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a) a √ a 2 − b 2 − ( 1 + a √ a 2 − b 2 ) : b a − √ a 2 − b 2 = a √ a 2 − b 2 − a + √ a 2 − b 2 √ a 2 − b 2 ⋅ a − √ a 2 − b 2 b = a √ a 2 − b 2 − a 2 − ( √ a 2 − b 2 ) 2 b √ a 2 − b 2 = a √ a 2 − b 2 − a 2 − ( a 2 − b 2 ) b √ a 2 − b 2 = a √ a 2 − b 2 − b 2 b ⋅ √ a 2 − b 2 = a √ a 2 − b 2 − b √ a 2 − b 2 = a − b √ a 2 − b 2 = √ a − b ⋅ √ a − b √ a − b ⋅ √ a + b (do a > b > 0 )$ = √ a − b √ a + b Vậy Q = √ a − b √ a + b . b) Thay a = 3 b vào Q = √ a − b √ a + b , ta được: Q = √ 3 b − b √ 3 b + b = √ 2 b √ 4 b = √ 2 b √ 2 ⋅ √ 2 b = 1 √ 2 = √ 2 2 .

) 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​1​

=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4 \sqrt{6}}{2}\right) \cdot \dfrac{1}{\sqrt{6}}=(26​​−246​​)⋅6​1​

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

=-\dfrac{3}{2}=-1,5=V P=−23​=−1,5=VP.
b) VT=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}VT=(1−2​14​−7​​+1−3​15​−5​​):7​−5​1​

=\left(\dfrac{\sqrt{7} \cdot \sqrt{2}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{5} \cdot \sqrt{3}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}=(1−2​7​⋅2​−7​​+1−3​5​⋅3​−5​​):7​−5​1​

=\left[\dfrac{\sqrt{7}(\sqrt{2}-1)}{1-\sqrt{2}}+\dfrac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}\right]: \dfrac{1}{\sqrt{7}-\sqrt{5}}=[1−2​7​(2​−1)​+1−3​5​(3​−1)​]:7​−5​1​

=(-\sqrt{7}-\sqrt{5})(\sqrt{7}-\sqrt{5})=(−7​−5​)(7​−5​)

=-(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=−(7​+5​)(7​−5​)

$=-(7-5)=-2=VP$.

c) V T=\dfrac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}VT=ab​ab​+ba​​:a​−b​1​

=\dfrac{\sqrt{a} \cdot \sqrt{a} \cdot \sqrt{b}+\sqrt{b} \cdot \sqrt{b} \cdot \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}=ab​a​⋅a​⋅b​+b​⋅b​⋅a​​:a​−b​1​

=\dfrac{\sqrt{a} \cdot \sqrt{a b}+\sqrt{b} \cdot \sqrt{a b}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}=ab​a​⋅ab​+b​⋅ab​​:a​−b​1​

=\dfrac{\sqrt{a b}(\sqrt{a}+\sqrt{b})}{\sqrt{a b}} \cdot(\sqrt{a}-\sqrt{b})=ab​ab​(a​+b​)​⋅(a​−b​)

=(\sqrt{a}+\sqrt{b}) \cdot(\sqrt{a}-\sqrt{b})=(a​+b​)⋅(a​−b​)

$=a-b=VP$.

d) VT=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)VT=(1+a​+1a+a​​)(1−a​−1a−a​​)

=\left(1+\dfrac{\sqrt{a} \cdot \sqrt{a}+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a} \cdot \sqrt{a}-\sqrt{a}}{\sqrt{a}-1}\right)=(1+a​+1a​⋅a​+a​​)(1−a​−1a​⋅a​−a​​)

=\left[1+\dfrac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}\right]\left[1-\dfrac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}\right]=[1+a​+1a​(a​+1)​][1−a​−1a​(a​−1)​]

=(1+\sqrt{a})(1-\sqrt{a})=(1+a​)(1−a​)

=1-(\sqrt{a})^{2}=1-a=V P=1−(a​)2=1−a=VP

a, \(\sqrt{\left(2x-1\right)^2}=3\Leftrightarrow\left|2x-1\right|=3\)

