a) `3\sqrt3=\sqrt(3^2 .3)=\sqrt27`

\sqrt12=\sqrt12`

`=> \sqrt27 > \sqrt12`

`=> 3\sqrt3 > \sqrt12`

b) `7=\sqrt(7^2)=\sqrt49`

`3\sqrt5=\sqrt(3^2 .5)=\sqrt45`

`=> \sqrt49>\sqrt45`

`=>7>3\sqrt5`

c) `1/3 \sqrt51 = \sqrt( (1/3)^2 .51) =\sqrt(17/3)`

`1/5 \sqrt150 =\sqrt( (1/5)^2 .150)=\sqrt6`

`=> \sqrt(17/3) < \sqrt6`

`=> 1/3 \sqrt51 < 1/5 \sqrt150`

d) `1/2 \sqrt6 = \sqrt(3/2)`

`6\sqrt(1/2) =\sqrt(18)`

`=> \sqrt(3/2) < \sqrt18`

`=> 1/2 \sqrt6 < 6\sqrt(1/2)`.

a) Ta có: 

+)√25+9=√34+)25+9=34.

+)√25+√9=√52+√32=5+3+)25+9=52+32=5+3

=8=√82=√64=8=82=64.

Vì 34<6434<64 nên √34<√6434<64

Vậy √25+9<√25+√925+9<25+9

b) Với a>0,b>0a>0,b>0, ta có

+)(√a+b)2=a+b+)(a+b)2=a+b.

+)(√a+√b)2=(√a)2+2√a.√b+(√b)2+)(a+b)2=(a)2+2a.b+(b)2

 =a+2√ab+b=a+2ab+b

 =(a+b)+2√ab=(a+b)+2ab. 

Vì a>0, b>0a>0, b>0 nên √ab>0⇔2√ab>0ab>0⇔2ab>0

⇔(a+b)+2√ab>a+b⇔(a+b)+2ab>a+b

⇔(√a+√b)2>(√a+b)2⇔(a+b)2>(a+b)2

⇔√a+√b>√a+b⇔a+b>a+b (đpcm)

a, Ta có : \(\sqrt{25+9}=\sqrt{34}\)

\(\sqrt{25}+\sqrt{9}=5+3=8=\sqrt{64}\)

mà 34 < 64 hay \(\sqrt{25+9}< \sqrt{25}+\sqrt{9}\)

b, \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)

bình phương 2 vế ta được : \(a+b< a+2\sqrt{ab}+b\)

\(\Leftrightarrow2\sqrt{ab}>0\)vì \(a;b>0\)nên đẳng thức này luôn đúng )

Vậy ta có đpcm 

a) \(\sqrt{25+9}=\sqrt{34}\)

\(\sqrt{25}+\sqrt{9}=5+3=8=\sqrt{64}\)

=> \(\sqrt{25+9}< \sqrt{25}+\sqrt{9}\)

b) Vì a,b > 0, bình phương hai vế ta có :

a + b < a + 2√ab + b

<=> -2√ab < 0 <=> 2√ab > 0 ( đúng vì a,b > 0 )

=> đpcm

a, Ta có  \(\sqrt{25-16}=\sqrt{9}=3\)

\(\sqrt{25}-\sqrt{16}=5-4=1\)

Do 3 > 1 nên \(\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)

a) căn 25 - 16  > căn 25 - căn 16

b)Với $a>b>0$ nên  \sqrt{a},\sqrt{b},\sqrt{a-b}a​,b​,− đều xác định

Để so sánh \sqrt{a}-\sqrt{b}a​−b​ và $-$ ta quy về so sánh \sqrt{a}a​ và \sqrt{a-b}+\sqrt{b}−+b​.

+) (\sqrt{a})^2=a(a​)2=a.

+) (\sqrt{a-b}+\sqrt{b})^2=(\sqrt{a-b})^2+2\sqrt{a-b}.\sqrt{b}+(\sqrt{b})^2=a-b+b+2\sqrt{a-b}.\sqrt{b}=a+2\sqrt{a-b}.\sqrt{b}(−+b​)2=(−)2+2−.b​+(b​)2=a−b+b+2−.b​=a+2−

.b​.

