a viết lộn hả bạn

b) 2019.(2020+1)-2019.2020

=2019.(2020+1-2020)\

=2019.1

=2019\

c)436+324.(234-1)-234.324+324

=436+324.(234-1-234+1)

=436+324.1

=436+324

=760

d)472.(36-1)+35.8

=472.35+35.8

=35.(472+8)

=35.480

=16800

Có tên giống mk quá!

1 Tính tổng

s = 9 + 99 + 999...+.....+9( 100 số 9)

s = 9 + 9 x 11+ 9 x 111 + 9 x 111 x11(100 chữ số 1)

s = 9 x ( 1+ 11 +111+1111+......+1111..11)(100 chữ số 1)

s = 9 x 111.....1111

s = 999...99

Bài 2 mk k bít làm thông cảm nha.Nhưng mk gửi lời mời kb.

1,

S=(10-1)+(10^2-1)+(10^3-1)+....+(10^100-1).

=(10+10^2+10^3+...+10^100)-10^2(chữ số 1)

10*S=10^2+10^3+10^4+...+10^101-10^3.

=>10*S-S=(10^2+10^3+...+10^101-10^3)-(10+10^2+10^3+...+10^100-10^2).

=>9S=10^101-10^3-10+10^2

9S=10^101-910

S=(10^101-910)/9.

2,

Tử:Mẫu=(2839+2162-1)*468/468=5000 (1)

Và:

1-1/2=1/2

1-1/3=2/3

1-1/4=3/4

1-1/5=4/5

.........

1-1/100=99/100 

Nhân lại với nhau,rút gọn ta được: 1/100 (2)

Từ (1),(2)=> a=5000*1/100=50.

Sửa lại cho mình số 2839 thành 2838 nha (đánh nhầm số 8 thành số 9).

Xin cảm ơn!

Bài này giống câu hỏi mình đưa ra quá.

Tích nha 100/100 đấy.

tu bam may tinh ra ik

Dễ mà bạn

bn sai đề bài ùi

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)

\(=>x+1=0\)

\(=>x=-1\)

b,

\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)

\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)

\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)

\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)

\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)

Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)

\(=>x+2021=0\)

\(=>x=-2021\)

c,

\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)

\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)

\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)

\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)

Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)

\(=>x+329=0\)

\(=>x=-329\)

\(A=1-3+5-7+......-2019+2021-2023\)

\(A=\left(1-3\right)+\left(5-7\right)+....+\left(2021-2023\right)\)

\(A=-2+\left(-2\right)+....+\left(-2\right)\left(506 cặp\right)\)

\(A=-2.506\)

\(A=-1012\)

*) A=(1-3)+(5-7)+....+(2021-2023)

<=> A=-2+(-2)+...+(-2)

Dãy A có (2023-1):2+1=1012 số số hạng 

=> Có 506 số (-2)

=> A=(-2).506=-1012

\(B=1+2-3-4+5+6-7-8+......+2017+2018-2019-2020\)

\(B=\left(1+2-3-4\right)+\left(5+6-7-8\right)+.....+\left(2017+2018-2019-2020\right)\)

\(B=-4+\left(-4\right)+.....+\left(-4\right)\left(505 cặp\right)\)

\(B=-4.505\)

\(B=-2020\)

(1+3+5+7+...+2019+2021)

A=1−3+5−7+......−2019+2021−2023

A=(1−3)+(5−7)+....+(2021−2023)A=(1−3)+(5−7)+....+(2021−2023)

A=−2+(−2)+....+(−2)(506)A=−2+(−2)+....+(−2)(506cặp)

a=−2.506A=−2.506

A=−1012A=−1012

(2+4+6+8+...+2020)

B=2+4+6+8+...+2018+2020
B = 2(1 + 2 + 3 + 4 + ... + 1009 + 1010)
B = 2 . (1011 . 1010 : 2)
B = 2 . 510555
B = 1 021 110

(1+3+5+7+......+2019+2021)-(2+4+6+8+.....+2020)

\(=\dfrac{\left(1+2021\right).\left[\left(2021-1\right):2+1\right]}{2}-\dfrac{\left(2+2020\right).\left[\left(2020-2\right):2+1\right]}{2}\)

\(=1011\)