right)\)

từ (1) zà (2) suy ra

\(\left(x+y+z\right)^2\le54-3\left(x+y+z\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2+3\left(x+y+z\right)-54\le0\)

\(\Leftrightarrow\left(x+y+z-6\right)\left(x+y+z+9\right)\le0\)

\(\Leftrightarrow0< x+y+z\le6\left(do\left(x+y+z>0;9>0\right)\right)\)

áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)ta có

\(P=\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}\ge\frac{9}{2\left(x+y+z\right)+3}\ge\frac{9}{2.6+3}=\frac{3}{5}\)

Dấu = xảy ra khi zà chỉ khi

\(\hept{\begin{cases}x+y+1=y+z+1=z+x+1\\x+y+z=6\end{cases}=>x=y=z=2}\)

zậy MinP= 3/5 khi x=y=z=2

Ta có : x(x + 1) + y (y+1 ) + z(z + 1) \(\le18\)

<=> x² + y² + z² + ( x + y + z ) \(\le18\)

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Rightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

=> 54 \(\ge\)( x + y+z)² + 3(x + y + z) 

<=> -9 \(\le\)x + y + z \(\le\)6

=> 0 \(\le\)x+y+z \(\le\)6 

\(\frac{1}{x+y+1}+\frac{x+y+1}{25}\ge\frac{2}{5}\)

\(\frac{1}{y+z+1}+\frac{y+z+1}{25}\ge\frac{2}{5}\)

\(\frac{1}{z+x+1}+\frac{z+x+1}{25}\ge\frac{2}{5}\)

=> \(P+\frac{2\left(x+y+z\right)+3}{25}\ge\frac{6}{5}\)

=> P \(\ge\frac{27}{25}-\frac{2}{25}\left(x+y+z\right)\ge\frac{15}{25}=\frac{3}{5}\)

Dấu " =" xảy ra khi :

\(\hept{\begin{cases}x=y=z>0;x+y+z=6\\\left(x+y+1\right)^2=\left(y+z+1\right)^2=\left(z+x+1\right)^2=25\end{cases}\Leftrightarrow x=y=z=2}\)

Vậy GTNN của P là \(\frac{3}{5}\)khi x = y =z =2

bij ngu af

             33333333333333  ddddddddddddddd

bị ngu à

kết bạn với mình

\(P=\frac{a^3+b^3+c^3}{2abc}+\frac{a^2c+b^2c}{c^3+abc}+\frac{b^2a+c^2a}{a^3+abc}+\frac{c^2b+a^2b}{b^3+abc}\)

\(\ge\frac{a^3}{2abc}+\frac{b^3}{2abc}+\frac{c^3}{2abc}+\frac{2abc}{c^3+abc}+\frac{2abc}{a^3+abc}+\frac{2abc}{b^3+abc}\)

\(=\left(\frac{a^3}{2abc}+\frac{2abc}{a^3+abc}\right)+\left(\frac{b^3}{2abc}+\frac{2abc}{b^3+abc}\right)+\left(\frac{c^3}{2abc}+\frac{2abc}{c^3+abc}\right)\)

Xét: \(\frac{a^3}{2abc}+\frac{2abc}{a^3+abc}=\frac{a^3}{2abc}+\frac{1}{2}+\frac{1}{\frac{a^3}{2abc}+\frac{1}{2}}-\frac{1}{2}\ge2\sqrt{\left(\frac{a^3}{2abc}+\frac{1}{2}\right).\frac{1}{\frac{a^3}{2abc}+\frac{1}{2}}}-\frac{1}{2}=\frac{3}{2}\)

Tương tự với 2 cặp còn lại

Vậy ta có: \(P\ge\frac{3}{2}+\frac{3}{2}+\frac{3}{2}=\frac{9}{2}\)

"=" xảy ra <=> a=b=c

Áp dụng bđt Cauchy :

\(\frac{1}{1+a}=\left(1-\frac{1}{1+b}\right)+\left(1-\frac{1}{1+c}\right)=\frac{b}{1+b}+\frac{c}{1+c}\ge2\sqrt{\frac{bc}{\left(b+1\right)\left(c+1\right)}}\)

