K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

18 tháng 8 2023

D=1/1.5+1/5.9+...+1/41.45

4D=4/1.5+4/5.9+...+4/41.45

4D=1-1/5+1/5-1/9+...+1/41-1/45

4D=1-1/45

D=44/45:4=11/45

15 tháng 8 2018

\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)

\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\frac{44}{45}\)

\(=\frac{11}{45}\)

15 tháng 8 2018

Đặt \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)  là A.

Ta có:

\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)

\(4A=4\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\right)\)

\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\)

\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\)

\(4A=1-\frac{1}{45}\)

\(4A=\frac{44}{45}\)

\(A=\frac{44}{45}:4\)

\(A=\frac{11}{45}\)

Vậy \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}=\frac{11}{45}\)

2 tháng 8 2023

\(\dfrac{1}{4}+\dfrac{1}{3}\times X=\dfrac{1}{2}\\ \dfrac{1}{3}\times X=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}\\ X=\dfrac{1}{4}:\dfrac{1}{3}=\dfrac{1}{4}\times3=\dfrac{3}{4}\)

2 tháng 8 2023
   

14+13×�=1213×�=12−14=14�=14:13=14×3=34

4 tháng 8 2023

\(\dfrac{3}{4}\times X-1=\dfrac{1}{2}\times X\\ \dfrac{3}{4}\times X-\dfrac{1}{2}\times X=1\\ \left(\dfrac{3}{4}-\dfrac{1}{2}\right)\times X=1\\ \dfrac{1}{4}\times X=1\\ X=1:\dfrac{1}{4}\\ X=4\)

\(\dfrac{3}{4}\times x-1=\dfrac{1}{2}\times x\\ \dfrac{3}{4}\times x-\dfrac{1}{2}\times x=1\\ x\times\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=1\\ x\times\dfrac{1}{4}=1\\ x=1:\dfrac{1}{4}\\x=4\)

20 tháng 5 2022

$#trúc$

`4/(1.5) + 4/(5.9) + 4/(9.13) + .... + 4/(2013 . 2017)`

=`1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/2013 - 1/2017`

= `1 - 1/2017`

= `2017/2017 - 1/2017`

= `2016/2017`

HQ
Hà Quang Minh
Giáo viên
4 tháng 8 2023

\(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{420}\\ \Rightarrow A=\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{20\times21}\\ \Rightarrow A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\\\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)

4 tháng 8 2023

Sau khi rút gọn phải còn:

\(A=\dfrac{1}{2}-\dfrac{1}{21}\) (chứ anh)

\(A=\dfrac{19}{42}\)

15 tháng 8 2019

\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)

\(4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\)

\(4B=\frac{44}{45}\)

\(B=\frac{11}{45}\)

15 tháng 8 2019

\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)

\(=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{41.45}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\frac{44}{45}\)

\(=\frac{11}{45}\)

3 tháng 7 2021

\(M=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\left|2-\sqrt{5}\right|}-\dfrac{2}{\left|2+\sqrt{5}\right|}\)

\(=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

\(=\dfrac{8}{1}=8\)

 

3 tháng 7 2021

Lm ơn giúp mik đii mà mik bt ơn bn đó nhiều lắm . Mik đang rất cần

28 tháng 8 2021

1) \(\left(2x+3\right)^2=4x^2+12x+9\)

\(\left(3x+2\right)^2=9x^2+12x+4\)

\(\left(2x+5\right)^2=4x^2+20x+25\)

\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)

\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)

2)  \(\left(2x-3\right)^2=4x^2-12x+9\)

\(\left(3x-2\right)^2=9x^2-12x+4\)

\(\left(2x-5\right)^2=4x^2-20x+25\)

\(\left(2x-\dfrac{1}{3}\right)^2=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)

\(\left(3x-\dfrac{1}{3}\right)^2=9x^2-2x+\dfrac{1}{9}\)

3) \(\left(2x-3\right)\left(2x+3\right)=4x^2-9\)

\(\left(3x-4\right)\left(3x+4\right)=9x^2-16\)

\(\left(2x-5\right)\left(2x+5\right)=4x^2-25\)

\(\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=x^2-\dfrac{1}{4}\)

\(\left(2x-\dfrac{1}{3}\right)\left(2x+\dfrac{1}{3}\right)=4x^2-\dfrac{1}{9}\)

1: \(\left(2x+3\right)^2=4x^2+12x+9\)

\(\left(3x+2\right)^2=9x^2+12x+4\)

\(\left(2x+5\right)^2=4x^2+20x+25\)

\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)

\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)

1 tháng 5 2022

sao khó dữ vậy chị

 

1 tháng 5 2022

em mới có lớp 1 thôi