I. [tex]\left(x-2\right)\left(x+2\right)\\=x^2-2^2=x^2-4=\left(-3\frac{1}{4}\right)^2-4=\left(-\frac{13}{4}\right)^2-4=\frac{169}{16}-4=\frac{169-64}{16}=\frac{105}{16}=\boxed{6\frac{9}{16}}[/tex]
II. [tex]x\left(x-2.2\right)=xx-x\cdot \:2.2=x^2-2.2x=\left(-3\frac{1}{4}\right)^2-2.2\left(-3\frac{1}{4}\right)=\left(-\frac{13}{4}\right)^2-2.2\left(-3\frac{1}{4}\right)=\left(-\frac{13}{4}\right)^2-2.2\left(-\frac{13}{4}\right)=\left(-\frac{13}{4}\right)^2+2.2\cdot \frac{13}{4}=\frac{13^2}{4^2}+7.15=10.5625+7.15=\boxed{17.7125}[/tex]
III. [tex]\left(x-2\right)^2=x^2-2x\cdot \:2+2^2=x^2-2x\cdot \:2+2^2=x^2-4x+4=\left(-3\frac{1}{4}\right)^2-4\left(-3\frac{1}{4}\right)+4=\left(-\frac{13}{4}\right)^2-4\left(-3\frac{1}{4}\right)+4=\left(-\frac{13}{4}\right)^2-4\left(-\frac{13}{4}\right)+4=\frac{169}{16}-\left(-4\cdot \frac{13}{4}\right)+4=\frac{169}{16}+4\cdot \frac{13}{4}+4=\frac{169}{16}+4\cdot \frac{13}{4}+\frac{64}{16}=\frac{233}{16}+4\cdot \frac{13}{4}=\frac{4\cdot \frac{13}{4}\cdot \:16+233}{16}=\frac{64\cdot \frac{13}{4}+233}{16}[/tex][tex]=\frac{441}{16}=\boxed{27\frac{9}{16}}[/tex]
IV. [tex]\left(\frac{2}{x}\right)^2=\frac{2^2}{x^2}=\frac{4}{x^2}=\left(-3\frac{1}{4}\right)=\frac{4}{\left(-\frac{13}{4}\right)^2}=\frac{4}{\frac{169}{16}}=\frac{4\cdot \:16}{169}=\boxed{\frac{64}{169}}[/tex]
Odpowiedź: Największą wartość spośród wypisanych przez Zbyszka wyrażeń dla [tex]x=\left(-3\frac{1}{4}\right)[/tex] ma wyrażenie C. III.
