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​

$=$

) \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​

20,24845673 : 16,9

1,198133534 x 121,55=145,6331311

) V T = ( 2 √ 3 − √ 6 √ 8 − 2 − √ 216 3 ) ⋅ 1 √ 6 = ( √ 2 ⋅ √ 2 ⋅ √ 3 − √ 6 √ 2 2 ⋅ 2 − 2 − √ 6 2 .6 3 ) ⋅ 1 √ 6 = ( √ 2 ⋅ √ 6 − √ 6 2 √ 2 − 2 − 6 . √ 6 3 ) ⋅ 1 √ 6 = [ √ 6 ( √ 2 − 1 ) 2 ( √ 2 − 1 ) − 6 √ 6 3 ] ⋅ 1 √ 6 = ( √ 6 2 − 2 √ 6 ) ⋅ 1 √ 6 = ( √ 6 2 − 4 √ 6 2 ) ⋅ 1 √ 6 = ( − 3 2 √ 6 ) ⋅ 1 √ 6 = − 3 2 = − 1 , 5 = V P . b) V T = ( √ 14 − √ 7 1 − √ 2 + √ 15 − √ 5 1 − √ 3 ) : 1 √ 7 − √ 5 = ( √ 7 ⋅ √ 2 − √ 7 1 − √ 2 + √ 5 ⋅ √ 3 − √ 5 1 − √ 3 ) : 1 √ 7 − √ 5 = [ √ 7 ( √ 2 − 1 ) 1 − √ 2 + √ 5 ( √ 3 − 1 ) 1 − √ 3 ] : 1 √ 7 − √ 5 = ( − √ 7 − √ 5 ) ( √ 7 − √ 5 ) = − ( √ 7 + √ 5 ) ( √ 7 − √ 5 ) = − ( 7 − 5 ) = − 2 = V P . c) V T = a √ b + b √ a √ a b : 1 √ a − √ b = √ a ⋅ √ a ⋅ √ b + √ b ⋅ √ b ⋅ √ a √ a b : 1 √ a − √ b = √ a ⋅ √ a b + √ b ⋅ √ a b √ a b : 1 √ a − √ b = √ a b ( √ a + √ b ) √ a b ⋅ ( √ a − √ b ) = ( √ a + √ b ) ⋅ ( √ a − √ b ) = a − b = V P . d) V T = ( 1 + a + √ a √ a + 1 ) ( 1 − a − √ a √ a − 1 ) = ( 1 + √ a ⋅ √ a + √ a √ a + 1 ) ( 1 − √ a ⋅ √ a − √ a √ a − 1 ) = [ 1 + √ a ( √ a + 1 ) √ a + 1 ] [ 1 − √ a ( √ a − 1 ) √ a − 1 ] = ( 1 + √ a ) ( 1 − √ a ) = 1 − ( √ a ) 2 = 1 − a = V P

a) VT=\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}VT=(8​−223​−6​​−3216​​)⋅6​1​

=\left(\dfrac{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3}-\sqrt{6}}{\sqrt{2^{2} \cdot 2}-2}-\dfrac{\sqrt{6^{2} .6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=(22⋅2​−22​⋅2​⋅3​−6​​−362.6​​)⋅6​1​

=\left(\dfrac{\sqrt{2} \cdot \sqrt{6}-\sqrt{6}}{2 \sqrt{2}-2}-\dfrac{6 . \sqrt{6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=(22​−22​⋅6​−6​​−36.6​​)⋅6​1​

=\left[\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}-\dfrac{6 \sqrt{6}}{3}\right] \cdot \dfrac{1}{\sqrt{6}}=[2(2​−1)6​(2​−1)​−366​​]⋅6​1​

=\left(\dfrac{\sqrt{6}}{2}-2 \sqrt{6}\right) \cdot \dfrac{1}{\sqrt{6}}=(26​​−26​)⋅6

a) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\dfrac{1}{2}-2=-\dfrac{3}{2}\)

 

  Chúc Bạn Học Tốt.

