\(P=2x+y+\dfrac{30}{x}+\dfrac{5}{y}\)

\(=\left(\dfrac{6x}{5}+\dfrac{30}{x}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)+\left(\dfrac{4x}{5}+\dfrac{4y}{5}\right)\)

\(\ge2.6+2+\dfrac{4}{5}.10=22\)

Vậy GTNN là P = 22 khi x = y = 5

Áp dụng BĐT cauchy ta có:\(\left\{{}\begin{matrix}x^2+y^2\ge2xy\\y^2+z^2\ge2yz\\x^2+z^2\ge2xz\end{matrix}\right.\)

\(P\le\dfrac{1}{4xy+4x+4}+\dfrac{1}{4yz+4y+4}+\dfrac{1}{4xz+4z+4}=\dfrac{1}{4}\left(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+x+1}\right)\)

xét biểu thức \(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{zx+z+1}=\dfrac{1}{xy+x+1}+\dfrac{x}{1+yx+x}+\dfrac{xy}{x+1+xy}=\dfrac{xy+x+1}{xy+x+1}=1\)do đó \(P\le\dfrac{1}{4}\)

dấu = xảy ra khi x=y=z=1

Trước tiên ta tính:

\(\dfrac{1}{x+xy+1}+\dfrac{1}{y+yz+1}+\dfrac{1}{z+zx+1}\)

Đặt: \(\left\{{}\begin{matrix}x=\dfrac{a}{b}\\y=\dfrac{b}{c}\\z=\dfrac{c}{a}\end{matrix}\right.\left(a,b,c\ne0\right)\)

Thì ta có: \(\dfrac{1}{\dfrac{a}{b}+\dfrac{a}{b}.\dfrac{b}{c}+1}+\dfrac{1}{\dfrac{b}{c}+\dfrac{b}{c}.\dfrac{c}{a}+1}+\dfrac{1}{\dfrac{c}{a}+\dfrac{c}{a}.\dfrac{a}{b}+1}\)

\(=\dfrac{bc}{ab+ac+bc}+\dfrac{ca}{ab+bc+ca}+\dfrac{ab}{ab+bc+ca}=1\)

Quay về bài toán ban đầu. Ta có:

\(P=\dfrac{1}{\left(x+2\right)^2+y^2+2xy}+\dfrac{1}{\left(y+2\right)^2+z^2+2yz}+\dfrac{1}{\left(z+2\right)^2+x^2+2xz}\)

\(=\dfrac{1}{x^2+4x+4+y^2+2xy}+\dfrac{1}{y^2+4y+4+z^2+2yz}+\dfrac{1}{z^2+4z+4+z^2+2xz}\)

\(=\dfrac{1}{\left(x-y\right)^2+4x+4xy+4}+\dfrac{1}{\left(y-z\right)^2+4y+4yz+4}+\dfrac{1}{\left(z-x\right)^2+4z+4zx+4}\)

\(\le\dfrac{1}{4x+4xy+4}+\dfrac{1}{4y+4yz+4}+\dfrac{1}{4z+4zx+4}\)

\(=\dfrac{1}{4}.\left(\dfrac{1}{x+xy+1}+\dfrac{1}{y+yz+1}+\dfrac{1}{z+zx+1}\right)=\dfrac{1}{4}\)

a: \(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)

\(=\dfrac{3}{x-3}\cdot\dfrac{-\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{-3}{x-3}\)

b: \(=\dfrac{x+1}{x+2}:\left(\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+3\right)^2}\right)\)

\(=\dfrac{x+1}{x+2}\cdot\dfrac{\left(x+3\right)^2}{\left(x+2\right)\left(x+1\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}\)

c: \(=\dfrac{x^2-2xy+y^2+x^2+2xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x^2+2xy+y^2}{2xy}\cdot\dfrac{xy}{x^2+y^2}\)

\(=\dfrac{2\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)^2}{x^2+y^2}\cdot\dfrac{1}{2}\)

