Câu 1:

1;\(\dfrac{49}{4}\);\(\dfrac{125}{8}\);\(\dfrac{16}{81}\)

Câu 2

\(3^9\)=\(3^{3.3}\)=\(\left(3^3\right)^3\)=\(27^3\)

2\(^{12}\)=\(2^{4.3}=\left(2^4\right)^3=16^3\)

Bài 5:
\(A+9=\frac{10}{1}+\frac{10}{2}+\frac{10}{3}+....+\frac{10}{9}\)

\(A=\frac{10}{2}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+1=\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)

\(A=10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)=10B\)

$\Rightarrow A:B=10$

em ơi, đây là câu hỏi từ bài đọc thì phải, nên cần phải dựa vào đề bài thì mới làm được?!

Bài 4:

\(a,\sqrt{64.\left(x-1\right)^2}=16\\ \Leftrightarrow8\sqrt{\left(x-1\right)^2}=16\\ \\ \Leftrightarrow\sqrt{\left(x-1\right)^2}=\dfrac{16}{8}=2\\ \left|x-1\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\\ b,\sqrt{4\left(x-2\right)}=8\\ \Leftrightarrow2\sqrt{x-2}=8\\ \Leftrightarrow\sqrt{x-2}=\dfrac{8}{2}=4\\ \Leftrightarrow x-2=4^2=16\\ \Leftrightarrow x=16+2=18\\ c,\dfrac{\sqrt{2x}}{\sqrt{8}}=5\\ \Leftrightarrow\dfrac{\sqrt{2}\sqrt{x}}{2\sqrt{2}}=5\\ \Leftrightarrow\dfrac{\sqrt{x}}{2}=5\\ \Leftrightarrow\sqrt{x}=5.2=10\\ \Leftrightarrow x=10^2=100\)

Bài 2:

\(a,\left(\sqrt{75}-2\sqrt{12}-\sqrt{27}\right).\sqrt{3}\\ =\left(\sqrt{3.5^2}-2.\sqrt{3.2^2}-\sqrt{3.3^2}\right).\sqrt{3}\\ =\left(5\sqrt{3}-2.2\sqrt{3}-3\sqrt{3}\right).\sqrt{3}\\ =-2\sqrt{3}.\sqrt{3}=-2.3=-6\\ b,\left(5\sqrt{2}-\sqrt{8}-\sqrt{98}\right):\sqrt{2}\\ =\left(5\sqrt{2}-\sqrt{2^2.2}-\sqrt{2.7^2}\right):\sqrt{2}\\ =\left(5\sqrt{2}-2\sqrt{2}-7\sqrt{2}\right):\sqrt{2}\\ =-4\sqrt{2}:\sqrt{2}=-4\)

\(c,\\ \dfrac{\sqrt{18}}{\sqrt{2}}=\sqrt{\dfrac{18}{2}}=\sqrt{9}=\sqrt{3^2}=3\\ d,\\ \sqrt{\dfrac{45}{7}}.\sqrt{\dfrac{28}{5}}=\sqrt{\dfrac{45.28}{7.5}}=\sqrt{\dfrac{9.5.4.7}{7.5}}=\sqrt{9.4}=\sqrt{36}=\sqrt{6^2}=6\)

3:

a: \(A=\sqrt{16\cdot a^2\cdot\left(a-1\right)^2}\)

=4|a(a-1)|

=4a(a-1)(Do a>=1)

=4a^2-4a

b: \(B=\dfrac{1}{a-5}\cdot7a^2\left|a-5\right|\)

\(=\dfrac{7a^2\cdot\left(a-5\right)}{a-5}\left(a>5\right)\)

=7a^2

c: \(=\sqrt{225a^2}-3a=15a-3a\left(a>=0\right)\)

=12a

d:D \(=\left(a-3\right)^2-\sqrt{36a^2}\)

=(a-3)^2-6|a|

TH1: a>=0

D=(a-3)^2-6a=a^2-12a+9

TH2: a<0

D=(a-3)^2+6a=a^2+9

e: \(C=\sqrt{225a^2}-8a=15a-8a=7a\)

a: Xét ΔOAI và ΔOBI có

OA=OB

AI=BI

OI chung

Do đó: ΔOAI=ΔOBI

b: Xét ΔOKI và ΔOHI có

OK=OH

\(\widehat{KOI}=\widehat{HOI}\)

OI chung

Do đó: ΔOKI=ΔOHI

Suy ra: KI=HI