\(\dfrac{360}{x}-\dfrac{400}{x+1}=1\) (ĐK: \(x\ne0,x\ne-1\))

\(\Leftrightarrow\dfrac{360\left(x+1\right)}{x\left(x+1\right)}-\dfrac{400x}{x\left(x+1\right)}=\dfrac{x\left(x+1\right)}{x\left(x+1\right)}\)

\(\Leftrightarrow360\left(x+1\right)-400x=x\left(x+1\right)\)

\(\Leftrightarrow360x+360-400x=x^2+x\)

\(\Leftrightarrow-40x+360=x^2+x\)

\(\Leftrightarrow x^2+40x+x-360=0\)

\(\Leftrightarrow x^2+41x-360=0\)

\(\Rightarrow\Delta=41^2-4\cdot1\cdot\left(-360\right)=3121>0\)

\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{-41+\sqrt{3121}}{2\cdot1}\approx7\left(tm\right)\\x_2=\dfrac{-41-\sqrt{3121}}{2\cdot1}\approx-48\left(tm\right)\end{matrix}\right.\)

\(\dfrac{360}{x}-\dfrac{400}{x+1}=1\)

Điều kiện: \(x\ne0;x\ne-1\)

PT \(\Leftrightarrow\dfrac{360\left(x+1\right)-400x}{x\left(x+1\right)}=1\)

\(\Rightarrow-40x+360=x\left(x+1\right)\)

\(\Leftrightarrow-40x+360=x^2+x\)

\(\Leftrightarrow x^2+41x-360=0\)

\(\Leftrightarrow x^2+2.\dfrac{41}{2}.x+\dfrac{1681}{4}=\dfrac{3121}{4}\)

\(\Leftrightarrow\left(x+\dfrac{41}{2}\right)^2=\left(\dfrac{\sqrt{3121}}{2}\right)^2\)

\(\Leftrightarrow x+\dfrac{41}{2}=\dfrac{\sqrt{3121}}{2}\) hoặc \(x+\dfrac{41}{2}=-\dfrac{\sqrt{3121}}{2}\)

\(\Leftrightarrow x=\dfrac{\sqrt{3121}}{2}-\dfrac{41}{2}\) hoặc \(x=-\dfrac{\sqrt{3121}}{2}-\dfrac{41}{2}\)

Vậy...

ĐKXĐ: \(x\ne0\)

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

=> \(360+2x-\dfrac{2160}{x}-12=360\)

<=> \(\dfrac{2x^2-12x-2160}{x}=0\)

=> \(x^2-6x-1080=0\)

<=> \(\left(x^2+30x\right)-\left(36x+1080\right)=0\)

<=> \(\left(x+30\right)\left(x-36\right)=0\)

<=> \(\left[{}\begin{matrix}x=-30\\x=36\end{matrix}\right.\) ( TM)

Vậy ....................................

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

\(ĐK:x\ne0\)

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

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

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

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

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

    \(=17424>0\)

`->` 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 \(S=\left\{-12;10\right\}\)

\(\left(x+6\right)\left(\frac{360}{x}+2\right)-12=360\) (\(x\ne0\))

\(\Leftrightarrow\left(x+6\right)\left(\frac{360}{x}+2\right)=372\)

\(\Leftrightarrow360+2x+\frac{2160}{x}+12=372\)

\(\Leftrightarrow360x+2x^2+2160+12x=372x\)

\(\Leftrightarrow x^2=-2160\)( vô lý )

=> phương trình vô nghiệm

\(x.y=360\)

e chỉ liệt kê vài Ư thôi nha !

\(\Rightarrow x;y\inƯ\left(360\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;...;\pm120;...\right\}\)

b, thực hiện như vậy !!!

=>(X+X+....+X)+(2+4+...+20)=360

=>10X+(20+2)*10:2=360

=> 10X+110=360

=> 10X=360-110=250

=>X=250:10

=>X=25

   study well

K NHA

 CẢM ƠN CẢM BẠN NHIỀU

(x + 2) + (x + 4) + ... + (x + 20) = 360 (10 cặp)

=> (x + x + ... + x) + (2 + 4 + ... + 20) = 360

     20 số hạng x        20 số hạng

=> 20x + 20.(20 + 2) : 2 = 360

=> 20x + 220 = 360

=> 20x           = 140

=>     x           = 7

Vậy x = 7

\(\left(x+2\right)+\left(x+4\right)+...+\left(x+20\right)=360\)

\(\left(x+x+x+...+x\right)+\left(2+4+6+...+20\right)=360\)   ( 10 số x )

\(10x+\left(20+2\right).10:2=360\)

\(\Rightarrow10x+120=360\Rightarrow10x=360-120=240\)

\(\Rightarrow x=240:10=24\)

360 : x - 126 : x = 6

(360 - 126) : x = 6

234 : x =6

x = 234 :6

x = 39

CÓ NGAY KQ NÈ EM YEW

360 : x - 126 : x = 6

(360 - 126) : x = 6

234 : x =6

x = 234 :6

x = 39