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 $\sqrt{a-b}>\sqrt{a}-\sqrt{b}$.

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) (\(\sqrt{3}\)-1)²=3-2\(\sqrt{3}\)+1= 4-2\(\sqrt{3}\) (ĐPCM)

b) \(\sqrt{4-2\sqrt{3}}\)=\(\sqrt{3}\)-1 >0

Bình phương 2 vế, ta có:

4-2\(\sqrt{3}\)=3-2\(\sqrt{3}\)+1= 4-2\(\sqrt{3}\) (ĐPCM)

a)  \(\left(\sqrt{3}-1\right)^2\)=\(\left(\sqrt{3}\right)^2\)- 2\(\sqrt{3}\) +1= 3- 2\(\sqrt{3}\) +1=4-2\(\sqrt{3}\)

b)  \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\) = \(\sqrt{\left(\sqrt{3}-1\right)^2}\) - \(\sqrt{3}\)= \(|\sqrt{3}-1|\)-\(\sqrt{3}\)=\(\sqrt{3}\)-1-\(\sqrt{3}\)=-1

a. ( √3 -1)^2 = 3- 2√3 +1 = 4-2√3 (=VP)

b. √(√3-1)^2 - √3 = |√3-1| - √3 = √3-1-√3 = -1 (=VP)

a, \(\sqrt{\frac{289}{25}}=\frac{\sqrt{289}}{\sqrt{25}}=\frac{17}{5}\)

b, \(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\frac{8}{5}\)

c, \(\sqrt{\frac{0,25}{9}}=\frac{\sqrt{0,25}}{\sqrt{9}}=\frac{0,5}{3}=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)

d, \(\sqrt{\frac{8,1}{16}}\)đề có sai ko cô ? 

a) căn 289 / 225 = 17/15

b) căn 64/ 25 = 8/5

c) căn 0,25 / 9 = 1/6

d) căn 8,1 / 1,6 = 9/4

289225=289225=1715.

1425=2.25+1425=6425=6425=85.

0,259=0,259=0,53=530=16.

8,11,6=8116=8116=94.

\(\sqrt{\dfrac{1}{600}}\)=\(\sqrt{\dfrac{1}{10^2\cdot6}}\)=\(\sqrt{\dfrac{1\cdot6}{10^2\cdot6\cdot6}}\)=\(\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}\)=\(\sqrt{\dfrac{11\cdot540}{540\cdot540}}\)=\(\dfrac{\sqrt{5940}}{540}\)=\(\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}\)=\(\sqrt{\dfrac{3\cdot50}{50\cdot50}}\)=\(\dfrac{\sqrt{150}}{50}\)=\(\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}\)=\(\sqrt{\dfrac{5\cdot98}{98\cdot98}}=\dfrac{\sqrt{490}}{98}=\dfrac{\sqrt{10}}{14}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)

\(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{10}}{14}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)

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.

a) Điều kiện: x≥0x≥0

√16x=816x=8⇔(√16x)2=82⇔(16x)2=82 ⇔16x=64⇔16x=64 ⇔x=6416⇔x=4⇔x=6416⇔x=4 (thỏa mãn điều kiện)

Vậy x=4x=4.

Cách khác: 

√16x=8⇔√16.√x=8⇔4√x=8⇔√x=2⇔x=22⇔x=416x=8⇔16.x=8⇔4x=8⇔x=2⇔x=22⇔x=4

b) Điều kiện: 4x≥0⇔x≥04x≥0⇔x≥0

 √4x=√54x=5 ⇔(√4x)2=(√5)2⇔4x=5⇔x=54⇔(4x)2=(5)2⇔4x=5⇔x=54 (thỏa mãn điều kiện) 

Vậy x=54x=54.

c) Điều kiện: 9(x−1)≥0⇔x−1≥0⇔x≥19(x−1)≥0⇔x−1≥0⇔x≥1

√9(x−1)=219(x−1)=21⇔3√x−1=21⇔3x−1=21⇔√x−1=7⇔x−1=7 ⇔x−1=49⇔x=50⇔x−1=49⇔x=50 (thỏa mãn điều kiện)

Vậy x=50x=50.

