5 căn a - 4b . căn 25a^3 + 5a . căn 16ab^2 - 2 căn 9a

Giải Toán 9 Bài 8: Rút gọn biểu thức căn bậc hai

Bài 59 trang 32 Toán lớp 9 Tập 1: 

a) $5\sqrt{a}-4b\sqrt{25a^3}+5a\sqrt{16a{b}^2}-2\sqrt{9a}$

b) $5a\sqrt{64a{b}^3}-\sqrt{3}.\sqrt{12a^3{b}^3}$$+2ab\sqrt{9ab}-5b\sqrt{81a^3.b}$

Lời giải:

a)

$5\sqrt{a}-4b\sqrt{25a^3}+5a\sqrt{16a{b}^2}-2\sqrt{9a}=5\sqrt{a}-4b\sqrt{25.a^2.a}+5a\sqrt{16.a.b^2}-2\sqrt{9.a}=5\sqrt{a}-4b\mathrm{.5.}\left.a\right|.\sqrt{a}+5a\mathrm{.4.}\left.b\right|.\sqrt{a}-\mathrm{2.3.}\sqrt{a}$$=5\sqrt{a}-20ab\sqrt{a}+20ab\sqrt{a}-6\sqrt{a}$ (vì $a>0;b>0\Rightarrow \left.a\right|=a;\left.b\right|=b$)

$=-\sqrt{a}$

b)

$5a\sqrt{64a{b}^3}-\sqrt{3}.\sqrt{12a^3{b}^3}+2ab\sqrt{9ab}$$-5b\sqrt{81a^3.b}$

$=5a\sqrt{64.ab.b^2}-\sqrt{3}.\sqrt{12.{\left(ab\right)}^2.ab}$$+2ab\sqrt{9.ab}-5b\sqrt{81.a^2.ab}$

$=5a.8\left.b\right|.\sqrt{ab}-\sqrt{3}.\sqrt{12}\left.ab\right|\sqrt{ab}$$+2ab\mathrm{.3.}\sqrt{ab}-5b\mathrm{.9.}\left.a\right|\sqrt{ab}$

$=40ab\sqrt{ab}-6ab\sqrt{ab}+6ab\sqrt{ab}-45ab\sqrt{ab}$ (vì a > 0; b > 0 nên |a| = a; |b| = b; |ab| = ab)

$=-5ab\sqrt{ab}$