[Toán 9] Rút gọn.
Ngày 24/7/2017 bạn Trần Chí gửi một số bài tập:
Bài 1. Rút gọn:
a) $\sqrt{\frac{2a^3}{3a^8}}$ với a $\geq$ 0 b) $\sqrt{13a}$$\sqrt{\frac{52}{a}}$ với a > 0
c) $\sqrt{5a}$.$\sqrt{45a}$ - 3a với a $\geq$ 0
d) $(3 - a)^2$ - $\sqrt{0,1}$.$\sqrt{208a^2}$ !?
Bài 2. Tính
a) ($\sqrt{12}$ + 3$\sqrt{15}$ - 4$\sqrt{35}$)$\sqrt{3}$
b) $\sqrt{252}$ - $\sqrt{700}$ + $\sqrt{1008}$ - $\sqrt{448}$
c) 2$\sqrt{40\sqrt{12}}$ - 2$\sqrt{\sqrt{75}}$ - 3$\sqrt{5\sqrt{48}}$
d) $\sqrt{3 + \sqrt{5 + 2\sqrt{3}}}$.$\sqrt{3 - 6\sqrt{6 + 2\sqrt{3}}}$ !?
e) $\sqrt{4 + \sqrt{8}}$.$\sqrt{2 + \sqrt{2 + \sqrt{2}}}$.$\sqrt{2 - \sqrt{2 + \sqrt{2}}}$
f) (4 + $\sqrt{15}$).($\sqrt{10}$ - $\sqrt{6}$).$\sqrt{4 - \sqrt{15}}$
g) (3 - $\sqrt{5}$).$\sqrt{3 + \sqrt{5}}$ + (3 + $\sqrt{5}$).$\sqrt{5 - \sqrt{5}}$
Trả lời cho bạn:
Bài 1
a) $\sqrt{\frac{2a^3}{3a^8}}$ = $\frac{\sqrt{2a.a^2}}{\sqrt{3.(a^4)^2}}$ = $\frac{\left | a \right |\sqrt{2a}}{a^4\sqrt{3}}$ = $\frac{\sqrt{2a}}{a^3\sqrt{3}}$ = $\frac{\sqrt{6a}}{a^3}$ với a > 0
b) $\sqrt{13a}$$\sqrt{\frac{52}{a}}$ = $\sqrt{13}$.$\sqrt{a}$.$\frac{\sqrt{52}}{\sqrt{a}}$ = $\sqrt{13}$.$\sqrt{4.13}$ = $\sqrt{13}$.2$\sqrt{13}$ = 2.13 = 26.
c) $\sqrt{5a}$.$\sqrt{45a}$ - 3a = $\sqrt{5a}$.$\sqrt{9.5a}$ - 3a = $\sqrt{5a}$.3$\sqrt{5a}$ - 3a = 3.5a - 3a = 12a.
Bài 2:
a) ($\sqrt{12}$ + 3$\sqrt{15}$ - 4$\sqrt{35}$)$\sqrt{3}$
= ($\sqrt{4.3}$ + 3$\sqrt{3}$$\sqrt{5}$ - $\sqrt{9}$$\sqrt{3}$$\sqrt{5}$)$\sqrt{3}$ = (2$\sqrt{3}$ + 3$\sqrt{3}$$\sqrt{5}$ - 12$\sqrt{3}$$\sqrt{5}$)$\sqrt{3}$
= [$\sqrt{3}$(2 + 3$\sqrt{5}$ - 12$\sqrt{5}$)]$\sqrt{3}$ = $\sqrt{3}$$\sqrt{3}$(2 - 9$\sqrt{5}$) = 6 - 27$\sqrt{5}$
b) $\sqrt{252}$ - $\sqrt{700}$ + $\sqrt{1008}$ - $\sqrt{448}$
= $\sqrt{7 . 36}$ - $\sqrt{7 . 100}$ + $\sqrt{7 . 144}$ - $\sqrt{7 . 64}$
= 6$\sqrt{7}$ - 10$\sqrt{7}$ + 12$\sqrt{7}$ - 8$\sqrt{7}$
= 6$\sqrt{7}$ + 12$\sqrt{7}$ - 10$\sqrt{7}$ - 8$\sqrt{7}$
= 18$\sqrt{7}$ - 18$\sqrt{7}$ = 0.
