\(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)

\(=1+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{7}\right)+...+\left(\dfrac{1}{512}+\dfrac{1}{513}+...+\dfrac{1}{1023}\right)+\dfrac{1}{1024}< 1+\dfrac{1}{2}.2+\dfrac{1}{2^2}.2^2+...+\dfrac{1}{2^9}.2^9+\dfrac{1}{1024}\)

\(=1+1+1+1+...+\dfrac{1}{1024}=10+\dfrac{1}{1024}< 11\left(đpcm\right)\)

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Lần đầu post, mình quên mất chưa nêu câu hỏi. Nhờ các bạn chứng minh dùm 3 câu trên với, cám ơn nhiều ah!

\(a.\)

\(1-\dfrac{1}{2}\left(\dfrac{3}{2}-2x\right)=4x-\dfrac{1}{4}\)

\(\Rightarrow1-\dfrac{3}{4}+x=4x-\dfrac{1}{4}\)

\(\Rightarrow1-\dfrac{3}{4}+\dfrac{1}{4}=4x-x\)

\(\Rightarrow3x=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{6}\)

\(b.\)

\(x^{10}=1024\)

\(\Rightarrow x^{10}=2^{10}\)

\(\Rightarrow x=2\)

\(c.\)

\(3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

a, \(\left(x+1\right)^2=169\)

\(\left(x+1\right)^2=13^2\)

\(x+1=13\)

\(x=13-1\)

\(x=12\)

1.

a) \(\left(x+1\right)^2=169\)

⇒ \(x+1=\pm13\)

⇒ \(\left[{}\begin{matrix}x+1=13\\x+1=-13\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=13-1\\x=\left(-13\right)-1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=12\\x=-14\end{matrix}\right.\)

Vậy \(x\in\left\{12;-14\right\}.\)

b) \(\left(x+3\right)^3=-\frac{1}{27}\)

⇒ \(\left(x+3\right)^3=\left(-\frac{1}{3}\right)^3\)

⇒ \(x+3=-\frac{1}{3}\)

⇒ \(x=\left(-\frac{1}{3}\right)-3\)

⇒ \(x=-\frac{10}{3}\)

Vậy \(x=-\frac{10}{3}.\)

c) \(\left(2x-4\right)^4=\frac{1}{625}\)

⇒ \(2x-4=\pm\frac{1}{5}\)

⇒ \(\left[{}\begin{matrix}2x-4=\frac{1}{5}\\2x-4=-\frac{1}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=\frac{1}{5}+4=\frac{21}{5}\\2x=\left(-\frac{1}{5}\right)+4=\frac{19}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{21}{5}:2\\x=\frac{19}{5}:2\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=\frac{21}{10}\\x=\frac{19}{10}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{21}{10};\frac{19}{10}\right\}.\)

Còn câu d) bạn làm tương tự như mấy câu trên.

Chúc bạn học tốt!

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