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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: ( y – 2 )^3 – ( y – 3 ) ( y^2 + 3y + 9 ) + 6 ( y + 1 )^2 = 49

Toán Lớp 8: ( y – 2 )^3 – ( y – 3 ) ( y^2 + 3y + 9 ) + 6 ( y + 1 )^2 = 49

Comments ( 2 )

  1. Giải đáp + giải thích các bước giải :
    ( y – 2 )^3 – ( y – 3 ) ( y^2 + 3y + 9 ) + 6 ( y + 1 )^2 = 49 $\\$ \Leftrightarrow y^3 – 6y^2 + 12y – 8- (y^3 – 27) + 6(y^2 + 2y + 1)=49 $\\$ \Leftrightarrow y^3 – 6y^2 + 12y – 8- y^3 + 27 + 6y^2 + 12y + 6 = 49 $\\$ \Leftrightarrow 24y + 25 = 49 $\\$ \Leftrightarrow 24y = 24 $\\$ \Leftrightarrow y = 1 $\\$ Vậy: y = 1

  2. (y-2)^3 – (y-3)(y^2 +3y +9) +6(y+1)^2 =49
    => (y-2)^3 – (y^3 – 3^3) +6(y^2 +2y +1) =49
    =>  y^3 -6y^2 +12y  -8 – y^3 +27 + 6y^2 + 12y +6 =49
    => (y^3 -y^3) + (-6y^2 +6y^2) +(12y +12y) + (-8 +27 +6) = 49
    => 24y +25 =49
    => 24y = 24
    => y =1
    Vậy y =1 . 

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222-9+11+12:2*14+14 = ? ( )

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