c) 2$\sqrt{40\sqrt{12}}$ - 2$\sqrt{\sqrt{75}}$ - 3$\sqrt{5\sqrt{48}}$
= 2$\sqrt{4.10\sqrt{4.3}}$ - 2$\sqrt{\sqrt{25.3}}$ - 3$\sqrt{5\sqrt{16.3}}$
= 2$\sqrt{4.10.2\sqrt{3}}$ - 2$\sqrt{5\sqrt{3}}$ - 3$\sqrt{5.4\sqrt{3}}$
= 2.2.2$\sqrt{5\sqrt{3}}$ - 2$\sqrt{5\sqrt{3}}$ - 6$\sqrt{5\sqrt{3}}$
= $\sqrt{5\sqrt{3}}$(8 - 2 - 6) = 0
e) $\sqrt{4 + \sqrt{8}}$.$\sqrt{2 + \sqrt{2 + \sqrt{2}}}$.$\sqrt{2 - \sqrt{2 + \sqrt{2}}}$
= $\sqrt{2(2 + \sqrt{2})}$.$\sqrt{(2 + \sqrt{2 + \sqrt{2}})(2 - \sqrt{2 + \sqrt{2}})}$
= $\sqrt{2(2 + \sqrt{2})}$.$\sqrt{4 - (2 + \sqrt{2})}$
= $\sqrt{2(2 + \sqrt{2})}$.$\sqrt{2 + \sqrt{2}}$
= $\sqrt{2(2 + \sqrt{2})(2 + \sqrt{2})}$
= $\sqrt{2(2 + \sqrt{2})^2}$
= $\sqrt{2}$(2 + $\sqrt{2}$) = 2$\sqrt{2}$ + 2
f) (4 + $\sqrt{15}$).($\sqrt{10}$ - $\sqrt{6}$).$\sqrt{4 - \sqrt{15}}$
= ($\sqrt{10}$ - $\sqrt{6}$).$\sqrt{(4 + \sqrt{15})^2.(4 - \sqrt{15})}$
= ($\sqrt{2}$$\sqrt{5}$ - $\sqrt{2}$$\sqrt{3}$).$\sqrt{(4 + \sqrt{15})(4 + \sqrt{15}).(4 - \sqrt{15})}$
= $\sqrt{2}$($\sqrt{5}$ - $\sqrt{3}$).$\sqrt{(4 + \sqrt{15}).[4^2 - (\sqrt{15})^2]}$
= $\sqrt{2}$($\sqrt{5}$ - $\sqrt{3}$)$\sqrt{4 + \sqrt{15}}$
= $\sqrt{2(\sqrt{5} + \sqrt{3})^2.(4 + \sqrt{15})}$
= $\sqrt{2(5 - 2\sqrt{15} + 3).(4 + \sqrt{15})}$
= $\sqrt{2(8 - 2\sqrt{15}).(4 + \sqrt{15})}$
= $\sqrt{2.2(4 - \sqrt{15}).(4 + \sqrt{15})}$
= $\sqrt{4[4^2 - (\sqrt{15})^2]}$ = 2
Mỗi bài toán có nhiều cách giải, đừng quên chia sẻ cách giải hoặc ý kiến đóng góp của bạn ở khung nhận xét bên dưới. Xin cảm ơn!
Bài 1. Rút gọn:
a) $\sqrt{\frac{2a^3}{3a^8}}$ với a $\geq$ 0 b) $\sqrt{13a}$$\sqrt{\frac{52}{a}}$ với a > 0
c) $\sqrt{5a}$.$\sqrt{45a}$ - 3a với a $\geq$ 0
d) $(3 - a)^2$ - $\sqrt{0,1}$.$\sqrt{208a^2}$ !?
Bài 2. Tính
a) ($\sqrt{12}$ + 3$\sqrt{15}$ - 4$\sqrt{35}$)$\sqrt{3}$
b) $\sqrt{252}$ - $\sqrt{700}$ + $\sqrt{1008}$ - $\sqrt{448}$
c) 2$\sqrt{40\sqrt{12}}$ - 2$\sqrt{\sqrt{75}}$ - 3$\sqrt{5\sqrt{48}}$
d) $\sqrt{3 + \sqrt{5 + 2\sqrt{3}}}$.$\sqrt{3 - 6\sqrt{6 + 2\sqrt{3}}}$ !?
e) $\sqrt{4 + \sqrt{8}}$.$\sqrt{2 + \sqrt{2 + \sqrt{2}}}$.$\sqrt{2 - \sqrt{2 + \sqrt{2}}}$
f) (4 + $\sqrt{15}$).($\sqrt{10}$ - $\sqrt{6}$).$\sqrt{4 - \sqrt{15}}$
g) (3 - $\sqrt{5}$).$\sqrt{3 + \sqrt{5}}$ + (3 + $\sqrt{5}$).$\sqrt{5 - \sqrt{5}}$
Trả lời cho bạn:
Bài 1
a) $\sqrt{\frac{2a^3}{3a^8}}$ = $\frac{\sqrt{2a.a^2}}{\sqrt{3.(a^4)^2}}$ = $\frac{\left | a \right |\sqrt{2a}}{a^4\sqrt{3}}$ = $\frac{\sqrt{2a}}{a^3\sqrt{3}}$ = $\frac{\sqrt{6a}}{a^3}$ với a > 0
b) $\sqrt{13a}$$\sqrt{\frac{52}{a}}$ = $\sqrt{13}$.$\sqrt{a}$.$\frac{\sqrt{52}}{\sqrt{a}}$ = $\sqrt{13}$.$\sqrt{4.13}$ = $\sqrt{13}$.2$\sqrt{13}$ = 2.13 = 26.
c) $\sqrt{5a}$.$\sqrt{45a}$ - 3a = $\sqrt{5a}$.$\sqrt{9.5a}$ - 3a = $\sqrt{5a}$.3$\sqrt{5a}$ - 3a = 3.5a - 3a = 12a.