Do $a>b>0$ nên 2\sqrt{a-b}.\sqrt{b}>02−.b​>0

$\Rightarrow$ a+2\sqrt{a-b}.\sqrt{b}>aa+2−.b​>a

$\Rightarrow$ (\sqrt{a-b}+\sqrt{b})^2>(\sqrt{a})^2(−+b​)2>(a​)2

Do \sqrt{a},\sqrt{a-b}+\sqrt{b}>0a​,−+b​>0 

$\Rightarrow$ \sqrt{a-b}+\sqrt{b}>\sqrt{a}−+b​>a​

$\iff$ \sqrt{a-b}>\sqrt{a}-\sqrt{b}−>a​−b​ (đpcm)

Vậy \sqrt{a-b}>\sqrt{a}-\sqrt{b}−>a​−b​.

a) +) $\sqrt{25-16}=\sqrt{9}=3$.

    +) $\sqrt{25}-\sqrt{16}=5-4=1$.

Vì $3>1$ nên $\sqrt{25-16}>\sqrt{25}-\sqrt{16}$.

Vậy $\sqrt{25-16}>\sqrt{25}-\sqrt{16}$.

b) Với $a>b>0$ nên $\sqrt{a},\sqrt{b},\sqrt{a-b}$ đều xác định. 

Để so sánh $\sqrt{a}-\sqrt{b}$ và $\sqrt{a-b}$ ta quy về so sánh $\sqrt{a}$ và $\sqrt{a-b}+\sqrt{b}$.

+) $(\sqrt{a}{)}^2=a$.

+) $(\sqrt{a-b}+\sqrt{b}{)}^2=(\sqrt{a-b}{)}^2+2\sqrt{a-b}.\sqrt{b}+(\sqrt{b}{)}^2=a-b+b+2\sqrt{a-b}.\sqrt{b}=a+2\sqrt{a-b}.\sqrt{b}$.

Do $a>b>0$ nên $2\sqrt{a-b}.\sqrt{b}>0$

$\Rightarrow$ $a+2\sqrt{a-b}.\sqrt{b}>a$

$\Rightarrow$ $(\sqrt{a-b}+\sqrt{b}{)}^2>(\sqrt{a}{)}^2$

Do $\sqrt{a},\sqrt{a-b}+\sqrt{b}>0$ 

$\Rightarrow$ $\sqrt{a-b}+\sqrt{b}>\sqrt{a}$

$\iff$ $\sqrt{a-b}>\sqrt{a}-\sqrt{b}$ (đpcm)

Vậy $\sqrt{a-b}>\sqrt{a}-\sqrt{b}$.

Trả lời:

a) ta có: 2 = √4

Vì 4 > 3 nên √4 > √3

Vậy 2 > √3

b) Ta có: 6 = √36

Vì 36 < 41 nên √36 < √41

Vậy 6 < √41

c)  ta có 7 = √49

Vì 49 > 47 nên √49 > √47

Vậy 7 > √47

>;<;>

a) 3\(\sqrt{3}\)=\(\sqrt{27}\)>\(\sqrt{12}\)

c) \(\frac{1}{3}\)\(\sqrt{51}\)=\(\sqrt{\frac{51}{9}}\)<\(\frac{1}{5}\)\(\sqrt{150}\)=\(\sqrt{\frac{150}{25}}\)=\(\sqrt{6}\)

b) 3\(\sqrt{5}\)=\(\sqrt{45}\)< 7=\(\sqrt{49}\)

d) \(\frac{1}{2}\sqrt{6}\)=\(\sqrt{\frac{6}{4}}\)=\(\sqrt{\frac{3}{2}}\)< 6\(\sqrt{\frac{1}{2}}\)=\(\sqrt{\frac{36}{2}}\)=\(\sqrt{18}\)

a) Ta có: $3\sqrt{3}=\sqrt{3^2.3}=\sqrt{9.3}=\sqrt{27}$

Vì $\sqrt{27}>\sqrt{12}$ nên $3\sqrt{3}>\sqrt{12}$

Vậy $3\sqrt{3}>\sqrt{12}$.
b) Ta có: $3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}$

$7=\sqrt{7^2}=\sqrt{49}$

Vì $\sqrt{49}>\sqrt{45}$ nên $7>3\sqrt{5}$

Vậy $7>3\sqrt{5}$.