Với \(x\ge\frac{1}{2}\)pt có dạng : \(2x-1=3\Leftrightarrow x=2\)( tm )

Với \(x< \frac{1}{2}\)pt có dạng : \(-2x+1=3\Leftrightarrow x=-1\)( tm ) 

Vậy tập nghiệm của pt là S = { -1 ; 2 } 

b, \(\frac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\)ĐK : \(x\ge0\)

\(\Leftrightarrow\frac{2}{3}\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\Leftrightarrow\frac{1}{3}\sqrt{15x}=2\)

\(\Leftrightarrow\sqrt{15x}=6\)bình phương 2 vế : \(\Leftrightarrow15x=36\Leftrightarrow x=\frac{36}{15}=\frac{12}{5}\)( tm ) 

Vậy tập nghiệm của pt là S = { 12/5 } 

) √ ( 2 x − 1 ) 2 = 3 ⇒ | 2 x − 1 | = 3 ⇔ 2 x − 1 = ± 3 +) TH1: 2 x − 1 = 3 ⇒ 2 x = 4 ⇒ x = 2 +) TH2: 2 x − 1 = − 3 ⇒ 2 x = − 2 ⇒ x = − 1 Vậy x = − 1 ; x = 2 . b) Điều kiện: x ≥ 0 5 3 √ 15 x − √ 15 x − 2 = 1 3 √ 15 x ⇔ 5 3 √ 15 x − √ 15 x − 1 3 √ 15 x = 2 ⇔ ( 5 3 − 1 − 1 3 ) √ 15 x = 2 ⇔ 1 3 √ 15 x = 2 ⇔ √ 15 x = 6 ⇔ 15 x = 36 ⇔ x = 12 5 Vậy x = 12 5 .

a) √ − 9 a − √ 9 + 12 a + 4 a 2 = √ − 9 a − √ 3 2 + 2.3 .2 a + ( 2 a ) 2 = √ 3 2 ⋅ ( − a ) − √ ( 3 + 2 a ) 2 = 3 √ − a − | 3 + 2 a | Thay a = − 9 ta được: 3 √ 9 − | 3 + 2 ⋅ ( − 9 ) | = 3.3 − 15 = − 6 . b) Điều kiện: m ≠ 2 1 + 3 m m − 2 √ m 2 − 4 m + 4 = 1 + 3 m m − 2 √ m 2 − 2.2 ⋅ m + 2 2 = 1 + 3 m m − 2 √ ( m − 2 ) 2 = 1 + 3 m | m − 2 | m − 2 +) m > 2 , ta được: 1 + 3 m m − 2 √ m 2 − 4 m + 4 = 1 + 3 m . ( 1 ) +) m < 2 , ta được: 1 + 3 m m − 2 √ m 2 − 4 m + 4 = 1 − 3 m . ( 2 ) Với m = 1 , 5 < 2 . Thay vào biểu thức ( 2 ) ta có: 1 − 3 m = 1 − 3.1 , 5 = − 3 , 5 Vậy giá trị biểu thức tại m = 1 , 5 là − 3 , 5 . c) √ 1 − 10 a + 25 a 2 − 4 a = √ 1 − 2.1 .5 a + ( 5 a ) 2 − 4 a = √ ( 1 − 5 a ) 2 − 4 a = | 1 − 5 a | − 4 a +) Với a < 1 5 , ta được: 1 − 5 a − 4 a = 1 − 9 a . ( 3 ) +) Với a ≥ 1 5 , ta được: 5 a − 1 − 4 a = a − 1 . ( 4 ) Vì a = √ 2 > 1 5 . Thay vào biểu thức ( 4 ) ta có: a − 1 = √ 2 − 1 . Vậy giá trị của biểu thức tại a = √ 2 là √ 2 − 1 . d) 4 x − √ 9 x 2 + 6 x + 1 = 4 x − √ ( 3 x ) 2 + 2.3 x + 1 = 4 x − √ ( 3 x + 1 ) 2 = 4 x − | 3 x + 1 | +) Với 3 x + 1 ≥ 0 ⇔ x ≥ − 1 3 , ta có: 4 x − ( 3 x + 1 ) = 4 x − 3 x − 1 = x − 1 . ( 5 ) +) Với 3 x + 1 < 0 ⇔ x < − 1 3 , ta có: 4 x + ( 3 x + 1 ) = 4 x + 3 x + 1 = 7 x + 1 . ( 6 ) Vì x = − √ 3 < − 1 3 . Thay vào biểu thức ( 6 ) , ta có: 7 x + 1 = 7 . ( − √ 3 ) + 1 = − 7 √ 3 + 1 . Giá trị của biểu thức tại x = − √ 3 là − 7 √ 3 + 1

