Toán Lớp 8: cho x+y=1 và xy=-6 tính x^2+y^2, x^3+y^3. x^5+y^5

Question

Toán Lớp 8:  cho x+y=1 và xy=-6  tính x^2+y^2, x^3+y^3. x^5+y^5

diepbich93 · Answer

x+y = 1 ; xy =-6   $ Rightarrow$ $x^{2}$ + $y^{2}$ =  ($x^{2}$ + $2xy$ + $y^2$) - $2xy$   = $(x+y)^{2}$ - $2xy$ = 1&sup2; - 2.( -6) = 1 +12 = 13     $x^{3}$ + $y^{3}$ = $(x+y)^{3}$ - $3x^{2}$y - $3xy^{2}$       = $(x+y)^{3}$ - $3xy(x+y)$  = 1&sup3; - 3 . -6 . 1 = 1 +18 = 19     $x^{5}$ + $y^{5}$ = $(x+y)^{5}$ - $5x^{4}y$ - $5xy^{4}$ - $10x^3y^2$ - $10x^2y^3$     = $(x+y)^{5}$ - $5xy(x^3 + y^{3})$ - $10(xy)^2(x + y)$     = 1 - 5.(-6). 19 - 10 . (-6)&sup2;.1     = 1 + 570 -360 = 211

quynhthanhnhu · Answer

+ x&sup2; + y&sup2;     Ta c&oacute; :   ( x + y )&sup2; = x&sup2; + 2xy + y&sup2; ( Hằng đẳng thức số 1 )   => 1&sup2; = x&sup2; + 2 . ( - 6 ) + y&sup2;   => x&sup2; + y&sup2; -12 = 1   => x&sup2; + y&sup2; = 13   Vậy x&sup2; + y&sup2; = 13   + x&sup3; + y&sup3;    Ta c&oacute; :    ( x + y )&sup3; = x&sup3; + 3x&sup2;y + 3xy&sup2; + y&sup3; ( Hằng đẳng thức số 4 )   1&sup3; = x&sup3; + y&sup3; + 3x&sup2;y + 3xy&sup2;    1 = x&sup3; + y&sup3; + 3xy ( x + y )   x&sup3; + y&sup3; + 3xy ( x + y ) = 1   x&sup3; + y&sup3; + 3xy = 1 ( Do x + y = 1 )   x&sup3; + y&sup3; + 3 . ( - 6 ) = 1   x&sup3; + y&sup3; - 18 = 1   x&sup3; + y&sup3; = 19   Vậy x&sup3; + y&sup3; = 19   + $x^{5}$ + $y^{5}$    Ta c&oacute; : $(x+y)^{5}$ = $x^{5}$ + 5$x^{4}$y + 10x&sup3;y&sup2; + 10x&sup2;y&sup3; + 5x$y^{4}$ + $y^{5}$               $1^{5}$ = $x^{5}$ + $y^{5}$ + 5xy ( x&sup3; + y&sup3; ) + 10x&sup2;y&sup2; ( x + y )                1 = $x^{5}$ + $y^{5}$ - 30.19 + 10 . (-6)&sup2; . 1               $x^{5}$ + $y^{5}$ - 570 + 360 = 1                $x^{5}$ + $y^{5}$ = 211 ( C&oacute; vẻ hơi lớn ạ )