\(\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+...=...+\left(\frac{x+6}{2014}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2016}+\frac{x+2020}{2015}+\frac{x+2020}{2014}\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy \(x=-2020\)

Ta có: \(\frac{13}{x^2}-\frac{36}{\left(x+6\right)^2}=1\left(x\ne\left\{0;-6\right\}\right)\)

\(\Leftrightarrow\frac{13\left(x+6\right)^2-36x^2}{x^2\left(x+6\right)^2}=1\)

\(\Leftrightarrow13\left(x^2+12x+36\right)-36x^2=x^2\left(x^2+12x+36\right)\)

\(\Leftrightarrow-23x^2+156x+468=x^4+12x^3+36x^2\)

\(\Leftrightarrow x^4+12x^3+59x^2-156x-468=0\)

\(\Leftrightarrow\left(x^4+2x^3\right)+\left(10x^3+20x^2\right)+\left(39x^2+78x\right)-\left(234x+468\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^3+10x^2+39x-234\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[\left(x^3-3x^2\right)+\left(13x^2-39x\right)+\left(78x-234\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x^2+13x+78\right)=0\)

Vì \(x^2+13x+78>0\left(\forall x\right)\)

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

Vậy x = -2 hoặc x = 3

\(\frac{x-2017}{2018}+\frac{x-2018}{2017}=\frac{2017}{x-2018}+\frac{2018}{x-2017}\)

\(\Leftrightarrow\frac{2017.\left(x-2017\right)+2018.\left(x-2018\right)}{2018.2017}=\frac{2017.\left(x-2017\right)+2018.\left(x-2018\right)}{\left(x-2018\right).\left(x-2017\right)}\)

\(\Leftrightarrow2018.2017=\left(x-2018\right).\left(x-2017\right)\)

\(\Leftrightarrow2018.2017=x^2-4035x+2018.2017\)

\(\Leftrightarrow x^2-4035x=2018.2017-2018.2017\)

\(\Leftrightarrow x.\left(x-4035\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4035=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=4035\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{0;4035\right\}\)

nhìn căng nhể :))

a) ( x - 1 )( x - 3 )( x + 5 )( x + 7 ) - 297 = 0

<=> [ ( x - 1 )( x + 5 ) ][ ( x - 3 )( x + 7 ) ] - 297 = 0

<=> ( x² + 4x - 5 )( x² + 4x - 21 ) - 297 = 0

Đặt t = x² + 4x - 5

pt <=> t( t - 16 ) - 297 = 0

<=> t² - 16t - 297 = 0

<=> t² - 27t + 11t - 297 = 0

<=> t( t - 27 ) + 11( t - 27 ) = 0

<=> ( t - 27 )( t + 11 ) = 0

<=> ( x² + 4x - 5 - 27 )( x² + 4x - 5 + 11 ) = 0

<=> ( x² + 4x - 32 )( x² + 4x + 6 ) = 0

<=> ( x² - 4x + 8x - 32 )( x² + 4x + 6 ) = 0

<=> [ x( x - 4 ) + 8( x - 4 ) ]( x² + 4x + 6 ) = 0

<=> ( x - 4 )( x + 8 )( x² + 4x + 6 ) = 0

Đến đây dễ rồi :)

Gọi vận tốc đi,là v₁  thời gian đi ; về lần lượt là t₁ ; t₂  (v₁ ; t₁ ; t₂ > 0)

=> vận tốc về v₁ - 5

Đổi 30 phút = 1/2 giờ

Ta có t₂ - t₁ = 1/2

<=> \(\frac{S}{v_1-5}-\frac{S}{v_1}=\frac{1}{2}\) 

<=> \(\frac{180}{v_1-5}-\frac{180}{v_1}=\frac{1}{2}\)

<=> \(\frac{1}{v_1-5}-\frac{1}{v_1}=\frac{1}{360}\)

 \(\Leftrightarrow\frac{5}{\left(v_1-5\right)v_1}=\frac{1}{360}\)

<=> (v₁ - 5).v₁ = 1800

<=> (v₁)² - 5.v₁ = 1800

<=> (v₁)² - 45.v₁ + 40v₁ - 1800 = 0

<=> v₁(v₁ - 45) + 40(v₁ - 45) = 0

<=> (v₁ + 40)(v₁ - 45) = 0

<=> \(\orbr{\begin{cases}v_1=-40\left(\text{loại}\right)\\v_1=45\left(\text{tm}\right)\end{cases}}\)

Vậy vận tốc lúc đi là 45 km/h

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

\(\Leftrightarrow2x^2-3x-9+x^2-2x=3\left(x^2-4x+4\right)\)

\(\Leftrightarrow3x^2-5x-9=3x^2-12x+12\)

\(\Leftrightarrow7x=21\Rightarrow x=3\)

b) \(\left(4x+7\right)\left(x-3\right)-x^2=3x\left(x+2\right)\)

\(\Leftrightarrow4x^2-5x-21-x^2=3x^2+6x\)

\(\Leftrightarrow11x=-21\Rightarrow x=-\frac{21}{11}\)

a,X=3

CBHT