Bài 2:
a) ($\sqrt{12}$ + 3$\sqrt{15}$ - 4$\sqrt{35}$)$\sqrt{3}$
= ($\sqrt{4.3}$ + 3$\sqrt{3}$$\sqrt{5}$ - $\sqrt{9}$$\sqrt{3}$$\sqrt{5}$)$\sqrt{3}$ = (2$\sqrt{3}$ + 3$\sqrt{3}$$\sqrt{5}$ - 12$\sqrt{3}$$\sqrt{5}$)$\sqrt{3}$
= [$\sqrt{3}$(2 + 3$\sqrt{5}$ - 12$\sqrt{5}$)]$\sqrt{3}$ = $\sqrt{3}$$\sqrt{3}$(2 - 9$\sqrt{5}$) = 6 - 27$\sqrt{5}$
b) $\sqrt{252}$ - $\sqrt{700}$ + $\sqrt{1008}$ - $\sqrt{448}$
= $\sqrt{7 . 36}$ - $\sqrt{7 . 100}$ + $\sqrt{7 . 144}$ - $\sqrt{7 . 64}$
= 6$\sqrt{7}$ - 10$\sqrt{7}$ + 12$\sqrt{7}$ - 8$\sqrt{7}$
= 6$\sqrt{7}$ + 12$\sqrt{7}$ - 10$\sqrt{7}$ - 8$\sqrt{7}$
= 18$\sqrt{7}$ - 18$\sqrt{7}$ = 0.
c) 2$\sqrt{40\sqrt{12}}$ - 2$\sqrt{\sqrt{75}}$ - 3$\sqrt{5\sqrt{48}}$
= 2$\sqrt{4.10\sqrt{4.3}}$ - 2$\sqrt{\sqrt{25.3}}$ - 3$\sqrt{5\sqrt{16.3}}$
= 2$\sqrt{4.10.2\sqrt{3}}$ - 2$\sqrt{5\sqrt{3}}$ - 3$\sqrt{5.4\sqrt{3}}$
= 2.2.2$\sqrt{5\sqrt{3}}$ - 2$\sqrt{5\sqrt{3}}$ - 6$\sqrt{5\sqrt{3}}$
= $\sqrt{5\sqrt{3}}$(8 - 2 - 6) = 0
e) $\sqrt{4 + \sqrt{8}}$.$\sqrt{2 + \sqrt{2 + \sqrt{2}}}$.$\sqrt{2 - \sqrt{2 + \sqrt{2}}}$
= $\sqrt{2(2 + \sqrt{2})}$.$\sqrt{(2 + \sqrt{2 + \sqrt{2}})(2 - \sqrt{2 + \sqrt{2}})}$
= $\sqrt{2(2 + \sqrt{2})}$.$\sqrt{4 - (2 + \sqrt{2})}$
= $\sqrt{2(2 + \sqrt{2})}$.$\sqrt{2 + \sqrt{2}}$
= $\sqrt{2(2 + \sqrt{2})(2 + \sqrt{2})}$
= $\sqrt{2(2 + \sqrt{2})^2}$
= $\sqrt{2}$(2 + $\sqrt{2}$) = 2$\sqrt{2}$ + 2
f) (4 + $\sqrt{15}$).($\sqrt{10}$ - $\sqrt{6}$).$\sqrt{4 - \sqrt{15}}$
= ($\sqrt{10}$ - $\sqrt{6}$).$\sqrt{(4 + \sqrt{15})^2.(4 - \sqrt{15})}$
= ($\sqrt{2}$$\sqrt{5}$ - $\sqrt{2}$$\sqrt{3}$).$\sqrt{(4 + \sqrt{15})(4 + \sqrt{15}).(4 - \sqrt{15})}$
= $\sqrt{2}$($\sqrt{5}$ - $\sqrt{3}$).$\sqrt{(4 + \sqrt{15}).[4^2 - (\sqrt{15})^2]}$
= $\sqrt{2}$($\sqrt{5}$ - $\sqrt{3}$)$\sqrt{4 + \sqrt{15}}$
= $\sqrt{2(\sqrt{5} + \sqrt{3})^2.(4 + \sqrt{15})}$
= $\sqrt{2(5 - 2\sqrt{15} + 3).(4 + \sqrt{15})}$
= $\sqrt{2(8 - 2\sqrt{15}).(4 + \sqrt{15})}$
= $\sqrt{2.2(4 - \sqrt{15}).(4 + \sqrt{15})}$
= $\sqrt{4[4^2 - (\sqrt{15})^2]}$ = 2
Mỗi bài toán có nhiều cách giải, đừng quên chia sẻ cách giải hoặc ý kiến đóng góp của bạn ở khung nhận xét bên dưới. Xin cảm ơn!
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