1351=(13)2.51=519

15150=(15)2.150=15025=6=6.99=549

549>519 nên 

126<612.

a) \(3\sqrt{3}=\sqrt{9}.\sqrt{3}=\sqrt{27}>\sqrt{12}\)

b) \(3\sqrt{5}=\sqrt{9}.\sqrt{5}=\sqrt{45}< \sqrt{49}=7\)

c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{1}{9}}.\sqrt{51}=\sqrt{\dfrac{51}{9}}=\sqrt{\dfrac{17}{3}}< \sqrt{6}=\dfrac{1}{5}\sqrt{150}\)

d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{3}{2}}< \sqrt{18}=6\sqrt{\dfrac{1}{2}}\)

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

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

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

Vậy $5>\sqrt[]{3123}$. 

b, Ta có :

+)53√6=3√53.6=3√125.6=3√750+)63√5=3√63.5=3√216.5=3√1080+)563=53.63=125.63=7503+)653=63.53=216.53=10803

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

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

Vậy $5\sqrt[]{36}<6\sqrt[]{35}$.

a) 5=\sqrt[3]{125}5=3125​ mà \sqrt[3]{125}>\sqrt[3]{123}3125​>3123​ suy ra 5>\sqrt[3]{123}5>3123​.

b) Đưa về so sánh 5^{3}.653.6 với 6^{3}.563.5 . Kết quả 6 \sqrt[3]{5}>5 \sqrt[3]{6}635​>536​.

Rút gọn ta được:

M=√a−1/√a

Viết M ở dạng M=1−1/√a

suy ra M<1

Với \(x>0;x\ne1\)

\(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

\(=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)

\(=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}\)

\(=1-\frac{1}{\sqrt{a}}< 1\)hay M < 1 

M = 1 - 1/√a < 1

Rút gọn các biểu thức sau với x≥0x≥0:

a) 2\(\sqrt{3x}\)-4\(\sqrt{3x}\)+27-3\(\sqrt{3x}\)=27-5\(\sqrt{3x}\)

b)3\(\sqrt{2x}\)-5\(\sqrt{8x}\)+7\(\sqrt{18x}\)+28

=3\(\sqrt{2x}\)-10\(\sqrt{2x}\)+21\(\sqrt{2x}\)+28

=14\(\sqrt{2x}\)+28=14(\(\sqrt{2x}\)+2)

a) \(2\sqrt{3x}-4\sqrt{3x}+27-3\sqrt{3x}\)

\(=\left(2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}\right)+27\)

\(=-5\sqrt{3x}+27\)

b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28\)

\(=3\sqrt{2x}-5\sqrt{4.2x}+7\sqrt{9.2x}+28\)

\(=3\sqrt{2x}-5\sqrt{2^2.2x}+7\sqrt{3^2.2x}+28\)

\(=3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}+28\)

\(=\left(3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}\right)+28\)

\(=\left(3-10+21\right)\sqrt{2x}+28\)

\(=14\sqrt{2x}+28\)

\(a,B=4\sqrt{x=1}-3\sqrt{x+1}+2\)\(\sqrt{x+1}+\sqrt{x+1}\)

\(=4\sqrt{x+1}\)

\(b,\)đưa về \(\sqrt{x+1}=4\Rightarrow x=15\)

a, Với \(x\ge-1\)

\(\Rightarrow B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)

\(=4\sqrt{x+1}\)

b, Ta có B = 16 hay 

\(4\sqrt{x+1}=16\Leftrightarrow\sqrt{x+1}=4\)bình phương 2 vế ta được 

\(\Leftrightarrow x+1=16\Leftrightarrow x=15\)

a) B = 4√x+1                                   b) x = 15