Question

7. \(\sqrt{x(x^2 + 2x + 3)} = \sqrt{x(x^2 + 1)}\).

8. \(\sqrt{(x + 1)^2(x^2 - x + 1)} = \sqrt{(x^2 + x)(x^2 + 3)}\).

9. \(\sqrt{(x - 1)^2(x^2 + x + 1)} = \sqrt{(x - 1)(x^3 + x^2 - 2)}\).

Answer 1

7:

ĐKXĐ: x>=0

Ta có: \(\sqrt{x\left(x^2+2x+3\right)}=\sqrt{x\left(x^2+1\right)}\)

=>\(x\left(x^2+2x+3\right)=x\left(x^2+1\right)\)

=>\(x\left(x^2+2x+3-x^2-1\right)=0\)

=>x(2x+2)=0

=>\(\left[\begin{array}{l}x=0\\ 2x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=-1\left(loại\right)\end{array}\right.\)

8: ĐKXĐ: \(\left[\begin{array}{l}x\ge0\\ x\le-1\end{array}\right.\)

Ta có: \(\sqrt{\left(x+1\right)^2\left(x^2-x+1\right)}=\sqrt{\left(x^2+x\right)\left(x^2+3\right)}\)

=>\(\left(x+1\right)^2\cdot\left(x^2-x+1\right)=x\left(x+1\right)\left(x^2+3\right)\)

=>\(\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)-\left(x+1\right)\left(x^3+3x^2\right)=0\)

=>\(\left(x+1\right)\left(x^3-1-x^3-3x^2\right)=0\)

=>\(\left(x+1\right)\left(-3x^2-1\right)=0\)

mà \(-3x^2-1\le-1<0\forall x\)

nên x+1=0

=>x=-1(nhận)

9: ĐKXĐ: \(\left(x-1\right)\left(x^3+x^2-2\right)\ge0\)

=>\(\left(x-1\right)\left(x^3-x^2+2x^2-2x+2x-2\right)\ge0\)

=>\(\left(x-1\right)\left(x-1\right)\left(x^2+2x+2\right)\ge0\)

=>\(\left(x-1\right)^2\cdot\left\lbrack\left(x+1\right)^2+1\right\rbrack\ge0\) (luôn đúng)

=>x∈R

Ta có: \(\sqrt{\left(x-1\right)^2\cdot\left(x^2+x+1\right)}=\sqrt{\left(x-1\right)\left(x^3+x^2-2\right)}\)

=>\(\left(x-1\right)^2\cdot\left(x^2+x+1\right)=\left(x-1\right)\left(x^3+x^2-2\right)\)

=>\(\left(x-1\right)^2\cdot\left(x^2+x+1\right)=\left(x-1\right)^2\cdot\left(x^2+2x+2\right)\)

=>\(\left(x-1\right)^2\left(x^2+2x+2-x^2-x-1\right)=0\)

=>\(\left(x-1\right)^2\cdot\left(x+1\right)=0\)

=>\(\left[\begin{array}{l}x-1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)