a) \(\sqrt{9\left(x-2\right)}=6\left(đk:x\ge2\right)\)

\(\Leftrightarrow9\left(x-2\right)=36\Leftrightarrow x-2=4\Leftrightarrow x=6\)(thỏa đk)

b) \(\sqrt{9\left(x-3\right)^2}=12\Leftrightarrow3\left|x-3\right|=12\Leftrightarrow\left|x-3\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\3-x=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

Không thừa đâu bạn. Từ \(2^0\rightarrow2^{99}\) có 100 số số hạng, mỗi cặp 5 số thì chia thành 20 cặp đủ nha bạn

\(A=2^0+2^1+2^2+2^3+2^4+...+2^{99}=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{95}\left(2^0+2^1+2^2+2^3+2^4\right)=31+31.2^5+...+31.2^{95}=31\left(1+2^5+...+2^{95}\right)⋮31\)

a) \(\left(9999+101\right):2=5050\)

b) \(\left(9998+101\right):2=\dfrac{10099}{2}=5049,5\)

c) \(\left(98765+1234\right):2=\dfrac{99999}{2}=49999,5\)

d) \(\left(100+1000+10000\right):3=3700\)

\(\sqrt{17+12\sqrt{2}}=\sqrt{3+12\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3+2\sqrt{2}\right)^2}=3+2\sqrt{2}\)

C.30 gam

Gọi tuổi Hà, tuổi ông, tuổi bố ban đầu lần lượt là a,b,c(tuổi)

Theo đề bài ta được:

      \(a+1=\dfrac{1}{7}b=\dfrac{1}{4}c\Rightarrow a+1=\dfrac{b}{7}=\dfrac{c}{4}=\dfrac{b-c}{7-4}=\dfrac{27}{3}=9\)

\(\Rightarrow a+1=9\Rightarrow a=8\)

Vậy tuổi Hà ban đầu là 8 tuổi

\(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)

\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)-\left(\dfrac{x-9}{10}+1\right)-\left(\dfrac{x-10}{11}+1\right)-\left(\dfrac{x-11}{12}+1\right)=0\)

\(\Leftrightarrow\dfrac{x+1}{7}+\dfrac{x+1}{8}+\dfrac{x+1}{9}-\dfrac{x+1}{10}-\dfrac{x+1}{11}-\dfrac{x+1}{12}=0\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}-\dfrac{1}{12}\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)(do \(\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}-\dfrac{1}{12}>0\))