\(148+144+140+...+56+52\)

\(=\left[\left(148-52\right):4+1\right]\left(148+52\right):2\)

\(=25.200:2=25.100=2500\)

$148+144+140+...+56+52$

$=\left[\left(148-52\right):4+1\right]\left(148+52\right):2$

$=25.200:2=25.100=2500$

Đề bài bạn yêu cầu gì?

\(\left(x-257\right)-921=2^5\)

\(x-257=32+921\)

\(\)\(x-257=953\)

\(x=953+257\)

\(x=1210\)

\(\left(x-257\right)-921=2^5\)

\(\Rightarrow x-1178=32\Rightarrow x=32+1178=1210\)

a) \(...=-246+301=55\)

b) \(...=-\dfrac{10}{9}+\dfrac{5}{12}=-\dfrac{40}{36}+\dfrac{15}{36}=-\dfrac{25}{36}\)

c) \(...=-\dfrac{3}{26}+\dfrac{24}{69}=-\dfrac{207}{1794}+\dfrac{624}{1794}=\dfrac{417}{1794}\)

\(A=248-242+240-234+232-226+...+216-210\)

\(=4+4+4+...4\)

Số số 4 là \(\left(248-216\right):8+1=5\)

\(\Rightarrow A=5.4=20\)

Bằng 20 bạn nhé

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

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

\(\Rightarrow2x-\dfrac{1}{3}=\dfrac{1}{2}\)

\(2x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(2x=\dfrac{5}{6}\)

\(x=\dfrac{5}{6}:2=\dfrac{5}{6}.\dfrac{1}{2}\)

\(x=\dfrac{5}{12}\)

\(\left(2x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=-\dfrac{1}{2}\\2x-\dfrac{1}{3}=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{6}\\2x=\dfrac{5}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{12}\\x=\dfrac{5}{12}\end{matrix}\right.\)

A

\(2400:500=4\) dư \(2400-500.4=400\Rightarrow C\)

\(\dfrac{\left(\dfrac{2}{7}\right)^7.7^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7.5^2+512}\)

\(=\dfrac{\dfrac{2^7}{7^7}.7^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3}{2^7.5^2+2.256}\)

\(=\dfrac{2^7+\left(\dfrac{9}{4}.\dfrac{16}{3}\right)^3}{2^7.5^2+2.2^8}=\dfrac{2^7+\left(12\right)^3}{2^7.5^2+2.2^8}\)

\(=\dfrac{2^7+\left(2^2.3\right)^3}{2^7.5^2+2^9}=\dfrac{2^7+2^6.3^3}{2^7.\left(5^2+2^2\right)}\)

\(=\dfrac{2^6\left(2+27\right)}{2^7.\left(25+4\right)}=\dfrac{29}{2.29}=\dfrac{1}{2}\)