Bài làm :

Ta có :

\(\left(x-120\right)\div35=5\)

\(x-120=5\times35\)

\(x-120=175\)

\(x=175+120=295\)

Vậy x=295

(x-120):35=5

(x-120)=5×35

(x-120)=175

x=175+120

x=295

vậy x=295

ĐK: ` x \ne 10; x \ne 0`

`120/(x-10)-3/5=120/x`

`<=>120/(x-10)-120/x=3/5`

`<=>1/(x-10) - 1/x= 1/200`

`<=> (x-x+10)/(x(x-10)) = 1/200`

`<=> 10/(x(x-10))= 1/200`

`<=> x^2-10=2000`

`<=>` \(\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)

Vậy `S={50;-40}`.

`120/(x-10)-3/5=120/x(x ne 0,x ne 10)`

`<=>40/(x-10)-1/5=40/x`

`<=>200x-x(x-10)=200(x-10)`

`<=>200x-200x+2000-x^2+10x=0`

`<=>x^2-10x-2000=0`

`Delta'=25+2000=2025`

`<=>x_1=50,x_2=-40`

Vậy `S={50,-40}`

\(\dfrac{120}{x}+\dfrac{120}{x-10}=\dfrac{3}{5}\left(dkxd:x>0,x\ne10\right)\)

\(\Leftrightarrow\dfrac{120}{x}+\dfrac{120}{x-10}-\dfrac{3}{5}=0\)

\(\Leftrightarrow\dfrac{120.5\left(x-10\right)+5.120x-3x\left(x-10\right)}{5x\left(x-10\right)}=0\)

\(\Leftrightarrow600x-6000+600x-3x^2+30x=0\)

\(\Leftrightarrow-3x^2+1230x-6000=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\approx405\\x\approx5\end{matrix}\right.\)\(\left(tmdk\right)\)

Vậy ...

ĐKXĐ: x ≠ 0; x ≠ 10 em ơi

xy + x + y = 120

x . ( y + 1 ) + y = 120

x . ( y + 1 ) + ( y + 1 ) = 121

( x + 1 ) . ( y + 1 ) = 121

Lập bảng ta có :

Vậy ...

tìm x,y biết: x×y+x+y=120

x×(y+1)+y=120

x×(y+1)+(y+1)×1=120+1=121

(y+1)×(x+1)=121

(y+1)×(x+1)=121×1=1×121=11×11

Ta có bảng:

Vậy (x;y):(120;0)(0;120)(10;10)

 

tháng nào mà chẳng có 28 ngày

tháng 2

a. ( x + 3 )( x - 3 ) = 16

⇔x²-9=16

⇔x²-16-9=0

⇔x²-25=0

⇔(x-5)(x+5)=0

⇔\(\left[{}\begin{matrix}x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=5\end{matrix}\right.\)

Câu 1: 

=>5x=15

hay x=3

Câu 2: 

\(\Leftrightarrow120-\left(3x+9\right)=12\)

=>3x+9=108

=>3x=99

hay x=33