Which of the following is not true?A. $( y+\frac{1}{y})^2=( y-\frac{1}{y})^2+4$B. $( y-\frac{1}{y})^2+2=y^2+\frac{1}{y^2}$C. $( y+\frac{1}{y})^2-2=( y-\frac{1}{y})^2$D. $( y-\frac{1}{y})^2-( y+\frac{1}{y})^2=-(2)^2$
Solution:
As known, $( y+\frac{1}{y})^2=y^2+\frac{1}{y^2}+2$
And $( y-\frac{1}{y})^2=y^2+\frac{1}{y^2}-2$
A. $( y+\frac{1}{y})^2=( y-\frac{1}{y})^2+4$
$y^2+\frac{1}{y^2}+2=y^2+\frac{1}{y^2}-2+4=y^2+\frac{1}{y^2}+2$
Therefore, $( y+\frac{1}{y})^2=( y-\frac{1}{y})^2+4$, is true.
B. $( y-\frac{1}{y})^2+2=y^2+\frac{1}{y^2}$
$y^2+\frac{1}{y^2}-2+2=y^2+\frac{1}{y^2}$
Therefore, $( y-\frac{1}{y})^2+2=y^2+\frac{1}{y^2}$, is true.
C. $( y+\frac{1}{y})^2-2=( y-\frac{1}{y})^2$
$y^2+\frac{1}{y^2}+2-2≠y^2+\frac{1}{y^2}-2$
Thus, $( y+\frac{1}{y})^2-2=( y-\frac{1}{y})^2$, is false.
D. $( y-\frac{1}{y})^2-( y+\frac{1}{y})^2=-(2)^2$
$y^2+\frac{1}{y^2}-2-y^2-\frac{1}{y^2}-2=-(2)^2$
$-4=-(2)^2$
Thus, $( y-\frac{1}{y})^2-( y+\frac{1}{y})^2=-(2)^2$, is true.
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