\(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

a, \(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
b, \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
c, \(x^2-10x+25=x^2-2.5x+5^2=\left(x-5\right)^2\)

1. \(25+10x+x^2\\ \Leftrightarrow5^2+2\cdot5\cdot x+x^2\\ \Leftrightarrow\left(5+x\right)^2\\ \Leftrightarrow\left(5+x\right)\left(5+x\right)\)

2. \(8x^3-\dfrac{1}{8}\\ \Leftrightarrow\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[4x^2+x+\dfrac{1}{4}\right]\)

3. \(x^2-10x+25\\ \Leftrightarrow x^2-2\cdot5\cdot x+5^2\\ \Leftrightarrow\left(x-5\right)^2\\ \Leftrightarrow\left(x-5\right)\left(x-5\right)\)

Số các số hạng (số ngoặc đơn) là

\(\dfrac{79-1}{2}+1=40\) số hạng

Biểu thức viết thành

\(40x+\left(1+3+5+...+79\right)=41x\)

Tổng của 1+3+5+...+79 là

\(\dfrac{40x\left(1+79\right)}{2}=1600\)

\(\Rightarrow40x+1600=41x\Rightarrow x=1600\)

25

23

25

\(a,\Rightarrow\dfrac{\left(-3\right)^x}{\left(-3\right)^4}=\left(-3\right)^3\\ \Rightarrow\left(-3\right)^{x-4}=\left(-3\right)^3\\ \Rightarrow x-4=3\Rightarrow x=7\\ b,Sửa:\left(x-\dfrac{1}{2}\right)^2=25\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=5\\x-\dfrac{1}{2}=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{5}\\x=-\dfrac{9}{5}\end{matrix}\right.\)

\(\dfrac{2}{5}:\dfrac{6}{25}\\ =\dfrac{2}{5}\times\dfrac{25}{6}\\ =\dfrac{2}{5}\times\dfrac{5\times5}{2\times3}=\dfrac{5}{3}\)

\(\dfrac{2}{5}:\dfrac{6}{25}\)

\(=\dfrac{2}{5}\times\dfrac{25}{6}\)

\(=\dfrac{2}{5}\times\dfrac{5\times5}{2\times3}\)

\(=\dfrac{5}{3}\)

kết quả:

27 x 9 + 1 = 243 + 1 = 244 

244 nha

học tốt

\(\left(x+\dfrac{1}{3}\right)\times\dfrac{9}{14}\times\dfrac{7}{3}-\dfrac{1}{3}=1:\dfrac{9}{5}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{5}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{5}{9}+\dfrac{1}{3}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{8}{9}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{8}{9}:\dfrac{3}{2}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{16}{27}\\ \Rightarrow x=\dfrac{16}{27}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{7}{27}\)