Chứng minh rằng (A ² + B ²)*(X ² + Y ²) = (AX – BY) ² + (AX + BY) ² 27/11/2021 Bởi Everleigh Chứng minh rằng (A ² + B ²)*(X ² + Y ²) = (AX – BY) ² + (AX + BY) ²
$\text{Ta có:}$ \[(ax-by)^2+(ax-by)^2\] \[=2(ax-by)^2\] \[=2(a^2x^2+b^2y^2-2axby)\] \[=2a^2x^2+2b^2y^2-4axby\] $\text{Ta lại có:}$ \[(a^2+b^2)(x^2+y^2)\] \[=a^2x^2+a^2y^2+b^2x^2+b^2y62\] \[=2a^2+x^2+2b^2x^2\] \[\text{Vì:} 2a^2x^2+2b^2y^2-4axby\neq2a^2+x^2+2b^2x^2\] \[⇔(ax-by)^2+(ax-by)^2\neq(a^2+b^2)(x^2+y^2)\] \[\text{Vậy:}(ax-by)^2+(ax-by)^2\neq(a^2+b^2)(x^2+y^2)\] Bình luận
$(a^2+b^2).(x^2+y^2)$ $ = a^2x^2+a^2y^2+b^2x^2+b^2y^2$ $(ax-by)^2+(ay+bx)^2$ $ = a^2x^2-2abxy+b^2y^2+a^2y^2+2aabxy+b^2x^2$ $ = = a^2x^2+a^2y^2+b^2x^2+b^2y^2$ Do đó : $(a^2+b^2).(x^2+y^2) = (ax-by)^2+(ay+bx)^2$ Bình luận
$\text{Ta có:}$
\[(ax-by)^2+(ax-by)^2\]
\[=2(ax-by)^2\]
\[=2(a^2x^2+b^2y^2-2axby)\]
\[=2a^2x^2+2b^2y^2-4axby\]
$\text{Ta lại có:}$
\[(a^2+b^2)(x^2+y^2)\]
\[=a^2x^2+a^2y^2+b^2x^2+b^2y62\]
\[=2a^2+x^2+2b^2x^2\]
\[\text{Vì:} 2a^2x^2+2b^2y^2-4axby\neq2a^2+x^2+2b^2x^2\]
\[⇔(ax-by)^2+(ax-by)^2\neq(a^2+b^2)(x^2+y^2)\]
\[\text{Vậy:}(ax-by)^2+(ax-by)^2\neq(a^2+b^2)(x^2+y^2)\]
$(a^2+b^2).(x^2+y^2)$
$ = a^2x^2+a^2y^2+b^2x^2+b^2y^2$
$(ax-by)^2+(ay+bx)^2$
$ = a^2x^2-2abxy+b^2y^2+a^2y^2+2aabxy+b^2x^2$
$ = = a^2x^2+a^2y^2+b^2x^2+b^2y^2$
Do đó : $(a^2+b^2).(x^2+y^2) = (ax-by)^2+(ay+bx)^2$