Question

La exercițiul 3 va rog​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

 

$\displaystyle\bf\\\sqrt[3]{7x+1}-x=1\\\\\sqrt[3]{7x+1}=x+1~~\Big|(ridicam~la~puterea~a~3-a)\\\\\Big(\sqrt[3]{7x+1}\Big)^3=(x+1)^3\\\\7x+1=x^3+3x^2+3x+1\\\\x^3+3x^2+3x+1=7x+1\\\\x^3+3x^2+3x+1-7x-1=0\\\\x^3+3x^2+3x-7x+1-1=0\\\\x^3+3x^2-4x=0\\\\x(x^2+3x-4)=0~~~\Big|~~3x=-x+4x\\\\x[x(x-1)+4(x-1)]=0\\\\x(x-1)(x+4)=0\\\\x=0~sau~(x-1)=0~sau~(x+4)=0\\\\\boxed{\bf x_1=0;~~x_2=1;~~x_3=-4}$