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\(S=\left(2.1\right)^2+\left(2.2\right)^2+\left(2.3\right)^2+....+\left(2.10\right)^2\)
\(\Rightarrow S=2^2.1^2+2^2.2^2+....+2^2.10^2\)
\(\Rightarrow S=2^2\left(1^2+2^3+3^2+.....+10^2\right)\)
Áp dụng giả thiết từ đề
\(\Rightarrow S=2^2.385\)
\(\Rightarrow S=4.384=1540\)
\(S=2^2+4^2+6^2+...+20^2\)
\(=1^2.4+2^2.4+3^2.4+...+10^2.4\)
\(=4.\left(1^2+2^2+3^2+...+10^2\right)\)
\(=4.385=1540\)
\(\left(12^2+14^2+16^2+18^2+20^2\right)-\left(1^2+3^2+5^2+7^2+9^2\right)\\\Rightarrow\left[\left(2^2.6^2\right)+\left(2^2.7^2\right)....+\left(2^2.10^2\right)\right]-\left(1^2+3^2+...+9^2\right)\\ \Rightarrow2^2.\left(6^2+7^2....+.10^2\right)-\left(1^2+3^2+5^2+7^2+9^2\right)\\ \Rightarrow4.330-165=1155\)
Ta có: S= 22+42+62+...+202
=>S=22(12+22+32+.......+102)
=>S=4.385=1540
a) A=\(\frac{2^{13}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^5-25^3.49^2}{\left(125.7\right)^3+5^9.14^3}\) =\(\frac{2^{13}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^5-\left(5^2\right)^3.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}\) =\(\frac{2^{13}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^5-5^6.7^4}{5^9.7^3+5^9.2^3.7^3}\) =\(\frac{2^{12}.3^4.\left(2.3-1\right)}{2^{12}.3^5.\left(6+1\right)}-\frac{5^6.7^4.\left(5^4.7-1\right)}{2^3.5^9.7^3\left(1+1.2^3\right)}\) =\(\frac{2^{12}.3^4.5}{2^{12}.3^5.7}-\frac{5^6.7^4.4374}{3^3.5^9.7^3.9}\) =\(\frac{5}{3.7}-\frac{7.4374}{3^3.5^3.3^2}\) =\(\frac{5}{21}-\frac{7.4374}{3^6.5^3}\) =\(\frac{5}{21}-\frac{7.4374}{729.125}\) =\(\frac{5}{21}-\frac{42}{125}\) =\(\frac{-257}{2625}\) b)S=\(2^2+4^2+....+20^2\) =\(\left(1.2\right)^2+\left(2.2\right)^2+....+\left(10.2\right)^2\) =\(2^2.\left(2^2+4^2+....+10^2\right)\) =\(2^2.385\) =4.385 =\(1540\)
Bài 2:
a: \(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮55\)
b: \(5A=5+5^2+...+5^{51}\)
\(\Leftrightarrow4A=5^{51}-1\)
hay \(A=\dfrac{5^{51}-1}{4}\)
Bài 3:
\(S=\left(1^2+2^3+3^3+...+10^2\right)\cdot2=385\cdot2=770\)
Bài 1
Nhân 2 vào biểu thức
Rút gọn và trừ đi 1 lần nó
còn lại \(\frac{1}{2}_{ }-\frac{1}{2^{10}}\)
\(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^{10}}\)
Ta có: \(S=\left(1.2\right)^2+\left(2.2\right)^2+\left(3.2\right)^2+...+\left(10.2\right)^2\)
\(\Rightarrow S=1.2^2+2^2.2^2+3^2.2^2+..+10^2.2^2\)
\(\Rightarrow S=2^2\left(1+2^2+3^2+..+10^2\right)\)
\(\Rightarrow S=4.385=1540\)
ta có : S=\(\left(2.1\right)^2+\left(2.2\right)^2+\left(2.3\right)^2+..+\left(2.10\right)^2\)
=\(2^2\left(1^2+2^2+3^2+...+10^2\right)\)
=4.385 =1540