a) (1314 - 808)  ÷ 23 + 1995 = 506  ÷ 23 + 1995

= 22 + 1995 = 2017

b) x  × 25 =  15 4

x =  15 4   ÷   25

x =  15 4   ÷   25 = 0,15.

số học sinh xếp loại văn hóa giỏi là

     50:10x3=15(h/sinh)

số học sinh còn lại là

    50-15=35(h/sinh)

số học sinh loại khá là

    35:8x3=

Bài 3: 

b: =>x-2=-4

hay x=-2

Bài 2:

$a$ là ước của $-20$ nên $a\in \left\{\pm 1; \pm 2; \pm 4; \pm 5; \pm 10; \pm 20\right\}$
Mà $a\leq -16$ nên $a=-20$

$b$ là ước của $-28$ nên $b\in \left\{\pm 1; \pm 2; \pm 4; \pm 7;\pm 14; \pm 28\right\}$
Mà $b>20$ nên $b=28$

Khi đó: $a+b=(-20)+28=8$

Bài 3:
a.

$-25x+4=79$

$-25x=79-4=75$

$x=75:(-25)=-3$
b.

$-5(x-2)=20$

$x-2=20:(-5)=-4$

$x=-4+2=-2$

c.

$-6^3.x^2=-1944$
$x^2=(-1944):(-6^3)=9=3^2=(-3)^2$

$\Rightarrow x=3$ hoặc $x=-3$

Đặt A=\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)

         =(\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\))+(\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\))+...+\(\dfrac{1}{70}\)

Nhận xét:

\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\)>\(\dfrac{1}{20}\)+\(\dfrac{1}{20}\)+...+\(\dfrac{1}{20}\)=\(\dfrac{10}{20}\)=\(\dfrac{1}{2}\)

\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\)>\(\dfrac{1}{30}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{30}\)=\(\dfrac{10}{30}\)=\(\dfrac{1}{3}\)

\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\)>\(\dfrac{1}{60}\)+\(\dfrac{1}{60}\)+...+\(\dfrac{1}{60}\)=\(\dfrac{30}{60}\)=\(\dfrac{1}{2}\)

=>A>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)+\(\dfrac{1}{61}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)=\(\dfrac{4}{3}\)

=>A>\(\dfrac{4}{3}\)

Vậy: \(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{4}{3}\) (ĐPCM)

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=2018^2*2018*2017

=2018^3*2017