\(\Leftrightarrow x\left(x-2017\right)-\left(x-2017\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2017\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2017=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2017\end{cases}}}\)

Vậy.....

\(x\left(x-2017\right)-x+2017=0\)

\(\Leftrightarrow x\left(x-2017\right)-\left(x-2017\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2017\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2017=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2017\end{cases}}\)

Ta có \(x^2\ge0\Rightarrow x^2+5>0\)

\(\Rightarrow x+3< 0\Leftrightarrow x< -3\)

Vậy x < -3 thì ( đề bài )

~ Học tốt ~

Bạn chứng minh cái này : a²ⁿ⁺¹ + b²ⁿ⁺¹ \(⋮\)a + b    ; aⁿ - bⁿ \(⋮\)a - b 

Ta có : 2018²⁰¹⁹ + 2020²⁰¹⁹ = ( 2018²⁰¹⁹ + 1 ) + ( 2020²⁰¹⁹ - 1 ) 

2018²⁰¹⁹ + 1 \(⋮\)( 2018 + 1 ) = 2019 ;  2020²⁰¹⁹ - 1 \(⋮\)( 2010 - 1 ) = 2019

\(\Rightarrow\) 2018²⁰¹⁹ + 2020²⁰¹⁹ \(⋮\) 2019 

\(\frac{x+19}{3}+\frac{x+13}{5}=\frac{x+7}{7}+\frac{x+1}{9}\)

\(=>\frac{x+19}{3}+3+\frac{x+13}{5}+3=\frac{x+7}{7}+3+\frac{x+1}{9}+3\)

\(=>\frac{x+28}{3}+\frac{x+28}{5}=\frac{x+28}{7}+\frac{x+28}{9}\)

\(=>\frac{x+28}{3}+\frac{x+28}{5}-\frac{x+28}{7}-\frac{x+28}{9}=0\)

\(=>\left(x+28\right)\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right)=0\)

Do :\(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\ne0\)

\(=>x+28=0\)

\(=>x=-28\)

Vậy nghiệm của phương trình trên là : -28

Thks nha

1) \(x\left(2^2-3\right)-x^2\left(5x+1\right)+x^2\)

\(=x-5x^3-x^2+x\)

\(=2x-5x^3-x^2\)

2) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

\(=\left(6x^2+23x-55\right)-\left(6x^2+23x+21\right)\)

\(=-76\)

Làm lại câu 1

\(x\left(2^2-3\right)-x^2\left(5x+1\right)+x^2\)

\(=x-5x^3-x^2+x^2\)

\(=x-5x^3\)

\(\left(x-1\right)^3-x\left(x-1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)

\(< =>\left(x-1+x\right)\left(x-1\right)^2=10x-5x^2-11x-22\)

\(< =>-x^2+x-1-10x+5x^2+11x+22=0\)

\(< =>4x^2+3x+21=0\)

\(< =>\left(2x\right)^2+2.2x.\frac{3}{4}+\left(\frac{3}{4}\right)^2+20\frac{9}{25}=0\)

\(< =>\left(2x+\frac{3}{4}\right)^2+20\frac{9}{25}=0\)

Do \(\left(2x+\frac{3}{4}\right)^2\ge0=>\left(2x+\frac{3}{4}\right)^2+20\frac{9}{25}\ge20\frac{9}{25}>0\)

Vậy phương trình vô nghiệm

Dòng 2 là (x-1-x) nha @@

\(ĐKXĐ:x\ne1;5;9\)

\(pt\Leftrightarrow\frac{2x-1}{\left(x-1\right)\left(x-5\right)}+\frac{\left(x-2\right)}{\left(x-1\right)\left(x-9\right)}=\frac{3x-12}{\left(x-9\right)\left(x+5\right)}\)

\(\Rightarrow\left(2x-1\right)\left(x-9\right)+\left(x-2\right)\left(x-9\right)=\left(3x-12\right)\left(x-1\right)\)

\(=>2x^2-x-18x+9+x^2-2x+5x-10=3x^2-12-3x+12\)

\(=>3x^2-16x-1=3x^2-15x+12\)

=>x=-13

ta có: \(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)

\(=\left(x+4\right)^2.\left(x^2-1\right)-\left(x^2-1\right)\)

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

\(=\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)

Cho mình nhé hihi!!!

x²(x+4)²-(x+4)²-(x²-1)

=(x+4)²  (x²-1)-(x²-1)

=(x²-1)(x²+8x+16-1)

=(x-1)(x+1)(x²+8x+15)