a) \(\sqrt{49}+\sqrt{25}-4\cdot0,25\)

\(=7+5-1=11\)

b) \(\sqrt{\frac{1}{9}}\cdot\sqrt{0,81}\cdot\sqrt{0,9}\)

\(=\frac{1}{3}\cdot\frac{9}{10}\cdot\frac{3\sqrt{10}}{10}\)

\(=\frac{9\sqrt{10}}{100}\)

c) \(\sqrt{6,4\cdot2400\cdot0,6}\)

\(=\sqrt{64\cdot36\cdot4}\)

\(=8\cdot6\cdot2=96\)

d) \(\sqrt{26^2-24^2}=\sqrt{\left(26-24\right)\left(26+24\right)}\)

\(=\sqrt{2\cdot50}=\sqrt{100}=10\)

\(\begin{array}{l}a)2.\sqrt 6 .( - \sqrt 6 )\\ =  - 2.\sqrt 6 .\sqrt 6 \\ =  - 2.{(\sqrt 6 )^2}\\ =  - 2.6\\ =  - 12\\b)\sqrt {1,44}  - 2.{(\sqrt {0,6} )^2}\\ = 1,2 - 2.0,6\\ = 1,2 - 1,2\\ = 0\\c)0,1.{(\sqrt 7 )^2} + \sqrt {1,69} \\ = 0,1.7 + 1,3 \\= 0,7 + 1,3 \\= 2\\d)( - 0,1).{(\sqrt {120} )^2} - \frac{1}{4}.{(\sqrt {20} )^2} \\= ( - 0,1).120 - \frac{1}{4}.20\\ =  - 12 - 5\\ =  - (12 + 5)\\ =  - 17\end{array}\)

a: \(=-2\sqrt{6}\cdot\sqrt{6}=-2\cdot\sqrt{6\cdot6}=-2\cdot6=-12\)

b: \(=1.2-2\cdot0.6=1.2-1.2=0\)

c: \(=0.1\cdot7+1.3=0.7+1.3=2\)

d: \(=-0.1\cdot120-\dfrac{1}{4}\cdot20=-12-5=-17\)

\(=4-\dfrac{12}{5}\sqrt{5}-6+\dfrac{18}{5}\sqrt{5}+2\sqrt{5}-6\)

\(=-8+\dfrac{16}{5}\sqrt{5}\)

\(i,\sqrt{12,1.360}=\sqrt{12,1}.6\sqrt{10}=6.\sqrt{12,1.10}=6.\sqrt{121}=6.\sqrt{11^2}=6.11=66\)

\(k,\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{\dfrac{64}{25}}=\dfrac{\sqrt{8^2}}{\sqrt{5^2}}=\dfrac{8}{5}\)

\(l,-0,4.\sqrt{\left(-0,4\right)^2}=-0,4.0,4=-0,16\)

\(m,\sqrt{2^4.\left(-7\right)^2}=\sqrt{4^2}.\sqrt{\left(-7\right)^2}=4.7=28\)

i, \(\sqrt{12,1\cdot360}=\sqrt{4356}=\sqrt{66^2}=66\)

k, \(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4\cdot6,4}=\sqrt{\dfrac{64}{25}}=\sqrt{\dfrac{2^6}{5^2}}=\dfrac{2^3}{5}=\dfrac{8}{5}\)

l, \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\cdot\left|-0,4\right|=-0,4\cdot0,4=-\dfrac{4}{25}\)

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

i)        \(\sqrt{12,1.360}=\sqrt{12,1}.\sqrt{360}=\sqrt{12,1}.\sqrt{36}.\sqrt{10}=\left(\sqrt{12,1}.\sqrt{10}\right)\sqrt{36}=\sqrt{121}.\sqrt{36}=11.6=66\)

k)       \(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{2,56}=1,6\)

l)        \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\left|-0,4\right|=-0,4.0,4=0,16\)

m)         \(\sqrt{2^4.\left(-7\right)^2}=\sqrt{2^2.2^2.\left(-7\right)^2}=\left|2.2.7\right|=28\)

bấm máy tính là ra nha!

\(\begin{array}{l}a)\sqrt {0,49}  + \sqrt {0,64}  = 0,7 + 0,8 = 1,5;\\b)\sqrt {0,36}  - \sqrt {0,81}  = 0,6 - 0,9 =  - 0,3;\\c)8.\sqrt 9  - \sqrt {64}  = 8.3 - 8 = 24 - 8 = 16;\\d)0,1.\sqrt {400}  + 0,2.\sqrt {1600}  = 0,1.20 + 0,2.40 = 2 + 8 = 10\end{array}\)

Ta thấy các số trong căn bậc hai đều lớn hơn 0, áp dụng \(\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\)

a) \(\sqrt{7}\cdot\sqrt{63}=\sqrt{7\cdot63}=21\)

b) \(\sqrt{2,5}\cdot\sqrt{30}\cdot\sqrt{48}=\sqrt{2,5\cdot30\cdot48}=60\)

c) \(\sqrt{0,4}\cdot\sqrt{6,4}=\sqrt{0,4\cdot6,4}=1,6\)

d) \(\sqrt{2,7}\cdot\sqrt{5}\cdot\sqrt{1,5}=\sqrt{2,7\cdot5\cdot1,5}=4,5\)

a. \(\sqrt{7}.\sqrt{63}=\sqrt{7.63}=\sqrt{441}=21\)

b.\(\sqrt{2,5}.\sqrt{30}.\sqrt{48}=\sqrt{2,5.30.48}=\sqrt{3600}=60\)

c.\(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{2,56}=1,6\)

d.\(\sqrt{2,7}.\sqrt{5}.\sqrt{1,5}=\sqrt{2,7.5.1,5}=\sqrt{20,25}=4,5\)

\(\sqrt{7}.\sqrt{63}=\sqrt{7.63}=\sqrt{441}=21\)

\(\sqrt{2,5}.\sqrt{30}.\sqrt{48}=\sqrt{2,5.30.48}=\sqrt{3600}=60\)

\(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{2,56}=1,6\)

\(\sqrt{2,7}.\sqrt{5}.\sqrt{1,5}=\sqrt{2,7.5.1,5}=\sqrt{20,25}=4,5\)

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) $\sqrt{7}.\sqrt{63}=\sqrt{7.63}=\sqrt{{\mathrm{7.7.3}}^2}=\sqrt{(7.3{)}^2}=7.3=21$.

b) $\sqrt{2,5}.\sqrt{30}.\sqrt{48}=\sqrt{2,\mathrm{5.30.48}}=\sqrt{\mathrm{25.3.48}}$

$\sqrt{5^2.3.3{.4}^2}=\sqrt{(\mathrm{5.3.4}{)}^2}=\mathrm{5.3.4}=60$.

c) $\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{0,04.64}$

$=\sqrt{0,2^2{.8}^2}=\sqrt{(0,2.8{)}^2}=0,2.8=1,6$.

d) $\sqrt{2,7}.\sqrt{5}.\sqrt{1,5}=\sqrt{2,\mathrm{7.5.1},5}$

$=\sqrt{3^2.0,\mathrm{3.5.5.0},3}=\sqrt{(3.0,3.5{)}^2}=4,5$.

\(\sqrt{0,4}+\sqrt{2,5}\)

\(=\sqrt{10}\left(\sqrt{0,04}+\sqrt{0,25}\right)\)

\(=\sqrt{10}\left(0,2+0,5\right)\)

\(=\sqrt{10}.\frac{7}{10}=\frac{7}{\sqrt{10}}\)