a) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}.\dfrac{196}{9}}=\sqrt{\dfrac{25}{81}}.\sqrt{\dfrac{16}{49}}.\sqrt{\dfrac{196}{9}}=\dfrac{5}{9}.\dfrac{4}{7}.\dfrac{14}{3}=\dfrac{40}{27}\)

b) \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{25}.2\dfrac{34}{81}}=\sqrt{\dfrac{49}{16}.\dfrac{64}{25}.\dfrac{196}{81}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{64}{25}}.\sqrt{\dfrac{196}{81}}=\dfrac{7}{4}.\dfrac{8}{5}.\dfrac{14}{9}=\dfrac{196}{45}\)

c) \(\dfrac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.49}{81}}=\dfrac{\sqrt{64}.\sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=\dfrac{56}{9}\)

d) \(\sqrt{21,6}.\sqrt{810}.\sqrt{11^2-5^2}=\sqrt{21,6.810.\left(11^2-5^2\right)}=\sqrt{216.81.\left(11+5\right)\left(11-5\right)}=\sqrt{36^2.9^2.4^2}=36.9.4=1296\)

a) \(\dfrac{40}{27}\)

b) \(\dfrac{196}{45}\)

c) \(\dfrac{56}{9}\)

d) 1296

a) \sqrt{\dfrac{25}{81} \cdot \dfrac{16}{49} \cdot \dfrac{196}{9}}8125​⋅4916​⋅9196​​

=\sqrt{\dfrac{25}{81}} \cdot \sqrt{\dfrac{16}{49}} \cdot \sqrt{\dfrac{196}{9}}=8125​​⋅4916​​⋅9196​​

=\sqrt{\left(\dfrac{5}{9}\right)^{2}} \cdot \sqrt{\left(\dfrac{4}{7}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{3}\right)^{2}}=(95​)2​⋅(74​)2​⋅(314​)2​

=\dfrac{5}{9} \cdot \dfrac{4}{7} \cdot \dfrac{14}{3}=\dfrac{40}{27}=95​⋅74​⋅314​=2740​.

b) \sqrt{3 \dfrac{1}{16} \cdot 2 \dfrac{14}{25} \cdot 2 \dfrac{34}{81}}3161​⋅22514​⋅28134​​

=\sqrt{\dfrac{49}{16} \cdot \dfrac{64}{25} \cdot \dfrac{196}{81}}=1649​⋅2564​⋅81196​​

=\sqrt{\dfrac{49}{16}} \cdot \sqrt{\dfrac{64}{25}} \cdot \sqrt{\dfrac{196}{81}}=1649​​⋅2564​​⋅81196​​

=\sqrt{\left(\dfrac{7}{4}\right)^{2}} \cdot \sqrt{\left(\dfrac{8}{5}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{9}\right)^{2}}=(47​)2​⋅(58​)2​⋅(914​)2​

=\dfrac{7}{4} \cdot \dfrac{8}{5} \cdot \dfrac{14}{9}=\dfrac{196}{45}=47​⋅58​⋅914​=45196​.

c) \dfrac{\sqrt{640} \cdot \sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.343}{567}}567​640​⋅34,3​​=567640.34,3​​=56764.343​​

=\sqrt{\dfrac{64.49 .7}{81.7}}=\sqrt{\dfrac{64.49}{81}}=81.764.49.7​​=8164.49​​

=\dfrac{\sqrt{64} \cdot \sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=81​64​⋅49​​=98.7​

=\dfrac{56}{9}=956​.

d) \sqrt{21,6} \cdot \sqrt{810} \cdot \sqrt{11^{2}-5^{2}}21,6​⋅810​⋅112−52​

=\sqrt{21,6.810 \cdot\left(11^{2}-5^{2}\right)}=21,6.810⋅(112−52)​

=\sqrt{216.81 .(11+5)(11-5)}=216.81.(11+5)(11−5)​

=\sqrt{36.6 .9^{2} \cdot 4^{2} .6}=36.6.92⋅42.6​

=\sqrt{36^{2} .9^{2} \cdot 4^{2}}=36.9 .4=1296=362.92⋅42​=36.9.4=1296.