\(a,\left(\sqrt{8}-3.\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)

\(=\sqrt{8}.\sqrt{2}-3\sqrt{2}.\sqrt{2}+\sqrt{10}.\sqrt{2}-\sqrt{5}\)

\(=\sqrt{16}-3.2+\sqrt{20}-\sqrt{5}\)

\(=\sqrt{4^2}-6+\sqrt{2^2.5}-\sqrt{5}\)

\(=2-6+2\sqrt{5}-\sqrt{5}\)

\(=-2+\sqrt{5}\)

\(b,\)

\(0,2\sqrt{\left(-10^2\right).3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)

\(=0,2.\left|-10\right|.\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|\)

\(=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)

\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)

\(=2\sqrt{5}\)

a) $xy-y\sqrt{x}+\sqrt{x}-1$

$=y\cdot \sqrt{x}\cdot \sqrt{x}-y\sqrt{x}+\sqrt{x}-1$

$=y\sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)$

$=(\sqrt{x}-1)(y\sqrt{x}+1)$.

b) $\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}$

$=(\sqrt{ax}+\sqrt{bx})-(\sqrt{ay}+\sqrt{by})$

$=(\sqrt{a}\cdot \sqrt{x}+\sqrt{b}$

a) x y − y √ x + √ x − 1 = y ⋅ √ x ⋅ √ x − y √ x + √ x − 1 = y √ x ( √ x − 1 ) + ( √ x − 1 ) = ( √ x − 1 ) ( y √ x + 1 ) . b) √ a x − √ b y + √ b x − √ a y = ( √ a x + √ b x ) − ( √ a y + √ b y ) = ( √ a ⋅ √ x + √ b ⋅ √ x ) − ( √ a ⋅ √ y + √ b ⋅ √ y ) = √ x ( √ a + √ b ) − √ y ( √ a + √ b ) = ( √ a + √ b ) ( √ x − √ y ) . c) √ a + b + √ a 2 − b 2 = √ a + b + √ ( a + b ) ( a − b ) = √ a + b + √ a + b ⋅ √ a − b = √ a + b ( 1 + √ a − b ) . d) 12 − √ x − x = 12 − 4 √ x + 3 √ x − x = 4 ( 3 − √ x ) + √ x ( 3 − √ x ) = ( 3 − √ x ) ( 4 + √ x

a) (√8 - 3√2 + √10)√2 - √5

= (√22.2 - 3√2 + √5.2)√2 - √5

= (2√2 - 3√2 + √5.√2)√2 - √5

= (2 - 3 + √5)√2.√2 - √5

= (-1 + √5).2 - √5

= -2 + 2√5 - √5

= -2 + √5

b) 0,2√((-10)2.3) + 2√(√3 - √52)

= 0,2.10√3 + 2|√3 - √5|

= 2√3 + 2(√5 - √3)

= 0,2.10.√3 + 2|√3 - √5|

= 2√3 + 2(√5 - √3)

= 2√3 + 2√5 - 2√3

 

= 2√5

Giải phần c và d

a) -2 + \(\sqrt{5}\)

b) 2\(\sqrt{5}\)

c) 54\(\sqrt{2}\)

d) 1 + \(\sqrt{2}\)

3a2b4=ab2.3a2b4

3a2.b4=ab2.3|a|.|b2|

3(−a).b2 (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên $b^2>0$ $\Rightarrow$  $\left|b^2\right|=b^2$)

$=-\sqrt{3}$.

27(a−3)248=9(a−3)216

9.(a−3)216=3.|a−3|4

3(a−3)4. 

(Do $a>3$ nên $|a-3|=a-3$)

9+12a+4a2b2=32+2.3.2a+(2a)2b2

(3+2a)2b2=|3+2a||b|

3a2b4=ab2.3a2b4

3a2.b4=ab2.3|a|.|b2|

3(−a).b2 (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên $b^2>0$ $\Rightarrow$  $\left|b^2\right|=b^2$)

$=-\sqrt{3}$.

27(a−3)248=9(a−3)216

9.(a−3)216=3.|a−3|4

3(a−3)4. 

(Do $a>3$ nên $|a-3|=a-3$)

9+12a+4a2b2=32+2.3.2a+(2a)2b2

(3+2a)2b2=|3+2a||b|

3a2b4=ab2.3a2b4

3a2.b4=ab2.3|a|.|b2|

3(−a).b2 (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên $b^2>0$ $\Rightarrow$  $\left|b^2\right|=b^2$)

$=-\sqrt{3}$.

27(a−3)248=9(a−3)216

9.(a−3)216=3.|a−3|4

3(a−3)4. 

(Do $a>3$ nên $|a-3|=a-3$)

9+12a+4a2b2=32+2.3.2a+(2a)2b2

(3+2a)2b2=|3+2a||b|