Tính nhanh:
248-242+240-234+232-226+...+216-210
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\(\left(2x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)
\(\left(2x-\dfrac{1}{3}\right)^2=\dfrac{1}{2}^2\)
\(\Rightarrow2x-\dfrac{1}{3}=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}+\dfrac{1}{3}\)
\(2x=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}:2=\dfrac{5}{6}.\dfrac{1}{2}\)
\(x=\dfrac{5}{12}\)
\(\left(2x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=-\dfrac{1}{2}\\2x-\dfrac{1}{3}=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{6}\\2x=\dfrac{5}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{12}\\x=\dfrac{5}{12}\end{matrix}\right.\)
\(\dfrac{\left(\dfrac{2}{7}\right)^7.7^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7.5^2+512}\)
\(=\dfrac{\dfrac{2^7}{7^7}.7^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3}{2^7.5^2+2.256}\)
\(=\dfrac{2^7+\left(\dfrac{9}{4}.\dfrac{16}{3}\right)^3}{2^7.5^2+2.2^8}=\dfrac{2^7+\left(12\right)^3}{2^7.5^2+2.2^8}\)
\(=\dfrac{2^7+\left(2^2.3\right)^3}{2^7.5^2+2^9}=\dfrac{2^7+2^6.3^3}{2^7.\left(5^2+2^2\right)}\)
\(=\dfrac{2^6\left(2+27\right)}{2^7.\left(25+4\right)}=\dfrac{29}{2.29}=\dfrac{1}{2}\)
\(\dfrac{3x-1}{3x^2+5x+2}=\dfrac{1}{x+2}\left(x\ne-2;x\ne\dfrac{1}{3}\right)\)
\(\Rightarrow\left(3x-1\right)\left(x+2\right)=3x^2+5x+2\)
\(\Rightarrow3x^2+6x-x-2=3x^2+5x+2\)
\(\Rightarrow3x^2+5x-2=3x^2+5x+2\)
\(\Rightarrow-2=2\) (vô lý)
Bạn xem lại đề bài
\(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}< 125^{12}\)
\(\Rightarrow5^{36}>11^{24}\)
\(A=248-242+240-234+232-226+...+216-210\)
\(=4+4+4+...4\)
Số số 4 là \(\left(248-216\right):8+1=5\)
\(\Rightarrow A=5.4=20\)
Bằng 20 bạn nhé