a) \sqrt{-9a}-\sqrt{9+12 a+4 a^{2}}−9a​−9+12a+4a2​

=\sqrt{-9 a}-\sqrt{3^{2}+2.3 .2 a+(2 a)^{2}}=−9a​−32+2.3.2a+(2a)2​

=\sqrt{3^{2} \cdot(-a)}-\sqrt{(3+2 a)^{2}}=32⋅(−a)​−(3+2a)2​

=3 \sqrt{-a}-|3+2 a|=3−a​−∣3+2a∣

Thay $a=-9$ ta được:

3 \sqrt{9}-|3+2 \cdot(-9)|=3.3-15=-639​−∣3+2⋅(−9)∣=3.3−15=−6.

b) Điều kiện: $m\ne 2$

1+\dfrac{3 m}{m-2} \sqrt{m^{2}-4 m+4}1+m−23m​m2−4m+4​

=1+\dfrac{3 m}{m-2} \sqrt{m^{2}-2.2 \cdot m+2^{2}}=1+m−23m​m2−2.2⋅m+22​

=1+\dfrac{3 m}{m-2} \sqrt{(m-2)^{2}}=1+m−23m​(m−2)2​

=1+\dfrac{3 m|m-2|}{m-2}=1+m−23m∣m−2∣​

+) $m>2$, ta được: 1+\dfrac{3 m}{m-2} \sqrt{m^{2}-4 m+4}=1+3 m1+m−23m​m2−4m+4​=1+3m. $(1)$

+) $m<2$, ta được: 1+\dfrac{3 m}{m-2} \sqrt{m^{2}-4 m+4}=1-3 m1+m−23m​m2−4m+4​=1−3m. $(2)$

Với $m=1,5<2$. Thay vào biểu thức $(2)$ ta có: $1-3m=1-3.1,5=-3,5$

Vậy giá trị biểu thức tại $m=1,5$ là $-3,5$.

c) \sqrt{1-10 a+25 a^{2}}-4a1−10a+25a2​−4a

=\sqrt{1-2.1 .5 a+(5 a)^{2}}-4 a=1−2.1.5a+(5a)2​−4a

=\sqrt{(1-5a)^{2}}-4 a=(1−5a)2​−4a

$=|1-5a|-4a$

+) Với a <\dfrac{1}{5}a<51​, ta được: $1-5a-4a=1-9a$. $(3)$

+) Với a \ge \dfrac{1}{5}a≥51​, ta được: $5a-1-4a=a-1$. $(4)$

Vì a=\sqrt{2}>\dfrac{1}{5}a=2​>51​. Thay vào biểu thức $(4)$ ta có: a-1=\sqrt{2}-1a−1=2​−1.

Vậy giá trị của biểu thức tại a=\sqrt{2}a=2​ là \sqrt{2}-12​−1.

d) 4 x-\sqrt{9 x^{2}+6 x+1}4x−9x2+6x+1​

=4 x-\sqrt{(3 x)^{2}+2.3 x+1}=4 x-\sqrt{(3 x+1)^{2}}=4x−(3x)2+2.3x+1​=4x−(3x+1)2​

$=4x-|3x+1|$

+) Với $3x+1\ge 0$ $\iff$ x \ge -\dfrac{1}{3}x≥−31​, ta có: $4x-(3x+1)=4x-3x-1=x-1$. $(5)$

+) Với $3x+1<0$ $\iff$ x <-\dfrac{1}{3}x<−31​, ta có: $4x+(3x+1)=4x+3x+1=7x+1$. $(6)$

Vì x=-\sqrt{3}<-\dfrac{1}{3}x=−3​<−31​. Thay vào biểu thức $(6)$, ta có: 7 x+1=7 .(-\sqrt{3})+1=-7 \sqrt{3}+17x+1=7 .(−3​)+1=−73​+1.

