Toán Lớp 8: Tìm x a)5x+x^2=0 b)2(x+5)-x^2-5x=0 c)5x(x-1)=x-1 d)x^2-10x=-25

Question

Toán Lớp 8:  Tìm x a)5x+x^2=0 b)2(x+5)-x^2-5x=0 c)5x(x-1)=x-1 d)x^2-10x=-25

truongthanhhung · Answer

a)5x+x^2=0         => x(1+5) = 0         =>$ left[ begin{matrix} x=0   1+5x=0 end{matrix} right.$          => $ left[ begin{matrix} x=0   x = -1/5  end{matrix} right.$         b)2(x+5)-x^2-5x=0       2x+10-x^2-5x=0       =>10-3x-x^2=0       =>10-5x+2x-x^2=0       =>5(2-x)+x(2-x)=0       =>(2-x)(5+x)=0       c) 5x(x-1)=x-1       =>5x=(x-1):(x-1)       =>5x = 1       =>x= 1/5       d)x^2-10x=-25       x&sup2;- 10x + 25 = 0        => (x-5) &sup2;  = 0       => x =5

kimlien · Answer

Giải đáp:       Lời giải và giải thích chi tiết:    a) 5x + x^2 = 0   &hArr; x(x + 5) = 0   &hArr;  ( left[  begin{array}{l}x=0  x+5=0 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=0  x=-5 end{array}  right. )    Vậy x &isin; {0 ; -5}   b) 2(x + 5) - x^2 - 5x = 0   &hArr; 2(x + 5) - (x^2 + 5x) = 0   &hArr; 2(x + 5) - x(x + 5) = 0   &hArr; (2 - x)(x + 5) = 0   &hArr;  ( left[  begin{array}{l}2-x=0  x+5=0 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=2  x=-5 end{array}  right. )    Vậy x &isin; {2 ; -5}   c) 5x(x - 1) = x - 1   &hArr; 5x(x - 1) -(x - 1) = 0   &hArr; (5x - 1)(x - 1) = 0   &hArr;  ( left[  begin{array}{l}5x-1=0  x-1=0 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=1/5  x=1 end{array}  right. )    Vậy x &isin; {1/5 ; 1}   d) x^2 - 10x = -25   &hArr; x^2 - 10x + 25 = 0   &hArr; (x - 5)^2 = 0   &hArr; x - 5 = 0   &hArr; x = 5   Vậy x = 5