M = 2² - ( 3 + 4 - 7 + 2020 )

= 4 - ( 7 - 7 + 2020 )

= 4 - 2020

= -2016

=> M = -2016

m=2²-(3+4-7+2020)

=4-2020

=-2016

1. \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)

\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)

\(=0+\dfrac{2020}{2021}=\dfrac{2020}{2021}\)

Giải:

1) \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)  

\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\) 

\(=\left(\dfrac{2019}{2020}-\dfrac{2019}{2020}\right)+\dfrac{2020}{2021}\) 

\(=0+\dfrac{2020}{2021}\) 

\(=\dfrac{2020}{2021}\) 

2) \(\dfrac{2}{9}+\dfrac{7}{9}:\left(\dfrac{42}{5}-\dfrac{7}{5}\right)\) 

\(=\dfrac{2}{9}+\dfrac{7}{9}:7\) 

\(=\dfrac{2}{9}+\dfrac{1}{9}\) 

\(=\dfrac{1}{3}\) 

3) \(\dfrac{3}{4}+\dfrac{x}{4}=\dfrac{5}{8}\) 

            \(\dfrac{x}{4}=\dfrac{5}{8}-\dfrac{3}{4}\) 

            \(\dfrac{x}{4}=\dfrac{-1}{8}\)  

\(\Rightarrow x=\dfrac{4.-1}{8}=\dfrac{-1}{2}\) 

4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\) 

            \(\left|3x-1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\) 

            \(\left|3x-1\right|=0\) 

             \(3x-1=0\) 

                    \(3x=0+1\) 

                    \(3x=1\) 

                      \(x=1:3\) 

                      \(x=\dfrac{1}{3}\) 

Chúc bạn học tốt!

4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\) 

            \(\left|3x+1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\) 

            \(\left|3x+1\right|=0\) 

              \(3x+1=0\) 

                    \(3x=0-1\) 

                    \(3x=-1\) 

                      \(x=-1:3\) 

                      \(x=\dfrac{-1}{3}\)

a) \(S=1+2+3+...+2021\)

\(=\left(2021+1\right).2021:2\)

\(=2043231\)

b) \(P=1+3+5+...+2021\)

\(=\left(2021+1\right).[\left(2021-1\right):2+1]:2\)

\(=2022.1011:2\)

\(=1022121\)

a) Số số hạng=(2020-5):5+1=404

=> A=(2020+5)×404 : 2= 409050

b) Từ 1 đến 2020 có 2020 số hạng=> có 1010 cặp,mỗi cặp = -1

=>B= 1010×(-1) +2021=1011

a) =[(2020-5):5+1].[(2020+5):5]

=403.405

=163215

SSHẠNG :(2020-5) : 5+1=404 ( 2021-1)+1=2021

= (2020+5). 404:2 (2021+1) :2.2021

A= 409050 B= 2043231

1 3/7-4/5=10/7-4/5=50/35-28/35=22/35

6/13x-3/10+2/5x4/13

=-9/65+8/65

=-1/65

câu đầu mik tính ra số to mà cx ko chắc là đúng nên mik ko viết

*chúc bn học tốt đạt nhiều điểm cao*

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019

\(\left(3x-\frac{1}{5}\right)^{2020}+\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}=0\)

Ta có: \(\left(3x-\frac{1}{5}\right)^{2020}\ge0\forall x\)

\(\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}\ge0\forall y\)

Mà \(\left(3x-\frac{1}{5}\right)^{2020}+\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(3x-\frac{1}{5}\right)^{2020}=0\\\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-\frac{1}{5}=0\\\frac{2}{5}\cdot y+\frac{4}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=\frac{1}{5}\\\frac{2}{5}\cdot y=\frac{-4}{7}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{15}\\y=\frac{-10}{7}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(\frac{1}{15};\frac{-10}{7}\right)\)

các cậu giúp mình với mai đi học rồi

\(3x\left(x-2020\right)-x+2020=0\)

\(3x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\left(3x-1\right)\left(x-2020\right)=0\)

