6.6..6 - 6=?

đặt \(a=x^2,b=y^2\left(a,b\ge0\right)\)thì \(P=\frac{\left(a-b\right)\left(1-ab\right)}{\left(1+a\right)^2\left(1+b\right)^2}\)

Zì \(a,b\ge0\)nên

\(\left(a-b\right)\left(1-ab\right)=a-a^2b-b+ab^2\le a+ab^2=a\left(1+b^2\right)\le a\left(1+2b+b^2\right)=a\left(1+b\right)^2\)

Lại có \(\left(1+a\right)^2=\left(1-a\right)^2+4a\ge4a\)

=>\(P\le\frac{a\left(1+b\right)^2}{4a\left(1+b\right)^2}=\frac{1}{4}\)

dấu "=" xảy ra khi zà chỉ khi\(\hept{\begin{cases}a=1\\b=0\end{cases}=>\hept{\begin{cases}x=\pm1\\y=0\end{cases}}}\)

zậy \(maxP=\frac{1}{4}khi\hept{\begin{cases}x=\pm1\\y=0\end{cases}}\)

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học giỏi

a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)

\(=2x^2-4xy+\dfrac{15}{4}y^2\)

b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)

\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)

\(=2x^2+2x+13-2x^2+2\)

=2x+15

a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)

b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)

\(=2x+15\)

a; \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

= \(x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)

= \(2x^2-4xy+\dfrac{15}{4}y^2\)

b; \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)

= \(x^2-4x+4+x^2+6x+9-2x^2+2\)

= \(2x+15\)

Áp dụng BĐT AM-GM: \(\left(x^2-y^2\right)\left(1-x^2y^2\right)\le\frac{1}{4}\left(x^2-y^2+1-x^2y^2\right)^2=\frac{1}{4}\left(1-y^2\right)^2\left(1+x^2\right)^2\)

\(P\le\frac{1}{4}\frac{\left(1-y^2\right)^2}{\left(1+y^2\right)^2}\)

mà theo BĐT AM-GM:\(\left(1-y\right)\left(1+y\right)\le\frac{1}{4}\left(1-y+1+y\right)^2=1\)

\(\Rightarrow P\le\frac{1}{4}.\frac{1}{\left(1+y^2\right)^2}\le\frac{1}{4}.\frac{1}{1}=\frac{1}{4}\)

Dấu = xảy ra khi x=1;y=0 wait : có gì đó sai sai. số thực

\(A=\dfrac{2\left(x^3+y^3\right)}{\left(x^4+y^2\right)\left(x^2+y^4\right)}=2.\dfrac{\left(x^3+y^3\right)}{x^4y^4+x^2y^2+x^6+y^6}\)

\(=2.\dfrac{\left(x^3+y^3\right)}{1+1+x^6+y^6}=2.\dfrac{x^3+y^3}{x^6+y^6+2x^3y^3}=2.\dfrac{x^3+y^3}{\left(x^3+y^3\right)^2}=\dfrac{2}{x^3+y^3}\left(1\right)\)

Áp dụng bất đẳng thức Cauchy ta có:

\(x^3+y^3+1\ge3\sqrt{xy.1}=3\)

\(\Rightarrow x^3+y^3\ge2\Rightarrow\dfrac{2}{x^3+y^3}\le1\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow A\le1\)

Dấu "=" xảy ra khi x=y=1.

Vậy MaxA là 1, đạt được khi x=y=1.

Ta có:

\(x^2+1=x^2+xy+yz+zx\)

           \(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)

Tương tự:

\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)

\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

TH1: x,y,z <0

\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)

TH2: x,y,z>0

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)

Ta có \(1+z^2=xy+yz+zx+z^2\)

\(=y\left(x+z\right)+z\left(x+z\right)\)

\(=\left(x+z\right)\left(y+z\right)\)

CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)

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

\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)

 Tương tự như thế, ta được

\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

 Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.