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Ta có: \(\left(x+z\right)\left(y+z\right)=1\)
\(\Rightarrow\left(x+z\right)^2\left(y+z\right)^2=1\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(y+z\right)^2}+\dfrac{1}{\left(z+x\right)^2}=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x+z\right)^2\left(y+z\right)^2}{\left(y+z\right)^2}+\dfrac{\left(x+z\right)^2\left(y+z\right)^2}{\left(z+x\right)^2}\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z\right)^2+\left(y+z\right)^2\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z\right)^2-2\left(x+z\right)\left(y+z\right)+\left(y+z\right)^2+2\) (Vì: (x+z)(y+z)=1 =>2(x+z)(y+z)=2 )
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z-y-z\right)^2+2\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2+2\)
Áp dụng bất đẳng thức Cauchy, ta có :
\(\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2\ge2\sqrt{\dfrac{1}{\left(x-y\right)^2}\cdot\left(x-y\right)^2}=2\cdot1=2\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2+2\ge2+2=4\)
Vậy \(MinP=4\) khi \(x-y=1\); \(y+z=\dfrac{\sqrt{5}-1}{2}\); \(x+z=\dfrac{2}{\sqrt{5}-1}\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a )
Sử dụng Cô-si , ta có :
\(x+y\ge2\sqrt{xy}\) (1)
\(\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}\) (2)
Nhân cả vế (1) vế (2) lại ta có :
\(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge2\sqrt{xy}.2\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}=4\)
\(\LeftrightarrowĐPCM.\)
\(A=\left(1-\dfrac{1}{x^2}\right)\left(1-\dfrac{1}{y^2}\right)=1+\dfrac{1}{x^2y^2}-\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)
Áp dụng bất đẳng thức Cauchy cho 2 số dương, ta có:
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{2}{xy}\) (1)
và \(x+y\ge2\sqrt{xy}\) (2)
TỪ (2) \(\Rightarrow\) \(\dfrac{1}{x^2y^2}\ge\dfrac{16}{\left(x+y\right)^4}\) và \(\dfrac{2}{xy}\ge\dfrac{8}{\left(x+y\right)^2}\)
Mặt khác, theo đề \(x+y\le1\)
=> \(\dfrac{1}{x+y}\ge1\)
=> A \(\ge1+\dfrac{16}{\left(x+y\right)^4}+\dfrac{2}{xy}\) \(\ge1+\dfrac{16}{\left(x+y\right)^4}-\dfrac{8}{\left(x+y\right)^2}\)
\(=1+16-8=9\)
Dấu ''='' xảy ra khi x = y = 0,5
Mình đánh nhầm, dòng 2 từ dưới lên phải là \(-\dfrac{2}{xy}\) nhá ! :))
a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
Ta có:
\(P=\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}-3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+5\)
\(=\dfrac{x^2}{y^2}-3\dfrac{x}{y}+\dfrac{9}{4}+\dfrac{y^2}{x^2}-3\dfrac{y}{x}+\dfrac{9}{4}+\dfrac{1}{2}\)
\(=\left(\dfrac{x}{y}-\dfrac{3}{2}\right)^2+\left(\dfrac{y}{x}-\dfrac{3}{2}\right)^2+\dfrac{1}{2}\)
Với \(x;y\ne0\) ta có:
\(\left(\dfrac{x}{y}-\dfrac{3}{2}\right)^2\ge0;\left(\dfrac{y}{x}-\dfrac{3}{2}\right)^2\ge0\)
\(\Rightarrow\left(\dfrac{x}{y}-\dfrac{3}{2}\right)^2+\left(\dfrac{y}{x}-\dfrac{3}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)
Vậy Min P = \(\dfrac{1}{2}\)
Để \(P=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{3}{2}=0\\\dfrac{y}{x}-\dfrac{3}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{3}{2}\\\dfrac{y}{x}=\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow x=y=0\)
Giải:
Có:
\(P=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\)
Vì \(\left(x+\dfrac{1}{x}\right)^2\ge0,\forall x\) và \(\left(y+\dfrac{1}{y}\right)^2\ge0,\forall y\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\ge0;\forall x,y\)
\(\Rightarrow Min_P=0\)
Chúc bạn học tốt!
Áp dụng BĐT \(x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\) và BĐT \(xy\le\dfrac{1}{4}\left(x+y\right)^2\), ta có:
\(\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\)
\(\ge\dfrac{1}{2}\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2\)\(=\dfrac{1}{2}\left(1+\dfrac{x+y}{xy}\right)^2\)
\(\ge\dfrac{1}{2}\left(1+\dfrac{1}{\dfrac{1}{4}\left(x+y\right)^2}\right)^2=\dfrac{25}{2}\left(x+y=1\right)\)
Dấu "=" xảy ra khi x = y = 0,5