) \sqrt{\dfrac{25}{81} \cdot \dfrac{16}{49} \cdot \dfrac{196}{9}}8125​⋅4916​⋅9196​​

=\sqrt{\dfrac{25}{81}} \cdot \sqrt{\dfrac{16}{49}} \cdot \sqrt{\dfrac{196}{9}}=8125​​⋅4916​​⋅9196​​

=\sqrt{\left(\dfrac{5}{9}\right)^{2}} \cdot \sqrt{\left(\dfrac{4}{7}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{3}\right)^{2}}=(95​)2​⋅(74​)2​⋅(314​)2​

=\dfrac{5}{9} \cdot \dfrac{4}{7} \cdot \dfrac{14}{3}=\dfrac{40}{27}=95​⋅74​⋅314​=2740​.

b) \sqrt{3 \dfrac{1}{16} \cdot 2 \dfrac{14}{25} \cdot 2 \dfrac{34}{81}}3161​⋅22514​⋅28134​​

=\sqrt{\dfrac{49}{16} \cdot \dfrac{64}{25} \cdot \dfrac{196}{81}}=1649​⋅2564​⋅81196​​

=\sqrt{\dfrac{49}{16}} \cdot \sqrt{\dfrac{64}{25}} \cdot \sqrt{\dfrac{196}{81}}=1649​​⋅2564​​⋅81196​​

=\sqrt{\left(\dfrac{7}{4}\right)^{2}} \cdot \sqrt{\left(\dfrac{8}{5}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{9}\right)^{2}}=(47​)2​⋅(58​)2​⋅(914​)2​

=\dfrac{7}{4} \cdot \dfrac{8}{5} \cdot \dfrac{14}{9}=\dfrac{196}{45}=47​⋅58​⋅914​=45196​.

c) \dfrac{\sqrt{640} \cdot \sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.343}{567}}567​640​⋅34,3​​=567640.34,3​​=56764.343​​

=\sqrt{\dfrac{64.49 .7}{81.7}}=\sqrt{\dfrac{64.49}{81}}=81.764.49.7​​=8164.49​​

=\dfrac{\sqrt{64} \cdot \sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=81​64​⋅49​​=98.7​

=\dfrac{56}{9}=956​.

d) \sqrt{21,6} \cdot \sqrt{810} \cdot \sqrt{11^{2}-5^{2}}21,6​⋅810​⋅112−52​

=\sqrt{21,6.810 \cdot\left(11^{2}-5^{2}\right)}=21,6.810⋅(112−52)​

=\sqrt{216.81 .(11+5)(11-5)}=216.81.(11+5)(11−5)​

=\sqrt{36.6 .9^{2} \cdot 4^{2} .6}=36.6.92⋅42.6​

=\sqrt{36^{2} .9^{2} \cdot 4^{2}}=36.9 .4=1296=362.92⋅42​=36.9.4=1296.

ở số chia, từ 1 tới 55 là các số mà 2 số liên tiếp hơn kém nhau 3 đơn vị nên từ 1 đến 55 có (55 – 1) :3 + 1 = 19 số).

a) Tìm được x ≠ -6 và x  ≠  0.

b) Gợi ý: x 3  + 4 x 2  - 6x + 36 = (x + 6) ( x 2  - 2x + 6)

Tìm được  P = x 2 − 2 x + 6 2 x

c) Ta có P = 3 2 ⇔ x 2 − 5 x + 6 = 0 . Từ đó tìm được x = 2 hoặc x = 3 (TMĐK).

d) Tương tự câu c, tìm được x = -6 (KTM) hoặc x = -1 (TM)

e) P = 1 Þ  x 2 ‑ - 4x + 6=  0 Û ( x -   2 ) 2 + 2 = 0 (vô nghiệm)

Vì  ( x -   2 ) 2  + 2 ≥ 2 > 0 với mọi x. Do vậy x ∈ ∅ .