\(D=\frac{4}{\left(2x-3\right)^2}+5\)

Để D đạt GTLN thì \(\frac{4}{\left(2x-3\right)^2}\) đạt GTLN

\(\Rightarrow2x-3=2\)

\(\Rightarrow2x=5\)

\(\Rightarrow x=2,5\)

Vậy GTNN của D = 6

Ta có: \(z^2=2\left(xz+yz-xy\right)=2xz+2yz-2xy\)

Xét:

\(x^2+\left(x-z\right)^2=x^2+z^2-z^2+\left(x-z\right)^2\)\(=\left(x-z\right)^2+2xz-\left(2xz+2yz-2xy\right)+\left(x-z\right)^2\)

\(=\left(x-z\right)^2+2xy-2yz+\left(x-z\right)^2=\left(x-z\right)^2+2y\left(x-z\right)+\left(x-z\right)^2\)

\(=\left(x-z\right)\left(x-z+2y+x-z\right)=\left(x-z\right)\left(2x+2y-2z\right)\)                                    (1)

Xét:

\(y^2+\left(y-z\right)^2=y^2+z^2-z^2+\left(y-z\right)^2\)\(=\left(y-z\right)^2+2yz-\left(2xz+2yz-2xy\right)\)

\(=\left(y-z\right)^2+2xy-2xz+\left(y-z\right)^2=\left(y-z\right)^2+2x\left(y-z\right)+\left(y-z\right)^2\)

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

Từ (1); (2) => \(\frac{x^2+\left(x-z\right)^2}{y^2+\left(y-z\right)^2}=\frac{\left(x-z\right)\left(2x+2y-2z\right)}{\left(y-z\right)\left(2x+2y-2z\right)}=\frac{x-z}{y-z}\) \(\left(ĐPCM\right)\)                    

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