1+3+5+...+97+99
Những câu hỏi liên quan
E=(1+1/3+1/5+........+1/97+1/99)/(1/1*99+1/3*97+1/5*95+.....+1/97*3+1/99*1)
4+3/5+3/7+...+3/95+3/97+3/99
1/99+1/3*97+1/5*95+..+1/95*5+1/97*3+1/99*1
tính A= (1+1/3+1/5+...+1/95+1/97+1/99) /(1/1*99+1/3*97+1/5*95+...+1/95*5+1/97*3+1/99*1)
tinh tong gia tri bieu thuc :
a)A=1+1/3+1/5+...+1/97+1/99/1/1*99+1/3*97+1/5*95+...+1/97*3+1/99*1
b)B=1/2+1/3+1/4+...+1/100/99/1+98/2+97/3+...+1/99
1+1/3+1/5+.....+1/97+1/99
1/1*99+1/3*97+1/5*95+...+1/97*3+1/99*1
A) 1×5+5×9+9×13+...89×93+93×97
B) 1×3+3×5+5×7+....95×97+97×99
C) 1×3+5×7+9×11+..93×95+97×99
1)
-1 +3 -5 +7-.....+97-99
2) 1+2-3-4+...+97+98-99-100
Tính tổng nhaa
Sửa đề: \(-1+3-5+7-...-97+99\)
1) Ta có: \(-1+3-5+7-...-97+99\)
\(=\left(-1+3\right)+\left(-5+7\right)+...+\left(-97+99\right)\)
\(=2+2+...+2=2\cdot50=100\)
2) Ta có: \(1+2-3-4+...+97+98-99-100\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=\left(-4\right)\cdot25=-100\)
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tính
\(P=\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1}.99+\frac{1}{3}.97+\frac{1}{5}.95+...+\frac{1}{97}.3+\frac{1}{99}.1}\)
Lời giải:
** Sửa đề: Chỗ $\frac{1}{1}$ ở mẫu chuyển thành $\frac{1}{2}$
$\frac{1}{1}.99+\frac{1}{3}.97+\frac{1}{5}.95+....+\frac{1}{97}.3+\frac{1}{99}.1$
$=50+(\frac{97}{3}+1)+(\frac{95}{5}+1)+....+(\frac{3}{97}+1)+(\frac{1}{99}+1)$
$=50+\frac{100}{3}+\frac{100}{5}+...+\frac{100}{97}+\frac{100}{99}$
$=100(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})$
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{100(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})}=\frac{1}{100}\)
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Tính B=1*3+5*7+9*11+...+97*101
C=1*3*5-3*5*7+5*7*9-....-97*99*101
D=1*99+3*97+5*95+...+49*51
E=1*3^3+3*5^3+5*7^3+...+49*51^3
F=1*99^2+2*98^2+3*97^2+...+49*51^2
cái này bạn mở sách bồi dưỡng toán ra trang gần cuối là thấy ngay ấy mà
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Giải pt : ( x-1/99 +x-99 )+( x-3/97 + x-97/3 )+ (x-5/95 + x-95/5 )=6
\(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-99}{1}-1\right)+\left(\dfrac{x-3}{97}-1\right)+\left(\dfrac{x-97}{3}-1\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-95}{5}-1\right)=0\)
=>x-100=0
=>x=100
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