Tương tự : \(\frac{1}{1+b}\ge2\sqrt{\frac{ac}{\left(a+1\right)\left(c+1\right)}}\)

\(\frac{1}{1+c}\ge2\sqrt{\frac{ab}{\left(a+1\right)\left(b+1\right)}}\)

Nhân theo vế : \(\frac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\frac{8abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)

\(\Rightarrow abc\le\frac{1}{8}\)

Vậy Max abc = 1/8 khi a = b = c = 1/2

7894561230++

Có: \(x^5+y^2=xy^2+1\)

<=> \(x^5-1=y^2\left(x-1\right)\)(1)

TH1: x = 1 

=> \(1^2+y^2=1.y^2+1\) đúng với mọi y

TH2: \(x\ne1\)

(1) <=> \(y^2=x^4+x^3+x^2+x+1\)

<=> \(4y^2=4x^4+4x^3+4x^2+4x+4\)

Có:

+)  \(4x^4+4x^3+4x^2+4x+4=4x^4+4x^3+x^2+2x^2+x^2+4x+4\)

\(=\left(2x^2+x\right)^2+2x^2+\left(x+2\right)^2>\left(2x^2+x\right)^2\)

=> \(\left(2y\right)^2>\left(2x^2+x\right)^2\)

+) \(4x^4+4x^3+4x^2+4x+4\le\left(2x^2+x+2\right)^2\)

=> \(\left(2y\right)^2\le\left(2x^2+x+2\right)^2\)

=> \(\left(2x^2+x\right)^2< \left(2y\right)^2\le\left(2x^2+x+2\right)^2\)

TH1: \(\left(2y\right)^2=\left(2x^2+x+2\right)^2\)

=> \(4x^4+4x^3+4x^2+4x+4=4x^4+x^2+4+4x^3+8x^2+4x\)

<=> x = 0 

=> \(y=\pm1\)

TH2: \(\left(2y\right)^2=\left(2x^2+x+1\right)^2\)

=> \(4x^4+4x^3+4x^2+4x+4=4x^4+x^2+1+4x^3+4x^2+2x\)

<=> \(2x+3-x^2=0\)

<=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Với x = -1 => \(y=\pm1\)

Với x = 3 => \(y=\pm11\)

Kết luận:...

Hạ sách : Nhân hết ra :)))

Ta có :

\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(xy+\frac{1}{xy}\right)^2-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(xy+\frac{1}{xy}\right)\)

\(=x^2+\frac{1}{x^2}+2+y^2+\frac{1}{y^2}+2+x^2y^2+\frac{1}{x^2y^2}+2-\left(xy+\frac{x}{y}+\frac{y}{x}+\frac{1}{xy}\right)\left(xy+\frac{1}{xy}\right)\)

\(=x^2+y^2+\frac{1}{x^2y^2}+x^2y^2+\frac{1}{x^2}+\frac{1}{y^2}+6-\left(x^2y^2+1+x^2+\frac{1}{y^2}+y^2+\frac{1}{x^2}+1+\frac{1}{x^2y^2}\right)\)

\(=6-1-1\)

\(=4\)

4655000

52266+

533333

minh bo tay

bó tay rùi

Ta có \(b\left(a^2-2\right)=a\left(ab+2\right)-2\left(a+b\right)\). Do \(a^2-2\vdots ab+2\) nên \(2\left(a+b\right)\vdots ab+2\to ab+2\le2a+2b\to\left(a-2\right)\left(b-2\right)\le2\). 

Với \(a=1\to-\frac{1}{b+2}\in Z\), loại
Với \(a=2\to\frac{4}{2b+2}\in Z\to2b+2=4\to b=1\)

Với \(a=3\to\frac{7}{3b+2}\in Z\to3b+2=7\to\)  loại
Với \(a=4\to\frac{14}{4b+2}\in Z\to4b+2=14\to b=3.\)
Với \(a\ge5\to b-2\le\frac{2}{a-2}

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