Bài giải

a, \(\frac{x+5}{2017}-\frac{x+5}{2018}+\frac{x+5}{2019}-\frac{x+5}{2020}=0\)

\(\left(x+5\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)=0\)

Do \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)\ne0\) 

\(\Rightarrow\text{ }x+5=0\)

\(x=0-5\)

\(=-5\)

                                                    Bài giải

a, \(\frac{x+5}{2017}-\frac{x+5}{2018}+\frac{x+5}{2019}-\frac{x+5}{2020}=0\)

\(\left(x+5\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)=0\)

Do \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)\ne0\) 

\(\Rightarrow\text{ }x+5=0\)

\(x=0-5\)

\(=-5\)

\(b,\text{ }x^2-3x>0\)

\(x\left(x-3\right)>0\)

\(TH1\text{ : }\hept{\begin{cases}x>0\\x-3>0\end{cases}}\)          \(\Rightarrow\hept{\begin{cases}x>0\\x>3\end{cases}}\)         \(\Rightarrow\text{ }x>3\)

\(TH2\text{ : }\hept{\begin{cases}x< 0\\x-3< 0\end{cases}}\)          \(\Rightarrow\hept{\begin{cases}x< 0\\x< 3\end{cases}}\)          \(\Rightarrow\text{ }x< 3\)

\(ĐKXĐ:x\ne-1;x\ne\frac{2}{3}\)

\(pt\Leftrightarrow\frac{7x-2\left(x+1\right)+\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=1\)

\(\Leftrightarrow7x-2\left(x+1\right)+\left(3x-2\right)=\left(3x-2\right)\left(x+1\right)\)

\(\Leftrightarrow8x-4=3x^2-2x+3x-2\)

\(\Leftrightarrow3x^2-7x+2=0\)

\(\Delta=7^2-4.3.2=25,\sqrt{\Delta}=5\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7+5}{6}=2\\x=\frac{7-5}{6}=\frac{1}{3}\end{cases}}\)

Tự cho đkxđ nha!!!

<=> \(\frac{x+1-x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2}{3x-2}\)

<=> \(\frac{3x-2}{\left(3x-2\right)\left(x+1\right)}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(3x-2\right)\left(x+1\right)}\)

<=> \(\frac{7x-2x-2-3x+2}{\left(3x-2\right)\left(x+1\right)}=0\)

<=> \(\frac{2x}{\left(3x-2\right)\left(x+1\right)}=0\)

=> 2x = 0

<=> x = 0 (TM)

Vậy ...

\(1-\frac{x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}+\frac{2}{2-3x}\)

\(\left(x+1\right)\left(3x-2\right)\left(2-3x\right)-x\left(3x-2\right)\left(2-3x\right)=7x\left(2-3x\right)+2\left(x+1\right)\left(3x-2\right)\)

\(-9x^2+12x-4=16x-15x^2-4\)

\(-9x^2+12x=16x-15x^2\)

\(9x^2-12x+16x-15x^2=0\)

\(-6x^2+4x=0\)

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

\(Th1:-2x=0\Leftrightarrow x=0\)

\(Th2:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)

\(2\frac{1}{3}+\left(x-\frac{3}{2}\right)=\left(3-\frac{3}{2}\right)x\)

\(\Rightarrow\frac{7}{3}+x-\frac{3}{2}=\frac{3}{2}x\)

\(\Rightarrow x-\frac{3}{2}x=\frac{3}{2}-\frac{7}{3}\)

\(\Rightarrow\frac{-1}{2}x=-\frac{5}{6}\)

\(\Rightarrow x=\frac{5}{3}\)

\(x=\frac{\left[6+3^2.2-6.3^2\right]^2}{\left[3^2+3.3^2-3^4\right]^2}=\frac{\left[6\left[1+3-9\right]\right]^2}{\left[9+27-81\right]^2}=\frac{\left[-30\right]^2}{\left[-45\right]^2}=\frac{900}{2025}=\frac{4}{9}\)

Bn tham khảo bài Lee Vincent nha!

ta có \(\left(1-\frac{1}{2010}.\right).\left(1-\frac{2}{2010}\right)....\left(1-\frac{2010}{2010}\right).\left(1-\frac{2011}{2010}\right)\)\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).......0.\left(1-\frac{2011}{2010}\right)=0\)

Hồ Ngọc Minh Châu Võ cho mình hỏi nhưng bài kia mỗi bài 1 dòng hay là cả một bài vậy bạn

mình chịu.

<=> \(x^2-25=10x+35-2x^2-7x\)

<=> \(3x^2-3x-60=0\)

<=> \(x^2-x-20=0\)

<=> \(\left(x-5\right)\left(x+4\right)=0\)

<=> \(\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)

Vay \(x\in\left\{-4;5\right\}\)

Chuc ban hoc tot 

\(x^2-25=\left(5-x\right)\left(2x+7\right)\)

\(\Leftrightarrow x^2-25=10x-2x^2+35-7x\)

\(\Leftrightarrow3x^2-3x-60=0\)

Ta có \(\Delta=3^2+4.3.60=729,\sqrt{\Delta}=27\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3+27}{6}=5\\x=\frac{3-27}{6}=-4\end{cases}}\)

Đặt S = 1x2 + 2x3 + 3x4 + 4x5 + ... + 98x99

3S = 1x2x3 + 2x3(4-1) + 3x4x(5-2) + 4x5x(6-3) ... + 98x99x(100 - 97)

3S = 1x2x3 + 2x3x4 - 1x3x4 + 3x4x5 - 2x3x4 + ... + 98x99x100 - 97x98x99

3S = 98x99x100 => S = 1/3x98x99x100.

Thay vào đề bài ta được:

\(\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{12}{\frac{6}{7}}:\frac{-3}{2}\Leftrightarrow\frac{33\cdot100\cdot x}{275}=-\frac{12}{\frac{6}{7}}\cdot\frac{2}{3}\)

\(\Leftrightarrow12x=-12\cdot\frac{7}{6}\cdot\frac{2}{3}\Leftrightarrow x=-\frac{7}{9}\)

/i 4 U 4 nothing but if U are nothing, nothing will come to U again. /i