Question

Tìm giá trị biểu thức:

a)\(\sqrt{4a^4-12a^2+9}-\sqrt{a^4-8a^2+16}\)Với \(a=\sqrt{3}\)

b)\(\sqrt{10a^2-12a\sqrt{10}+36}\)Với \(a=\sqrt{\frac{5}{2}}-\sqrt{\frac{2}{5}}\)

c)\(\sqrt{16\left(1+4x+4x^2\right)^2}\)Với \(x=-1\)

Answer 1

a/ \(\sqrt{4a^4-12a^2+9}-\sqrt{a^4-8a^2+16}\)

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

= \(|2a^2-3|-|a^2-4|\)

= \(2a^2-3+a^2-4\)
= \(3a^2-7\)

Thay a=\(\sqrt{3}\).Ta có:

\(3.\left(\sqrt{3}\right)^2-7\)

= 3.3-7=2

Answer 2

b/ \(\sqrt{10a^2-12a\sqrt{10}+36}\)

= \(\sqrt{\left(a\sqrt{10}\right)^2-2.a\sqrt{10}.6+6^2}\)

= \(\sqrt{\left(a\sqrt{10}-6\right)^2}\)

= \(|a\sqrt{10}-6|\)

=  \(-a\sqrt{10}+6\)

Thay  a= \(\sqrt{\frac{5}{2}}-\sqrt{\frac{2}{5}}\)=\(\frac{3}{\sqrt{10}}\),Ta có:

\(-\frac{3}{\sqrt{10}}.\sqrt{10}+6\)

= -3+6 =3