4/1x3+4/3x5+..............+9/99x101
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\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+...+\frac{4}{99\cdot101}-x-\frac{200}{101}=1\)
\(\frac{4}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)-x=1+\frac{200}{101}\)
\(\frac{4}{2}\cdot\left(1-\frac{1}{101}\right)-x=\frac{301}{101}\)
\(\frac{4}{2}\cdot\frac{100}{101}-x=\frac{301}{101}\)
\(\frac{200}{101}-x=\frac{301}{101}\)
\(\Rightarrow x=\frac{301}{101}-\frac{200}{101}=1\)
Ta có : \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}-x-\frac{200}{101}=1\)
\(\Rightarrow\)\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}=1+\frac{200}{101}+x\)
=> \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}=\frac{301}{101}+x\)
=> \(2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\right)=\frac{301}{101}+x\)
=> \(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)=\frac{301}{101}+x\)
=> \(2\left(1-\frac{1}{101}\right)=\frac{301}{101}+x\)
=> \(2.\frac{100}{101}=\frac{301}{101}+x\)
=> \(\frac{200}{101}=\frac{301}{101}+x\)
\(\Rightarrow x=\frac{301}{101}-\frac{200}{101}=1\)
A=1x3x(5+1) + 3x5x(7-1) +5x7x(9-3) +...+ 99x101x(103-97)
6A=3+ 1x3x5 +3x5x7-1x3x5 + 5x7x9 -3x5x7 +....+99x101x103 - 97x99x101
6A=3+99x101x103=1019703
vậy = 1019703
nếu sai chỗ nào thì sửa hộ mk vs
=1(2+1)+2(3+1)+3(4+1)+...+100(101+1)
=1.2+1+2.3+2+3.4+3+...+100.101+100
=(1.2+2.3+3.4+..+100.101)+(1+2+3+...+100)
=333300+5000
=338300
A=1x3 +3x5 +5x7 +....+99x101
6A=1x3x(5+1) + 3x5x(7-1) +5x7x(9-3) +...+ 99x101x(103-97)
6A=3+ 1x3x5 +3x5x7-1x3x5 + 5x7x9 -3x5x7 +....+99x101x103 - 97x99x101
6A=3+99x101x103=1019703
\(S=1.3+2.4+3.5+...+99.101\)
\(\Rightarrow S=1\left(2+1\right)+2\left(3+1\right)+...+99\left(100+1\right)\)
\(\Rightarrow S=\left(1.2+2.3+...+99.100\right)+\left(1+2+3+...+99\right)\)
Đặt \(A=1.2+2.3+...+99.100\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)
\(\Rightarrow3S=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)
\(\Rightarrow S=\frac{99.100.101}{3}\)
Đặt \(B=1+2+3+...+99\)
\(\Rightarrow B=\frac{\left(99+1\right)\left[\left(99-1\right):2+1\right]}{2}\)
\(\Rightarrow B=\frac{100.50}{2}=2500\)
\(\Rightarrow S=A+B=\frac{99.100.101}{3}+2500\)
S = 1 x 3 + 2 x 4 + 3 x 5 + ... + 99 x 101
S = ( 1 x 3 + 3 x 5 + ...+ 99 x 101) + ( 2 x 4 + ...+ 98 x 100)
Đặt A = 1 x 3 + 3 x 5 + ...+ 99 x 101
=> 6 A = 1 x 3 x 6 + 3 x 5 x 6 + ...+ 99 x 101 x 6
6 A = 1 x 3 x ( 5+1) + 3 x 5 x ( 7-1) + ...+ 99 x 101 x ( 103 - 97)
6A = 1 x 3 x 5 + 1 x 3 + 3 x 5 x 7 - 1 x 3 x 5 + ...+ 99 x 101 x 103 - 97 x 99 x 101
6A = ( 1 x 3 + 1 x 3 x 5 + 3 x 5 x 7 +...+ 99 x 101 x 103) - ( 1 x 3 x 5 + ...+ 97 x 99 x 101)
6A = 1 x 3 + 99 x 101 x 103
\(\Rightarrow A=\frac{1.3+99.101.103}{6}=171650\)
Đặt B = 2 x 4 + ...+ 98 x 100
=> 6B = 2 x 4 x 6 + 4 x 6 x 6 + ...+ 98 x 100 x 6
6B = 2 x 4 x 6 + 4 x 6 x ( 8-2) + ...+ 98 x 100 x ( 102 - 96)
6B = 2 x 4 x 6 + 4 x6 x8 - 2x4x6 + ...+ 98x100x102 - 96x98x100
6B = ( 2 x 4 x 6 + 4 x 6 x 8 +...+98x100x102) - ( 2x4x6+...+96x98x100)
6B = 98 x 100 x 102
\(\Rightarrow B=\frac{98.100.102}{6}=166600\)
Thay A;B vào S, có
S = 171 650 + 166 600
S = 338 250
Sửa đề: \(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{99.101}\)
Đặt: \(A=\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{99.101}\)
\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\left(1-\dfrac{1}{101}\right)=\dfrac{200}{101}\)
*Lưu ý: Dấu ".'' trong bài là dấu nhân nhé, lên lớp 6 bạn sẽ được học
\(\dfrac{200}{101}\)