Bài 13 trang 7 SBT toán 9 tập 1

Đề bài

Rút gọn rồi tính:

a) \(5\sqrt {{{( - 2)}^4}} \)

b) \( - 4\sqrt {{{( - 3)}^6}} \)

c) \(\sqrt {\sqrt {{{( - 5)}^8}} } \)

d) \(2\sqrt {{{( - 5)}^6}}  + 3\sqrt {{{( - 2)}^8}} \).

a)

\(\eqalign{
& 5\sqrt {{{( - 2)}^4}} = 5\sqrt {{{\left[ {{{( - 2)}^2}} \right]}^2}} \cr 
& = 5.\left| {{{( - 2)}^2}} \right| = 5.4 = 20 \cr} \)

b)

\(\eqalign{
& - 4\sqrt {{{( - 3)}^6}} = - 4\sqrt {{{\left[ {{{\left( { - 3} \right)}^3}} \right]}^2}} \cr 
& = - 4.\left| {{{\left( { - 3} \right)}^3}} \right| = - 4.\left| { - 27} \right| \cr 
& = - 4.27 = - 108 \cr} \)

c)

\(\eqalign{
& \sqrt {\sqrt {{{( - 5)}^8}} } = \sqrt {\sqrt {{{\left[ {{{\left( { - 5} \right)}^4}} \right]}^2}} } \cr 
& = \sqrt {{{( - 5)}^4}} = \sqrt {{{\left[ {{{\left( { - 5} \right)}^2}} \right]}^2}} \cr 
& = \left| {{{( - 5)}^2}} \right| = 25 \cr} \)

d)

\(\eqalign{
& 2\sqrt {{{( - 5)}^6}} + 3\sqrt {{{( - 2)}^8}} \cr 
& = 2.\sqrt {{{\left[ {{{\left( { - 5} \right)}^3}} \right]}^2}} + 3.\sqrt {{{\left[ {{{\left( { - 2} \right)}^4}} \right]}^2}} \cr} \)

\(\eqalign{
& = 2.\left| {{{( - 5)}^3}} \right| + 3.\left| {{{( - 2)}^4}} \right| \cr 
& = 2.\left| { - 125} \right| + 3.\left| {16} \right| \cr 
& = 2.125 + 3.16 = 298 \cr} \)