Question

Answer 1

h: ĐKXĐ: x>=0

TA có: \(\sqrt{x+2\sqrt{x}+1}-\sqrt{x-2\sqrt{x}+1}=2\)

=>\(\sqrt{\left(\sqrt{x}+1\right)^2}-\sqrt{\left(\sqrt{x}-1\right)^2}=2\)

=>\(\sqrt{x}+1-\left|\sqrt{x}-1\right|=2\)

=>\(\left|\sqrt{x}-1\right|=\sqrt{x}+1-2=\sqrt{x}-1\)

=>\(\sqrt{x}-1\ge0\)

=>\(\sqrt{x}\ge1\)

=>x>=1

i: ĐKXĐ: x>=0

Ta có: \(\sqrt{x+4\sqrt{x}+4}+\sqrt{x-4\sqrt{x}+4}=4\)

=>\(\sqrt{\left(\sqrt{x}+2\right)^2}+\sqrt{\left(\sqrt{x}-2\right)^2}=4\)

=>\(\sqrt{x}+2+\left|\sqrt{x}-2\right|=4\)

=>\(\left|\sqrt{x}-2\right|=4-\sqrt{x}-2=2-\sqrt{x}\)

=>\(\sqrt{x}-2\le0\)

=>\(\sqrt{x}\le2\)

=>0<=x<=4