a,35x34+35x86+65x75+65x45

=35x(34+86)+65x(75+45)

=35x120+65x120

=120x(35+65)

=120x100

=12000

b,3x25x8+4x37x6+2x38x12

=(3x8)x25+(4x6)x37+(2x12)x38

=24x25+24x37+24x38

=24x(25+37+38)

=24x(62+38)

=24x100

=2400

c,12x53+53x172-53x84

=53x(12+172-84)

=53x(184-84)

=53x100

=5300

a.35x(34+86)+65x(75+45)=35x120+65x120=120x(35+65)=120x100=12000

b.3x25x8+4x37x6x2x38x12=24x25+24x37+24x38=24x(25+37+38)=24x100=2400

c.53x(12+172-84)=53x100=5300

= 1512

chúc bạn học tốt

c) 3 x 25 x 8 + 4 x 37 x 6+ 2 x 12

= 3 x 8 x 25 + 4 x 6 x 37 + 2 x 12

= 24 x 25 + 24 x 37 + 24

= 24 x 25 + 24 x 37 + 24 x 1

= 24 x ( 25 + 37 + 1 )

= 24 x 63

= 1 512

~~ Hok T ~~

a) = 54

b) = ( - 315 )

c) = 132

d) = 2400

Bạn có thể giải cách làm đc k ?

136 × 48 + 16 × 272 + 68 × 20 × 2

= 136 × 48 + 32 × 136 + 136 × 20

= 136 × ( 48 + 32 + 20 )

= 136 × 100

= 13600

3 × 25 × 8 + 4 × 37 × 6 + 2 × 38 × 12

= ( 3 × 8 ) × 25 + ( 4 × 6 ) × 37 + ( 2 × 12 ) × 38

= 24 × 25 + 24 × 37 + 24 × 38

= 24 × ( 25 + 37 + 38 )

= 24 × 100 = 2400

1 + 6 + 11 + 16 + ... + 46 + 51

Số số hạng của dãy số trên là :

  ( 51 - 1 ) ÷ 5 + 1 = 11 ( số )

Tổng dãy số đó là :

   ( 1 + 51 ) × 11 ÷ 2 = 286

\(=2^{2020}\left(1+2+2^2\right)=7\cdot2^{2020}\)

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)

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

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

\(\Rightarrow N=328350\)

=(2^78+79+80):(2^77+76+75)

=2²³⁷:2²²⁸

=2^237-228

=2⁹

=512