Question

Đã trả lời bởi chuyên gia

Answer 1

Đáp án+Giải thích các bước giải:

x3+y3+1=3xy�3+�3+1=3��

⇔x3+y3−3xy+1=0⇔�3+�3-3��+1=0

⇔(x3+3x2y+3xy2+y3)−3x2y−3xy2−3xy+1=0⇔(�3+3�2�+3��2+�3)-3�2�-3��2-3��+1=0

⇔(x+y)3−3xy(x+y+1)+1=0⇔(�+�)3-3��(�+�+1)+1=0

⇔(x+y+1)[(x+y)2−(x+y)+13]−3xy(x+y+1)=0⇔(�+�+1)[(�+�)2-(�+�)+13]-3��(�+�+1)=0

⇔(x+y+1)(x2+2xy+y2−x−y+1)−3xy(x+y+1)=0⇔(�+�+1)(�2+2��+�2-�-�+1)-3��(�+�+1)=0

⇔(x+y+1)(x2+2xy+y2−x−y+1−3xy)=0⇔(�+�+1)(�2+2��+�2-�-�+1-3��)=0

⇔(x+y+1)(x2−xy+y2−x−y+1)=0⇔(�+�+1)(�2-��+�2-�-�+1)=0

x;y>0�;�>0

⇒x+y+1>0⇒�+�+1>0

⇒x2−xy+y2−x−y+1=0⇒�2-��+�2-�-�+1=0

⇔2x2−2xy+2y2−2x−2y+2=0⇔2�2-2��+2�2-2�-2�+2=0

⇔(x2−2xy+y2)+(x2−2x+1)+(y2−2y+1)=0⇔(�2-2��+�2)+(�2-2�+1)+(�2-2�+1)=0

⇔(x−y)2+(x−1)2+(y−1)2=0⇔(�-�)2+(�-1)2+(�-1)2=0

Nhận xét:

(x−y)2≥0∀x;y(�-�)2≥0∀�;�

(x−1)2≥0∀x(�-1)2≥0∀�

(y−1)2≥0∀y(�-1)2≥0∀�

⇒(x−y)2+(x−1)2+(y−1)2≥0∀x;y⇒(�-�)2+(�-1)2+(�-1)2≥0∀�;�

⇔⎧⎪ ⎪⎨⎪ ⎪⎩(x−y)2=0(x−1)2=0(y−1)2=0⇔{(�-�)2=0(�-1)2=0(�-1)2=0

≤<⎧⎪⎨⎪⎩x−y=0x−1=0y−1=0≤<{�-�=0�-1=0�-1=0

⇔⎧⎪⎨⎪⎩x=yx=1y=1⇔{�=��=1�=1 (thỏa mãn)

⇒Q=x2017+y2018⇒�=�2017+�2018

=12017+12018=12017+12018

=1+1=1+1

=2=2

Vậy Q=2�=2 tại x3+y3+1=3xy

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Answer 2