Ответ:
х=4-у х=4-у х=4-у х=4-у х=4-1 х=3
(4-у)²-у²=8 16-8у+у²-у²=8 -8у=8-16 у=1 у=1 у=1
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[tex]\displaystyle \bf \left \{ {{x^2-y^2=8} \atop {x+y=4 \ \ \ }} \right. \ \ \ \left \{ {{x^2-y^2=8} \atop {x=4-y \ \ \ }} \right. \ \ \ \left \{ {{(4-y)^2-y^2=8} \atop x=4-y \ \ \ \ \ \ \ \ \ \ }} \right. \\\\\\\left \{ {{y=1 \ \ \ \ \ } \atop {x=4-1}} \right. \ \ \ \left \{ {{y=1} \atop {x=3}} \right.[/tex]
[tex]\displaystyle \bf \Big( 4-y\Big)^2-y^2=8\\\\16-8y+y^2-y^2=8\\\\16-8y=8\\\\-8y=8-16\\\\-8y=-8\\\\y=1[/tex]
[tex]\boxed{ \displaystyle \bf \Big(3 \ ; 1\Big)}[/tex]
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Answers & Comments
Ответ:
х=4-у х=4-у х=4-у х=4-у х=4-1 х=3
(4-у)²-у²=8 16-8у+у²-у²=8 -8у=8-16 у=1 у=1 у=1
Verified answer
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[tex]\displaystyle \bf \left \{ {{x^2-y^2=8} \atop {x+y=4 \ \ \ }} \right. \ \ \ \left \{ {{x^2-y^2=8} \atop {x=4-y \ \ \ }} \right. \ \ \ \left \{ {{(4-y)^2-y^2=8} \atop x=4-y \ \ \ \ \ \ \ \ \ \ }} \right. \\\\\\\left \{ {{y=1 \ \ \ \ \ } \atop {x=4-1}} \right. \ \ \ \left \{ {{y=1} \atop {x=3}} \right.[/tex]
[tex]\displaystyle \bf \Big( 4-y\Big)^2-y^2=8\\\\16-8y+y^2-y^2=8\\\\16-8y=8\\\\-8y=8-16\\\\-8y=-8\\\\y=1[/tex]
[tex]\boxed{ \displaystyle \bf \Big(3 \ ; 1\Big)}[/tex]