Giá trị của biểu thức tại x=-\sqrt{3}x=−3​ là -7 \sqrt{3}+1−73​+1.

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

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

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

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

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

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

\(b,\)

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

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

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

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

\(=2\sqrt{5}\)

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

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

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

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

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

c) $\sqrt{a+b}+\sqrt{a^2-b^2}$

$=\sqrt{a+b}+\sqrt{(a+b)(a-b)}$

$=\sqrt{a+b}+\sqrt{a+b}\cdot \sqrt{a-b}$

$=\sqrt{a+b}(1+\sqrt{a-b})$.

d) $12-\sqrt{x}-x$

$=12-4\sqrt{x}+3\sqrt{x}-x$

$=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})$

$=(3-\sqrt{x})(4+\sqrt{x})$.

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) \(\dfrac{40}{27}\)

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.

a) Ta có: $5=\sqrt[3]{35^3}=\sqrt[3]{3125}$

Vì $125>123\iff \sqrt[3]{3125}>\sqrt[3]{3123}$   

                        $\iff 5>\sqrt[3]{3123}$

Vậy $5>\sqrt[3]{3123}$. 

b, Ta có :

$\begin{array}{c}+)5\sqrt[3]{36}=\sqrt[3]{35^3.6}=\sqrt[3]{3125.6}=\sqrt[3]{3750}\\ +)6\sqrt[3]{35}=\sqrt[3]{3{}^{}.5}=\sqrt[3]{3216.5}=\sqrt[3]{31080}\end{array}$

Vì $750<1080\iff \sqrt[3]{3750}<\sqrt[3]{31080}$

                          $\iff 5\sqrt[3]{36}<6\sqrt[3]{35}$.

Vậy $5\sqrt[3]{36}<6\sqrt[3]{35}$.

LG a

3√27−3√−8−3√125273−−83−1253

Phương pháp giải:

Tính từng căn bậc ba rồi thực hiện phép tính

Lời giải chi tiết:

3√27−3√−8−3√125=3√33−3√(−2)3−3√53273−−83−1253=333−(−2)33−533

=3−(−2)−5=3−(−2)−5

=3+2−5=0=3+2−5=0.

LG b

 3√1353√5−3√54.3√4135353−543.43

Phương pháp giải:

Sử dụng các công thức:

3√a.b=3√a.3√ba.b3=a3.b3.

3√ab=3√a3√bab3=a3b3,  với b≠0b≠0.

Lời giải chi tiết:

3√1353√5−3√54.3√4=3√27.53√5−3√54.4135353−543.43=27.5353−54.43

=3√5.3√273√5−3√216=53.27353−2163

=3√27−3√216=273−2163

=3√33−3√63=333−633

=3−6=−3=3−6=−3.

a) \sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}=3-(-2)-5=0327​−3−8​−3125​=3−(−2)−5=0.

b) \dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54} \cdot \sqrt[3]{4}=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54.4}=3-6=-335​3135​​−354​⋅34​=35135​​−354.4​=3−6=−3.

Ta có:

+ 3√512=3√83=8;5123=833=8;

+ 3√−729=3√(−9)3=−9;−7293=(−9)33=−9;

+ 3√0,064=3√0,43=0,4;0,0643=0,433=0,4;

+ 3√−0,216=3√(−0,6)3=−0,6;−0,2163=(−0,6)33=−0,6;

+ 3√−0,008=3√(−0,2)3=−0,2.

Đáp án:

( lần lượt như trên nhé!!! Ko viết lại đề)

8 ; - 9 ; 0,4 ; - 0,6 ; - 0,2