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\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
100-3(x-1)2=52
3(x-1)2=100-52
3(x-1)2=48
(x-1)2=48:3
(x-1)2=16
(x-1)2=42=(-4)2
=> x-1=4 hoặc x-1=-4
TH1:
x-1=4
x=4+1
x=5
TH2:
x-1=-4
x=-4+1
x=-3
Vậy x=5 hoặc x=-3
100 - 3(x - 1)2 = 52
<=> 3(x - 1)2 = 48
<=> (x - 1)2 = 16
<=> (x - 1)2 = 42 = (-4)2
<=> \(\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}}\)
<=> \(\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)
\(2A=1-\frac{1}{3^8}\)
\(A=\frac{1}{2}-\frac{1}{2.3^8}\)
Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
Tính nhanh:
\(A=\frac{2}{1+2}+2+\frac{3}{12+3}+...+2+3+\frac{20}{1+2+3+...+20}\)
Đặt \(A=\frac{2}{1+2}+2+\frac{3}{12+3}+...+2+3+\frac{20}{1+2+3+...+20}\)
\(=2-1+2+\frac{3}{12+3}+...+2+3+\frac{20}{1+2+3+...+20}\)
\(=\) Không biết! Nhờ Doraeiga với At the speed of light - Trang của At the speed of light - Học toán với OnlineMath giải nhé! Tui mới lớp 6 thôi! Chưa học tới bài này
\(A=\frac{2}{1+2}+\frac{2+3}{1+2+3}+....+\frac{2+3+...+20}{1+2+3+...+20}\)
\(A=\frac{2}{3}+\frac{5}{6}+...+\frac{209}{210}\)
\(A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{210}\right)\)
\(A=\left(1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{6}+....+\frac{1}{210}\right)\)
\(A=19-\left(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{420}\right)\)
\(A=19-\left(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\right)\)
\(A=19-\left[2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\right]\)
\(A=19-\left[2\cdot\left(\frac{1}{2}-\frac{1}{21}\right)\right]\)
\(A=19-\left[2\cdot\frac{19}{42}\right]=19-\frac{19}{21}=\frac{380}{21}\)
Vậy A = .....