Bài 1:

a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)

= \(\dfrac{29}{10}\)

b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)

= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)

= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)

= \(\dfrac{19}{20}\)

c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)

= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)

= \(\dfrac{29}{6}\)

Bài 2:

   3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\) 

= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)

= \(\dfrac{28}{5}\)

b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)

= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)

= \(\dfrac{43}{34}\)

Bài 3:

a, 800 kg 5g > 8,005 kg

b, 9 ha 4 dam² < 9\(\dfrac{5}{100}\) ha

c, 5m 5mm > 5,0005 m

d, 250 kg > \(\dfrac{1}{5}\) tấn

Lời giải chi tiết:

5 - 1 - 1 = 3     4 - 1 - 1 = 2     3 - 1 - 1 = 1

5 - 1 - 2 = 2     5 - 2 - 1 = 2     5 - 2 - 2 = 1

1/

\(N=1.\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)=\)

\(=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)=\)

Đặt 

\(A=1.2+2.3+3.4+...+99.100\)

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-98.99.100+99.100.101=\)

\(=99.100.101\Rightarrow A=\dfrac{99.100.101}{3}=33.100.101\)

Đặt

\(B=1+2+3+...+99=\dfrac{99.\left(1+99\right)}{2}=4950\)

\(\Rightarrow N=A-B\)

2/

Số hạng cuối cùng là 10000 hoặc 1000000 mới làm được

\(A=1^2+2^2+3^2+...+100^2\) 

Tính như câu 1

3/ Làm như bài 4

4/

\(S=1^2+3^2+5^2+...+99^2=\)

\(=1.\left(3-2\right)+3\left(5-2\right)+5\left(7-2\right)+...+99\left(101-2\right)=\)

\(=\left(1.3+3.5+5.7+...+99.101\right)-2\left(1+3+5+...+99\right)\)

Đặt

\(B=1+3+5+...+99=\dfrac{50.\left(1+99\right)}{2}=2500\) 

Đặt

\(A=1.3+3.5+5.7+...+99.101\)

\(6A=1.3.6+3.5.6+3.7.6+...+99.101.6=\)

\(=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+99.101.\left(103-97\right)=\)

\(=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-97.99.101+99.101.103=\)

\(=3+99.101.103\Rightarrow A=\dfrac{3+99.101.103}{6}\)

\(\Rightarrow S=A-2B\)

Bài 1:

\(N=1^2+2^2+3^3+...+99^2\)

\(N=1.1+2.2+3.3+...+99.99\)

\(N=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+99.\left(100-1\right)\)

\(N=1.2-1+2.3-2+3.4-3+...+99.100-99\)

\(N=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)\)

Đặt \(\left\{{}\begin{matrix}A=1.2+2.3+3.4+...+99.100\\B=1+2+3+...+99\end{matrix}\right.\)

+) Tính \(A=1.2+2.3+3.4+...+99.100\)

Ta có:

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(3A=99.100.101\)

\(\Rightarrow A=\dfrac{99.100.101}{3}=333300\)

+) Tính \(B=1+2+3+...+99\)

\(B\) có số số hạng là: \(\dfrac{99-1}{1}\) + 1 = 99 (số hạng)

\(\Rightarrow B=\dfrac{\left(99+1\right).99}{2}=4950\)

\(\Rightarrow N=A-B=333300-4950=328350\)

\(\Rightarrow N=328350\)

xin loi mik danh nham nhe bai do la 10000 nhe

5 - 1 = 4     1 + 4 = 5     2 + 3 = 5

5 - 2 = 3     4 + 1 = 5     3 + 2 = 5

5 - 3 = 2     5 - 1 = 4     5 - 2 = 3

5 - 4 = 1     5 - 4 = 1     5 - 3 = 2

5-1=4     1+4=5     2+3=5

5-2=3     4+1=4     3+2=5

5-3=2     5-1=4      5-2=3

5-4=1     5-4=1      5-3=2

Lời giải chi tiết: 

5-1=4     4-1=3    3-1=2   2+3=5     

5-2=3     4-2=2    3-2=1   3+2=5

5-3=2     4-3=1    2-1=1   5-2=3

5-4=1                               5-3=2

Tính: