Toán Lớp 8: Tim x: a) $x^{2}$ . (x+1) + x + 1 = 0 b) $x^{2}$ (x-1) + $x^{2}$ =x

Question

Toán Lớp 8:  Tim x:  a) $x^{2}$ . (x+1) + x + 1 = 0 b) $x^{2}$ (x-1) + $x^{2}$ =x

kymmailien · Answer

Giải đáp+Lời giải và giải thích chi tiết: a) x² (x+1) + x+1= 0 <=> x²(x+1) +(x+1) = 0 <=> (x+1)(x²+1) = 0 <=>x + 1 = 0 ( x² + 1 > 0 ∀ x) <=> x = -1 b) x²(x-1) + x² = x <=> x²(x-1) + x² - x = 0 <=> x²(x-1) + (x²-x) = 0 <=> x²(x-1) + x(x-1) = 0 <=> (x-1)(x²+x) = 0 <=> x(x-1)(x+1) = 0 <=> ( left[ begin{array}{l}x=0 x=1 x=-1 end{array} right. )

dantam45 · Answer

a) x^2(x+1)+x+1=0   &hArr;x^2(x+1)+(x+1)=0   &hArr;(x+1)(x^2+1)=0   &hArr; ( left[  begin{array}{l}x+1=0  x^2+1=0 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=-1  x^2=-1(v&ocirc; l&yacute;) end{array}  right. )    Vậy S={-1}   b) x^2(x-1)+x^2=x   &hArr;x^2(x-1)+(x^2-x)=0   &hArr;x^2(x-1)+x(x-1)=0   &hArr;x(x-1)(x+1)=0   &hArr; ( left[  begin{array}{l}x=0  x-1=0  x+1=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=0  x=1  x=-1 end{array}  right. )    Vậy S={0,1,-1}