cho `a^3/(a^2+ab+b^2) + b^3/(b^2+bc+c^2) +c^3/(c^2+ca+a^2)=2019 `.
tìm giá trị`P=b^3/(a^2+ab+b^2) + c^3/(b^2+bc+c^2) +a^3/(c^2+ca+a^2)`
cho `a^3/(a^2+ab+b^2) + b^3/(b^2+bc+c^2) +c^3/(c^2+ca+a^2)=2019 `.
tìm giá trị`P=b^3/(a^2+ab+b^2) + c^3/(b^2+bc+c^2) +a^3/(c^2+ca+a^2)`
Đáp án:
$P=2019$
Giải thích các bước giải:
Ta có: $P=\dfrac{b^3}{a^2+ab+b^2}+\dfrac{c^3}{b^2+bc+c^2}+\dfrac{a^3}{c^2+ca+a^2}$
$⇔ P-2019=\dfrac{b^3}{a^2+ab+b^2}+\dfrac{c^3}{b^2+bc+c^2}+\dfrac{a^3}{c^2+ca+a^2}-\dfrac{a^3}{a^2+ab+b^2}-\dfrac{b^3}{b^2+bc+c^2}-\dfrac{c^3}{c^2+ca+a^2}$
$⇔ P-2019=\dfrac{b^3-a^3}{a^2+ab+b^2}+\dfrac{c^3-b^3}{b^2+bc+c^2}+\dfrac{a^3-c^3}{c^2+ca+a^2}$
$⇔ P-2019=\dfrac{(b-a)(a^2+ab+b^2)}{a^2+ab+b^2}+\dfrac{(c-b)(b^2+bc+c^2)}{b^2+bc+c^2}+\dfrac{(a-c)(c^2+ca+a^2)}{c^2+ca+a^2}$
$⇔ P-2019=b-a+c-b+a-c=0$
$⇔ P=2019$
Đáp án:
Giải thích các bước giải:
`P = b^3/( a^2 + ab + b^2 ) + c^3/( b^2 + bc + c^2 ) + a^3/( c^2 + ca + a^2 )`
`<=> – P = – b^3/( a^2 + ab + b^2 ) – c^3/( b^2 + bc + c^2 ) – a^3/( c^2 + ca + a^2 )`
`<=> 2019 – P = ( a^3/( a^2 + ab + b^2 ) + b^3/( b^2 + bc + c^2 ) + c^3/( c^2 + ca + a^2 )) + ( b^3/( a^2 + ab + b^2 ) – c^3/( b^2 + bc + c^2 ) – a^3/( c^2 + ca + a^2 ))`
`<=> 2019 – P = a^3/( a^2 + ab + b^2 ) + b^3/( b^2 + bc + c^2 ) + c^3/( c^2 + ca + a^2 ) – b^3/( a^2 + ab + b^2 ) – c^3/( b^2 + bc + c^2 ) – a^3/( c^2 + ca + a^2 )`
`<=> 2019 – P = ( a^3/( a^2 + ab + b^2 ) – b^3/( a^2 + ab + b^2 )) + ( b^3/( b^2 + bc + c^2 ) – c^3/( b^2 + bc + c^2 )) + ( c^3/( c^2 + ca + a^2 ) – a^3/( c^2 + ca + a^2 ))`
`<=> 2019 – P = ( a^3 – b^3 )/( a^2 + ab + b^2 ) + ( b^3 – c^3 )/( b^2 + bc + c^2 ) + ( c^3 – a^3 )/( c^2 + ca + a^2 )`
`<=> 2019 – P = (( a – b )( a^2 + ab + b^2 ))/( a^2 + ab + b^2 ) + (( b – c )( b^2 + bc + c^2 ))/( b^2 + bc + c^2 ) + (( c – a )( c^2 + ca + a^2 ))/( c^2 + ca + a^2 )`
`<=> 2019 – P = ( a – b ) + ( b – c ) + ( c – a )`
`<=> 2019 – P = a – b + b – c + c – a`
`<=> 2019 – P = 0`
`<=> P = 2019`