Cách khác:

√9(x−1)=21⇔9(x−1)=212⇔9(x−1)=441⇔x−1=49⇔x=509(x−1)=21⇔9(x−1)=212⇔9(x−1)=441⇔x−1=49⇔x=50

d) Điều kiện: x∈Rx∈R (vì 4.(1−x)2≥04.(1−x)2≥0 với mọi x)x)

√4(1−x)2−6=04(1−x)2−6=0⇔2√(1−x)2=6⇔2(1−x)2=6 ⇔|1−x|=3⇔|1−x|=3 ⇔[1−x=31−x=−3⇔[1−x=31−x=−3 ⇔[x=−2x=4⇔[x=−2x=4 

Vậy x=−2;x=4.

a, \(\sqrt{16x}=8\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

b, \(\sqrt{4x}=\sqrt{5}\)ĐK : x \(\ge0\)

bình phương 2 vế ta được : \(4x=5\Leftrightarrow x=\frac{5}{4}\)

c, \(\sqrt{9\left(x-1\right)}=21\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\)

bình phương 2 vế ta được : \(x-1=49\Leftrightarrow x=50\)

d, \(\sqrt{4\left(1-x\right)^2}-6=0\Leftrightarrow2\left|1-x\right|=6\Leftrightarrow\left|1-x\right|=3\)

TH1 : \(1-x=3\Leftrightarrow x=-2\)

TH2 : \(1-x=-3\Leftrightarrow x=4\)

a) Điều kiện: $x\ge 0$.

$\sqrt{16x}=8$

$\Rightarrow$ $(\sqrt{16x}{)}^2=8^2$ 

$\Rightarrow$ $16x=64$

$\Rightarrow$ $x=64:16$

$\Rightarrow$ $x=4$ (thỏa mãn điều kiện).

Vậy $x=4$.

b) Điều kiện: $4x\ge 0$ $\iff$ $x\ge 0$.

$\sqrt{4x}=\sqrt{5}$ 

$\Rightarrow$ $(\sqrt{4x}{)}^2=(\sqrt{5}{)}^2$

$\Rightarrow$ $4x=5$

54 (thỏa mãn điều kiện).

54.

c) Điều kiện: $9(x-1)\ge 0$ $\iff$ $x-1\ge 0$ $\iff$ $x\ge 1$.

$\sqrt{9(x-1)}=21$ 

$\Rightarrow$ $\sqrt{9}.\sqrt{x-1}=21$ (do $x-1\ge 0$)

$\Rightarrow$ $3\sqrt{x-1}=21$

$\Rightarrow$ $\sqrt{x-1}=7$

$\Rightarrow$ $x-1=49$

$\Rightarrow$ $x=50$ (thỏa mãn điều kiện).

Vậy $x=50$.

d) Điều kiện: $x$$\in$ $R$ (vì $4.(1-x{)}^2\ge 0$ với mọi $x$$)$

$\sqrt{4(1-x{)}^2}-6=0$

$\Rightarrow$  $2\sqrt{(1-x{)}^2}=6$ 

$\Rightarrow$ $|1-x|=3$ 

$\iff$ 

Vậy $x=-2$ hoặc $x=4$.

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

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

\(b,\)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

a) \(\sqrt{x^2}\)=7

=> x²=49

=> x={-7;7}

b) \(\sqrt{x^2}\)=|-8|=8

=> x²=64

=>x={-8;8}

c) \(\sqrt{4x^2}\)=6

4x²=36

=>x²=9

=> x={-3;3}

d)\(\sqrt{9x^2}\)=|-12|=12

=> 9x²=144

=> x²=16

=> x={-4;4}

a)x=+7 hoặc x= -7

b) x=8 hoặc x= -8

c)x=3 hoặc x =-3

d) x=4 hoặc x= -4

a)  \(\sqrt{x^2}=7\)⇒\(|x|\)=7⇒xϵ{7:-7}

b)  \(\sqrt{x^2}\) = \(|-8|\)⇒xϵ{8:-8}

c)  \(\sqrt{4x^2}\)=6⇒xϵ{3:-3}

d)  \(\sqrt{9x^2}\) = \(|\)-12\(|\)⇒xϵ{4:-4}