a) ĐS: 2.4.

b) ĐS: 28.

c) HD: Đổi 12,1.360 thành 121.36. ĐS: 66

d) ĐS: 18.

a) \(\sqrt{0,09.64}\)

\(=\sqrt{0,09}.\sqrt{64}\)

\(=0,3.8=2,4\)

b) \(\sqrt{2^4.\left(-7\right)^2}\)

\(=\sqrt{2^4}.\sqrt{\left(-7\right)^2}\)

\(=2^2.7=4.7=28\)

c) \(\sqrt{12,1.360}\)

\(=\sqrt{121.36}\)

\(=\sqrt{121}.\sqrt{36}\)

\(=11.6=66\)

d) \(\sqrt{2^2.3^4}\)

\(=\sqrt{2^2}.\sqrt{3^4}\)

\(=2.3^2=2.9=18\)

a) \(\sqrt{0,09.64}=\sqrt{\left(0,3\right)^2.8^2}=0,3.8=2,4\)

b) \(\sqrt{2^4.\left(-7\right)^2}=\sqrt{\left(2^2\right)^2.\left(-7\right)^2}=2^2.\left|-7\right|=7.4=28\)

c) \(\sqrt{12,1.360}=\sqrt{12,1.10.36}=\sqrt{121.36}=\sqrt{11^2.6^2}=11.6=66\)

d) \(\sqrt{2^2.3^4}=\sqrt{2^2.\left(3^2\right)^2}=2.3^2=9.2=18\)

a) \(\sqrt{0,09\cdot64}=\sqrt{0,09}\cdot\sqrt{64}=0,3\cdot8=2,4\)

b) \(\sqrt{2^4\cdot\left(-7\right)^2}=\sqrt{2^4}\cdot\sqrt{\left(-7\right)^2}=2^2\cdot7=4\cdot7=28\)

c) \(\sqrt{12,1\cdot360}=\sqrt{12,1\cdot10\cdot36}=\sqrt{121\cdot36}=\sqrt{121}\cdot\sqrt{36}=11\cdot6=66\)

d) \(\sqrt{2^2\cdot3^4}=\sqrt{2^2}\cdot\sqrt{3^4}=2\cdot3^2=2\cdot9=18\)

Học tốt nhé :)

a)\(\sqrt{7.63}\)=21

b)\(\sqrt{2,5.30.48}\)=60

c)\(\sqrt{0,4.6,4}\)=1,6

d)\(\sqrt{2,7.5.1,5}\)=4,5

a) 2 \sqrt{6}, \sqrt{29}, 4 \sqrt{2}, 3 \sqrt{5} ;26​,29​,42​,35​;

b) \sqrt{38}, 2 \sqrt{14}, 3 \sqrt{7}, 6 \sqrt{2}38​,214​,37​,62

a) \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)

b) \(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)

a, \(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)do \(0,1>0\)

b, \(\sqrt{\left(-0,3\right)^2}=\sqrt{\left(0,3\right)^2}=\left|0,3\right|=0,3\)do \(0,3>0\)

c, \(-\sqrt{\left(-1,3\right)^2}=-\sqrt{\left(1,3\right)^2}=-\left|1,3\right|=-1,3\)do \(1,3>0\)

d, \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\sqrt{\left(0,4\right)^2}=-0,4.\left|0,4\right|=-0,4.0,4=-0,14\)

do \(0,4>0\)

\(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)

\(\sqrt{\left(-0,3\right)^2}=\left|-0,3\right|=0,3\)

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

\(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\cdot\left|-0,4\right|=-0,16\)

Em mới lớp 7 nên em chỉ làm những câu em biết thôi nhé:

\(a,\sqrt{x}=15\)

\(\Rightarrow x=15^2\)

\(\Rightarrow x=225\)

\(b,2\sqrt{x}=14\)

\(\sqrt{x}=14:2\)

\(\sqrt{x}=7\)

\(x=7^2\)

\(x=49\)

\(c,\sqrt{x}< \sqrt{2}\)

\(\Rightarrow x< 2\)

Còn ý d em không biết làm ạ ! 

\(a)\sqrt{x}=15\)

Vì \(x\ge0\) nên bình phương hai vế ta được:

\(x=15^2\Leftrightarrow x=225\)

Vậy \(x=225\)

\(b)2\sqrt{x}=14\Leftrightarrow\sqrt{x}=7\)

Vì  \(x\ge0\) nên bình phương hai vế ta được:

\(x=7^2\Leftrightarrow x=49\)

Vậy \(x=49\)

\(c)\sqrt{x}< \sqrt{2}\)

Vì \(x\ge0\) nên bình phương hai vế ta được: \(x< 2\)

Vậy \(0\le x\le2\)

\(d)\sqrt{2x}< 4\)

Vì \(x\ge0\)nên bình phương hai vế ta được:

\(2x< 16\Leftrightarrow x< 8\)

Vậy \(0\le x< 8\)

a) \(\sqrt{16}\).\(\sqrt{25}\)+\(\sqrt{196}\):\(\sqrt{49}\)

=4.5+14/7

=20+2

=22

a) \(\sqrt{16}\).\(\sqrt{25}\) + \(\sqrt{196}\) : \(\sqrt{49}\) = 4.5+14:9=22

b) 36:\(\sqrt{2.3^2.18}\) - \(\sqrt{169}\)= 36 :  \(\)18 - 13 = -11

c) \(\sqrt{\sqrt{81}}\) = 3

d) \(\sqrt{3^2+4^2}\)= \(\sqrt{25}\)=5

Áp dụng quy tắc khai phương một thương, hãy tính :

a) 9169−−−−√ = \(\sqrt{\dfrac{3^2}{13^2}}\) = \(\left|\dfrac{3}{13}\right|\) = \(\dfrac{3}{13}\)

b) 25144−−−−√ = \(\sqrt{\dfrac{5^2}{12^2}}\) = \(\left|\dfrac{5}{12}\right|\) = \(\dfrac{5}{12}\)

c) 1916−−−−√ = \(\sqrt{\dfrac{25}{16}}\) = \(\sqrt{\dfrac{5^2}{4^2}}\) = \(\left|\dfrac{5}{4}\right|\) = \(\dfrac{5}{4}\)

d) 2781−−−−√ = \(\sqrt{\dfrac{169}{81}}\) = \(\sqrt{\dfrac{13^2}{9^2}}\) = \(\left|\dfrac{13}{9}\right|\) = \(\dfrac{13}{9}\)

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