\(=\dfrac{\left(x+y\right)}{x-y}\)

rút gọn A

\(A=\dfrac{4xy}{y^2-y^2}:\left(\dfrac{x+y+\left(y-x\right)}{\left(y-x\right)\left(x+y\right)^2}\right)=\dfrac{4xy\left[\left(y-x\right)\left(x+y\right)^2\right]}{2y\left(y-x\right)\left(x+y\right)}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|x\right|\ne\left|y\right|\\A=2x\left(x+y\right)=2x^2+2xy\end{matrix}\right.\)

\(B=3x^2+y^2+2x-2y\)

\(B-A+1=x^2+y^2+2x-2y-2xy+1=\left(x+1-y\right)^2\)

\(\Rightarrow A\le1\Rightarrow A=1\)\(\Rightarrow x+1-y=0\) thay lại ra được x,y

Lời giải:

Cách 1:

Áp dụng BĐT S.Vacxo ta có:

\(\frac{1}{xy+1}+\frac{1}{1+yz}+\frac{1}{1+xz}\geq \frac{9}{1+xy+1+yz+1+xz}=\frac{9}{3+xy+yz+xz}(1)\)

Theo BĐT Cauchy ta có bổ đề quen thuộc:

\(xy+yz+xz\leq x^2+y^2+z^2\leq 3(2)\)

Từ \((1);(2)\Rightarrow \frac{1}{xy+1}+\frac{1}{yz+1}+\frac{1}{xz+1}\geq \frac{9}{3+xy+yz+xz}\geq \frac{9}{3+3}=\frac{3}{2}\)

Vậy \(P_{\min}=\frac{3}{2}\Leftrightarrow x=y=z=1\)

Cách 2:

Áp dụng BĐT Cauchy cho các số dương:

\(\frac{1}{xy+1}+\frac{xy+1}{4}\geq 2.\sqrt{\frac{1}{xy+1}.\frac{xy+1}{4}}=1\)

\(\frac{1}{yz+1}+\frac{yz+1}{4}\geq 2.\sqrt{\frac{1}{yz+1}.\frac{yz+1}{4}}=1\)

\(\frac{1}{xz+1}+\frac{xz+1}{4}\geq 2.\sqrt{\frac{1}{xz+1}.\frac{xz+1}{4}}=1\)

Cộng tất cả các BĐT trên theo vế và rút gọn:

\(\Rightarrow \frac{1}{xy+1}+\frac{1}{yz+1}+\frac{1}{xz+1}\geq \frac{9-(xy+yz+xz)}{4}\geq \frac{9-3}{4}=\frac{3}{2}\)

Vậy \(P_{\min}=\frac{3}{2}\)

1) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\)

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

\(=\dfrac{x^2}{x+1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x-1}+1\) MTC: \(\left(x-1\right)\left(x+1\right)\)

\(=\dfrac{x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^2\left(x-1\right)+2x+\left(x+1\right)+\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^3-x^2+2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\)

b) \(\dfrac{1}{x^3-x}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{x^2-1}\)

\(=\dfrac{1}{x\left(x^2-1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\) MTC: \(x\left(x-1\right)\left(x+1\right)\)

\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1-\left(x+1\right)+2x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1-x-1+2x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)

b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)

A=\(\dfrac{4xy}{\left(y-x\right)\left(y+x\right)}\):(\(\dfrac{1}{\left(y-x\right)\left(y+x\right)}\)+\(\dfrac{1}{\left(y+x\right)^2}\) ) với ĐKXĐ là y≠(x,-x)

A=\(\dfrac{4xy}{\left(y-x\right)\left(y+x\right)}\):\(\dfrac{x+y+y-x}{\left(y-x\right)\left(y+x\right)^2}\)

A=\(\dfrac{4xy}{\left(y-x\right)\left(y+x\right)}\)\(\times\)\(\dfrac{\left(y-x\right)\left(y+x\right)^2}{2y}\)

A=2x(y+x)

A=2xy+2\(x^2\)