\(\orbr{\begin{cases}x=\frac{1}{3}\left(TM\right)\\x=2020\left(TM\right)\end{cases}}\)

\(b,4-9x^2=0\)

\(2^2-\left(3x\right)^2=0\)

\(\left(2-3x\right)\left(2+3x\right)=0\)

\(\orbr{\begin{cases}2-3x=0\\2+3x=0\end{cases}\orbr{\begin{cases}x=\frac{2}{3}\left(TM\right)\\x=-\frac{2}{3}\left(TM\right)\end{cases}}}\)

\(c,x^2-x+\frac{1}{4}=0\)

\(x^2-x+\left(\frac{1}{2}\right)^2=0\)

\(\left(x-\frac{1}{2}\right)^2=0\)

\(x-\frac{1}{2}=0\)

\(x=\frac{1}{2}\)

\(d,x\left(x-3\right)+\left(x-3\right)=0\)

\(\left(x-3\right)\left(x+1\right)=0\)

\(\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\orbr{\begin{cases}x=3\left(TM\right)\\x=-1\left(TM\right)\end{cases}}}\)

\(e,9x\left(x-7\right)-x+7=0\)

\(9x\left(x-7\right)-\left(x-7\right)=0\)

\(\left(9x-1\right)\left(x-7\right)=0\)

\(\orbr{\begin{cases}9x-1=0\\x-7=0\end{cases}\orbr{\begin{cases}x=\frac{1}{9}\left(TM\right)\\x=7\left(TM\right)\end{cases}}}\)

a) 3x(x - 2020) - x + 2020 = 0 

<=> 3x(x - 2020) - (x - 2020) = 0

<=> (3x - 1)(x - 2020) = 0

<=> \(\orbr{\begin{cases}3x-1=0\\x-2020=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2020\end{cases}}\)

Vậy tập nghiệm phương trình là \(S=\left\{\frac{1}{3};2020\right\}\)

b) \(4-9x^2=0\)

<=> \(\left(2-3x\right)\left(2+3x\right)=0\)

<=> \(\orbr{\begin{cases}2-3x=0\\2+3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{2}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{2}{3};-\frac{2}{3}\right\}\)là nghiệm phương trình 

c) \(x^2-x+\frac{1}{4}=0\)

<=> \(\left(x-\frac{1}{2}\right)^2=0\)

<=> \(x-\frac{1}{2}=0\)

<=> \(x=\frac{1}{2}\)

d) x(x - 3) + (x - 3) = 0

<=> (x + 1)(x - 3) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Vậy \(x\in\left\{-1;3\right\}\)là nghiệm phương trình

e) 9x(x - 7) - x + 7 = 0

<=> (9x - 1)(x - 7) = 0

<=> \(\orbr{\begin{cases}9x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{9}\\x=7\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{9};7\right\}\)là nghiệm phương trình

1, \(3x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow\)  \(3x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left(3x-1\right)=0\)  

  \(\Leftrightarrow\)\(\orbr{\begin{cases}x-2020=0\\3x-1=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2020\\3x=1\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2020\\x=\frac{1}{3}\end{cases}}\)

Vậy phương trình có nghiệm x=2020 hoặc x=\(\frac{1}{3}\)

2, \(4-9x^2=0\)

\(\Leftrightarrow4=9x^2\)

\(\Leftrightarrow\frac{4}{9}=x^2\)

\(\Leftrightarrow x=\pm\frac{2}{3}\)

Vậy phương trình có nghiệm x=\(\pm\frac{2}{3}\)

3, \(x^2-x+\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy phương trình có nghiệm x=\(\frac{1}{2}\)

4, \(x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Vậy phương trình có nghiệm x=3 hoặc x= -1

5, \(9x\left(x-7\right)-x+7=0\)

\(\Leftrightarrow9x\left(x-7\right)-\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(9x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\9x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\9x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{1}{9}\end{cases}}\)

Vậy phương trình có nghiệm x=7 hoặc x=\(\frac{1}{9}\)