Đáp án:
Bài 5: $x=-\dfrac53$
Bài 7: $x\in\{-2,-6,-8,-4\}$
Giải thích các bước giải:
Bài 5:
Ta có:
$(\dfrac1{24.25}+\dfrac{25.26}+...+\dfrac1{29.30}).120+x:\dfrac13=-4$
$\to (\dfrac{25-24}{24.25}+\dfrac{26-25}{25.26}+...+\dfrac{30-29}{29.30}).120+3x=-4$
$\to (\dfrac1{24}-\dfrac1{25}+\dfrac1{25}-\dfrac1{26}+...+\dfrac1{29}-\dfrac1{30}).120+3x=-4$
$\to (\dfrac1{24}-\dfrac1{30}).120+3x=-4$
$\to \dfrac1{120}.120+3x=-4$
$\to 1+3x=-4$
$\to 3x=-5$
$\to x=-\dfrac53$
Bài 7:
Ta có:
$xy+5x+5y+28=0$
$\to (xy+5x)+(5y+25)+3=0$
$\to x(y+5)+5(y+5)+3=0$
$\to (x+5)(y+5)+3=0$
$\to (x+5)(y+5)=-3$
$\to (x+5,y+5)$ là cặp ước của $-3$ vì $x,y\in Z$
$\to (x+5,y+5)\in\{(3,-1), (-1,3), (-3,1), (1,-3)\}$
$\to (x,y)\in\{(-2,-6), (-6,-2), (-8,-4), (-4,-8)\}$
$\to x\in\{-2,-6,-8,-4\}$
