C = \(\frac{2}{3}\sqrt{144}-\left(-\frac{3}{4}\right)\div\sqrt{\frac{225}{144}}\)

C = \(\frac{2}{3}.12+\frac{3}{4}\div\frac{5}{4}\)

C = \(8+\frac{3}{5}\)

C = \(8\frac{3}{5}\)

D = \(\frac{4^6.25^5-2^{12}.25^4}{2^{12}.5^8-10^8.64}\)

D = \(\frac{\left(2^2\right)^6.\left(5^2\right)^5-2^{12}.\left(5^2\right)^4}{2^{12}.5^8-\left(2.5\right)^8.2^6}\)

D = \(\frac{2^{12}.5^{10}-2^{12}.5^8}{2^{12}.5^8-2^8.5^8.2^6}\)

D = \(\frac{2^{12}.5^8.\left(25-1\right)}{2^{12}.5^8.\left(1-2^2\right)}\)

D = \(\frac{24}{-3}\)

D = \(-8\)

\(C=\frac{2}{3}\sqrt{144}-\left(\frac{-3}{4}\right):\sqrt{\frac{225}{144}}\)

\(=\frac{2}{3}\cdot12+\frac{3}{4}:\frac{5}{4}\)

\(=8+\frac{3}{4}\cdot\frac{4}{5}\)

\(=8+\frac{3}{5}\)

\(=\frac{40}{5}+\frac{3}{4}=\frac{43}{5}\)

\(D=\frac{4^6\cdot25^5-2^{12}\cdot25^4}{2^{12}\cdot5^8-10^8\cdot64}=\frac{\left(2^2\right)^6\cdot\left(5^2\right)^5-2^{12}\cdot\left(5^2\right)^4}{2^{12}\cdot5^8-\left(2\cdot5\right)^8\cdot2^6}\)

\(=\frac{2^{12}\cdot5^{10}-2^{12}\cdot5^8}{2^{12}\cdot5^8-2^{14}\cdot5^8}=\frac{5^8\left(2^{12}\cdot5^2-2^{12}\right)}{5^8\left(2^{12}-2^{14}\right)}\)

\(=\frac{2^{12}\cdot5^2-2^{12}}{2^{12}-2^{14}}=\frac{2^{12}\left(5^2-1\right)}{2^{12}\left(1-2^2\right)}=\frac{24}{-3}=-8\)

a)\(\sqrt{9.4}=\sqrt{36}=6;\sqrt{9}.\sqrt{4}=3.2=6\Rightarrow\sqrt{9.4}=\sqrt{9}.\sqrt{4}\)

b)\(\sqrt{169-144}=\sqrt{25}=5;\sqrt{169}-\sqrt{144}=13-12=1\Rightarrow\sqrt{169-144}>\sqrt{169}-\sqrt{144}\)

tra loi ho mik lun di mai ik hoc roi !chut chut chuit chut

\(\frac{1}{2}.\sqrt{144}+\sqrt{\frac{1}{9}}\\ =\frac{1}{2}.12+\frac{1}{3}\\ =6+\frac{1}{3}\\ =\frac{19}{3}\)

\(\frac{1}{2}.\sqrt{144}+\sqrt{\frac{1}{9}}\)

\(=\frac{1}{2}.12+\frac{1}{3}\)

\(=6+\frac{1}{3}\)

\(=\frac{19}{3}.\)

Chúc bạn học tốt!

Câu a:

\(\sqrt{26}\) + \(\sqrt{17}\) và 9

\(\sqrt{26}\) > \(\sqrt{25}\) = 5

\(\sqrt{17}\) > \(\sqrt{16}\) = 4

Cộng vế với vế ta có:

\(\sqrt{26}\) + \(\sqrt{17}\) > 5+ 4 = 9

Vậy \(\sqrt{26}\) + \(\sqrt{17}\) > 9

Câu b:

\(\sqrt8\) - \(\sqrt5\) và 1

\(\sqrt5\) > \(\sqrt4\) = 2

- \(\sqrt5\) < - 2 (nhân cả hai vế của bất đẳng thức với cùng một số âm thì bất đẳng thức đổi chiều)

- \(\sqrt5\) < - 2 (chứng minh trên)

\(\sqrt8\) < \(\sqrt9\) = 3

Cộng vế với vế ta có:

\(\sqrt8\) - \(\sqrt5\) < - 2 + 3

\(\sqrt8\) - \(\sqrt5\) < 1

Vậy \(\sqrt8\) - \(\sqrt5\) < 1

1)

a) = 15

1 a)15 b)\(2\) c)\(\sqrt{7}-1\)

\(b,\sqrt{225}-\frac{1}{\sqrt{5}}-1=15-\frac{1}{\sqrt{5}}-1\)

\(=\frac{15\sqrt{5}}{\sqrt{5}}-\frac{1}{\sqrt{5}}-\frac{\sqrt{5}}{\sqrt{5}}=\frac{15\sqrt{5}-1-\sqrt{5}}{\sqrt{5}}\)

\(=\frac{14\sqrt{5}-1}{\sqrt{5}}=\frac{\sqrt{5}\left(14\sqrt{5}-1\right)}{5}=\frac{70-\sqrt{5}}{5}\)

Bài 1:

a) Ta có: \(\left(0.125\right)\cdot\left(-3\cdot7\right)\cdot\left(-2\right)^3\)

\(=\frac{1}{8}\cdot\left(-21\right)\cdot\left(-8\right)\)

\(=\frac{1}{8}\cdot168\)

\(=21\)

b) Ta có: \(\sqrt{36}\cdot\sqrt{\frac{25}{16}}+\frac{1}{4}\)

\(=\sqrt{36\cdot\frac{25}{16}}+\frac{1}{4}\)

\(=\sqrt{\frac{225}{4}}+\frac{1}{4}\)

\(=\frac{15}{2}+\frac{1}{4}\)

\(=\frac{31}{4}\)

c) Ta có: \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)

\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)

\(=\frac{2}{5}-\frac{7}{5}=-1\)

d) Ta có: \(0,1\cdot\sqrt{225}\cdot\sqrt{\frac{1}{4}}\)

\(=0,1\cdot15\cdot\frac{1}{2}=\frac{3}{4}\)

hay quá  cảm ơn bạn nhé 

\(a,\sqrt{64}-\sqrt{16}+\sqrt{\left(-5\right)^2}\)

\(=8+4+5\)

\(=17\)

\(b,\sqrt{49}+\sqrt{4}-\sqrt{9}.\sqrt{144}\)

\(=7+2-3.12\)

\(=9-36\)

\(=-27\)