\(\left(x+2\right)\left(\dfrac{360}{x}-6\right)=360\)

\(ĐK:x\ne0\)

\(\Leftrightarrow\left(x+2\right)\left(\dfrac{360-6x}{x}\right)=360\)

\(\Leftrightarrow360-6x+\dfrac{720-12x}{x}=360\)

\(\Leftrightarrow360x-6x^2+720-12x=360x\)

\(\Leftrightarrow6x^2+12x-720=0\)

\(\Delta=12^2-4.6.\left(-720\right)\)

    \(=17424>0\)

`->` pt có 2 nghiệm

\(\left\{{}\begin{matrix}x_1=\dfrac{-12-\sqrt{17424}}{12}=-12\\x_2=\dfrac{-12+\sqrt{17424}}{12}=10\end{matrix}\right.\) ( tm )

Vậy \(S=\left\{-12;10\right\}\)

\(\left(x^2+1\right)\left(x-2\right)+2x=4\Leftrightarrow x^3-2x^2+x-2+2x-4=0\Leftrightarrow x^3-2x^2+3x-6=0\Leftrightarrow\left(x-2\right)\left(x^2+3\right)=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)(do \(x^2+3\ge3>0\))

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

\((x^3+x)(x-2)+2x-4=0\)

\(x^4-2x^3+x^2-2x+2x-4=0\)

\(x^4-2x^3+x^2-4=0\)

\((x^4-2x^3)+(x^2-4)=0\)

\(x^3(x-2)+(x-2)(x+2)=0\)

\((x^3+x+2)(x-2)=0\)

th1:\(x^3+x+2=0 \)

              x=-1

th2 x-2=0

         x=2

ptrình cs tập nghiệm S={-1,0}

Ta có: \(\dfrac{x+1}{29}+\dfrac{x+3}{27}=\dfrac{x-3}{33}+\dfrac{x-7}{37}\)

\(\Leftrightarrow\dfrac{x+30}{29}+\dfrac{x+30}{27}-\dfrac{x+30}{33}-\dfrac{x+30}{37}=0\)

\(\Leftrightarrow x+30=0\)

hay x=-30

`2x-2/3=1/2`

`2x=1/2+2/3`

`2x=7/6`

`x=7/6:2=7/12`

\(2x-\dfrac{2}{3}=\dfrac{1}{2}\Leftrightarrow2x=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\Leftrightarrow x=\dfrac{7}{6}:2=\dfrac{7}{12}\)

2x - 2/3 = 1/2

2x = 1/2 + 2/3 = 7/6

x = 7/6 : 2 = 7/12

vậy x = 7/12

`48/[x+4]+48/[x-4]=5`           `ĐK: x \ne +-4`

`<=>[48(x-4)+48(x+4)]/[(x-4)(x+4)]=[5(x+4)(x-4)]/[(x-4)(x+4)]`

   `=>48x-192+48x+192=5x^2-80`

`<=>5x^2-96x-80=0`

`<=>5x^2-100+4x-80=0`

`<=>5x(x-20)+4(x-20)=0`

`<=>(x-20)(5x+4)=0`

`<=>` $\left[\begin{matrix} x=20\\ x=\dfrac{-4}{5}\end{matrix}\right.$   (t/m)

Vậy `S={-4/5;20}`

ĐK : \(x\ne\pm4\)

\(\Leftrightarrow\cdot\dfrac{48\left(x+4\right)+48\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5\left(x+4\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\)

\(\Leftrightarrow48x+192+48x-192==5x^2-80\)

\(\Leftrightarrow96x=5x^2-80\)

\(\Leftrightarrow5x^2-96x-80=0\)

\(\Leftrightarrow5x^2+4x-100-80=0\)

\(\Leftrightarrow4\left(x-20\right)+5x\left(x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-20=0\\5x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Một đội đồng diễn thể dục có 108 học sinh, trong đó có  số học sinh là học sinh nam. Hỏi đội đó có bao nhiêu học sinh nữ ?

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+.....+\left(x+100\right)=5750.\)

\(=100x+\left(1+2+3+...+100\right)=5750\)

\(=100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(100.x=700\)

\(x=700:100\)\

\(x=7\)

Vậy x = 7

Bài 1:

\(x+11,2=12,6\times2\\ x+11,2=25,2\\ x=25,2-11,2=14\)

\(2,5\times x=24,4:2\\ 2,5\times x=12,2\\x=\dfrac{12,2}{2,5}=4,88 \)

\(12,76-x=9,5+1,8\\ 12,76-x=11,3\\ x=12,76-11,3=1,46\)