\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{9}+2\sqrt{21}\)

=\(\left(\sqrt{4}\sqrt{7}-\sqrt{7}-\sqrt{12}\right).3+2\sqrt{21}\)

=\(\left(2\sqrt{7}-\sqrt{7}-\sqrt{4}\sqrt{3}\right).3+2\sqrt{21}\)

=\(\left(\sqrt{7}-2\sqrt{3}\right).3+2\sqrt{21}\)

=\(3\sqrt{7}-6\sqrt{3}+2\sqrt{21}\)

đề có sai ko nhưng kết quả ra thế

\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{9}+2\sqrt{21}=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right).3+2\sqrt{21}=\left(\sqrt{7}-2\sqrt{3}\right).3+2\sqrt{21}=3\sqrt{7}-6\sqrt{3}+2\sqrt{21}\)

\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\frac{16a^4b^6}{128a^6b^6}}=\sqrt{\frac{1}{8a^2}}=\frac{\sqrt{1}}{\sqrt{8a^2}}=\frac{1}{\sqrt{2}\sqrt{4}\sqrt{a}}\)

=\(\frac{1}{2\sqrt{2}a}\)

\(x^2-2\left(M+1\right)x+m^2+5m-2=0\text{ (1)}\)

Để (1) có nghiệm thì \(\Delta'=\left(m+1\right)^2-\left(m^2+5m-2\right)\ge0\Leftrightarrow-3m+3\ge0\Leftrightarrow m\le1\)

(1) có nghiệm x = -1 \(\Rightarrow\left(-1\right)^2-2\left(M+1\right).\left(-1\right)+m^2+5m-2=0\)

\(\Leftrightarrow m^2+7m+1=0\Leftrightarrow m=\frac{-7+3\sqrt{5}}{2}\left(tm\right)\text{ hoặc }m=\frac{-7-3\sqrt{5}}{2}\left(tm\right)\)

\(+m=\frac{-7+3\sqrt{5}}{2}\text{ thì }x_2=-\frac{b}{a}-x_1=2\left(\frac{-7+3\sqrt{5}}{2}+1\right)-\left(-1\right)=-4+3\sqrt{5}\)

\(+m=\frac{-7-3\sqrt{5}}{2}\text{ thì }x_2=-\frac{b}{a}-x_1=2\left(\frac{-7-3\sqrt{5}}{2}+1\right)-\left(-1\right)=-4-3\sqrt{5}\)

miku ???

đặt A=\(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

=>\(\sqrt{2}A=\sqrt{2}\sqrt{2+\sqrt{3}}-\sqrt{2}\sqrt{2-\sqrt{3}}\)

=\(\sqrt{2\left(2+\sqrt{3}\right)}-\sqrt{2\left(2-\sqrt{3}\right)}=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

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

=\(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\left(\sqrt{3}-1\right)\)

=\(\sqrt{3}+1-\sqrt{3}+1=2\)

=>A=\(\frac{2}{\sqrt{2}}=\frac{\sqrt{2}\sqrt{2}}{\sqrt{2}}=\sqrt{2}\)

vậy \(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\sqrt{2}\)

Viết đề rõ hơn một chút