\(\dfrac{60}{x}-\dfrac{60}{x+20}=\dfrac{1}{2}\left(đk:x>0\right)\)

\(\Leftrightarrow\dfrac{120\left(x+20\right)-120x-x\left(x+20\right)}{2x\left(x+20\right)}=0\)

\(\Leftrightarrow120x+2400-120x-x^2-20x=0\)

\(\Leftrightarrow-x^2-20x+2400=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=40\left(n\right)\\x_2=-60\left(l\right)\end{matrix}\right.\)

Vậy \(S=\left\{40\right\}\)

\(\dfrac{120}{7}\)

\(\dfrac{300}{7}\)

60:7x2=120/7

60:7x5=300/7

60 : 7 x 2 = 120/7

60 : 7 x 5 = 300/7

a) 60 : (2 x 5) = 60 : 10 = 6

   60 : 2 : 5 = 30 : 5 = 6             

  60 : 5 : 2 = 12 : 2 = 6

Vậy 60 : (2 x 5) = 60 : 2 : 5 = 60 : 5 : 2

b) (24 x 48) : 12 = 1 152 : 12 = 96      

   (24 : 12) x 48 = 2 x 48 = 96            

   24 x (48 : 12) = 24 x 4 =  96

Vậy (24 x 48) : 12 = (24 : 12) x 48 = 24 x (48 : 12)

\(\dfrac{60}{x}-\dfrac{60}{x+2}=\dfrac{1}{20}\left(đk:x\ne0,x\ne-2\right)\)

\(\Leftrightarrow\dfrac{60x+120-60x}{x\left(x+2\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{120}{x^2+2x}=\dfrac{1}{20}\Leftrightarrow x^2+2x=2400\)

\(\Leftrightarrow\left(x+1\right)^2=2401\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=49\\x+1=-49\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=48\\x=-50\end{matrix}\right.\)(thỏa đk)

Ta có: \(\dfrac{60}{x}-\dfrac{60}{x+2}=\dfrac{1}{20}\)

\(\Leftrightarrow x\left(x+2\right)=1200x+2400-1200x\)

\(\Leftrightarrow x^2+2x-2400=0\)

\(\Delta=2^2-4\cdot1\cdot\left(-2400\right)=9604\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-2-98}{2}=-50\left(nhận\right)\\x_2=\dfrac{-2+98}{2}=48\left(nhận\right)\end{matrix}\right.\)

1)x=60-15=45

2) x=180-60=120 Mà ảnh đại diện của bn đẹp z

1) x + 15 = 60

           x = 60 - 15 

           x = 45

2) x + 60 = 180

          x  = 180 - 60 

          x = 120

1 ) x + 15 = 60

             x  = 60 - 15

              x  = 45 .

2 ) x + 60 = 180 

              x = 180 - 60

               x = 120 .

sin^2x+sin^2(60-x)+sinx*sin(60 độ-x)

\(=sin^2x+\left[sin60\cdot cosx-sinx\cdot cos60\right]^2+sinx\cdot\left[sin60\cdot cosx-sinx\cdot cos60\right]\)

\(=sin^2x+\left[-\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right]^2+sinx\left[\dfrac{-1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right]\)

\(=sin^2x+\dfrac{1}{4}sin^2x-\dfrac{\sqrt{3}}{2}\cdot sinx\cdot cosx+\dfrac{3}{4}\cdot cos^2x-\dfrac{1}{2}\cdot sin^2x+\dfrac{\sqrt{3}}{2}\cdot sinx\cdot cosx\)

\(=\dfrac{5}{4}sin^2x+\dfrac{3}{4}\cdot cos^2x-\dfrac{1}{2}\cdot sin^2x\)

=3/4*(sin^2x+cos^2x)=3/4

đặt x² + 16x + 60 = t thì PT đã cho trở thành :

t ( t + x ) - 6x² = 0 \(\Leftrightarrow\)t² + xt - 6x² = 0

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

+) t = 2x thì x² + 16x + 60 = 2x \(\Leftrightarrow\)x² + 14x +  60 = 0 ( vô nghiệm )

+) t = -3x thì x² + 16x + 60 = -3x \(\Leftrightarrow\)x² + 19x + 60 = 0 

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-4\\x=-15\end{cases}}\)

Vậy ....

1/ Ta có xy=-6

Với x=-6 => y=1

x=-3 => y=2 

x= -2 => y=3

x=-1 => y=6

2/ Ta có x=y+4 

Thay x=y+4 vào bt, ta được 

<=> y+4-3/y-2 =3/2

<=> y+1/y-2=3/2

<=> 2(y+1)=3(y-2)

<=> 2y +2 = 3y - 6

<=> 3y - 2y= 2+ 6

<=> y= 8 <=> x= 12

3/ -4/8 = x/-10 <=> x= (-4)*(-10)/8=5

-4/8 = -7/y <=> y=(-7)*8/(-4) =14

-4/8 = z/-24 <=> z= (-4)*(-24)/8=12