a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)

\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)

\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)

\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)

\(A=\dfrac{119}{720}\)

b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)

\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)

\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)

\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)

\(B=1\)

a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)

   A  = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)

   A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)

   A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)

  A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)

 A = \(\dfrac{119}{720}\)

b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]

    B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]

    B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]

   B  = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))

B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)

B = 1

Bài 1:

a. $=\frac{25-4}{12}=\frac{21}{12}=\frac{7}{4}$

b. $=\frac{-5}{4}+\frac{15}{4}=\frac{15-5}{4}=\frac{10}{4}=\frac{5}{2}$

c. $=\frac{-15}{12}+\frac{15}{12}+\frac{3}{-4}=\frac{-15+15}{12}+\frac{3}{-4}=\frac{3}{-4}$

d.

$=3+\frac{1}{4}-1-\frac{1}{3}=(3-1)-(\frac{1}{3}-\frac{1}{4})$

$=2-\frac{1}{12}=\frac{23}{12}$

e.

$=3,1-2,5+2,5-3,1=(3,1-3,1)-(2,5-2,5)=0-0=0$

g.

$=\frac{1}{18}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}$

$=\frac{1}{18}-\frac{2}{5}=\frac{-31}{90}$

Bài 2:

a.

$x+\frac{1}{5}=\frac{3}{7}$

$x=\frac{3}{7}-\frac{1}{5}=\frac{8}{35}$

b.

$x=\frac{1}{2}+\frac{3}{4}=\frac{2}{4}+\frac{3}{4}=\frac{5}{4}$

c.

$\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}$
$x=\frac{1}{4}-\frac{2}{5}=\frac{-3}{20}$

d.

$2x(x-\frac{1}{7})=0$
$\Rightarrow 2x=0$ hoặc $x-\frac{1}{7}=0$

$\Rightarrow x=0$ hoặc $x=\frac{1}{7}$

e.

$\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=\frac{-7}{20}$

$x=\frac{1}{4}: \frac{-7}{20}=\frac{-5}{7}$

f.

$x=\frac{5}{12}-\frac{1}{3}=\frac{1}{12}$

`2x-3y-z` hay `2x+3y-z` ạ?

\(S_{35}=1-2+3-4+...+35\)

\(\Rightarrow S_{35}=\left(-1\right)+\left(-1\right)+...+35=17.\left(-1\right)+35=18\)

\(S_{60}=1-2+3-4+...+60\)

\(\Rightarrow S_{60}=\left(-1\right)+\left(-1\right)+...+59-60=30.\left(-1\right)=-30\)

\(\Rightarrow S_{35}+S_{60}=18-30=-12\)

\(A=1999^{2000}+\dfrac{1}{1999^{1999}}+1\)

\(B=1999^{1999}+\dfrac{1}{1999^{1998}}+1\)

\(\Rightarrow A-B=1999^{1999}.\left(1999-1\right)+\dfrac{1}{1999^{1998}}.\left(\dfrac{1}{1998}-1\right)\)

\(\Rightarrow A-B=1998.1999^{1999}-\dfrac{1997}{1998}.\dfrac{1}{1999^{1998}}\)

mà \(0< \dfrac{1997}{1998}.\dfrac{1}{1999^{1998}}< 1;1998.1999^{1999}>0\)

\(\Rightarrow A-B=1998.1999^{1999}-\dfrac{1997}{1998}.\dfrac{1}{1999^{1998}}>0\)

\(\Rightarrow A>B\)

Kq là: (-32)⁹ < (-18)¹³ .

Trình bày rõ ràng và chi tiết vào ạ? Kết quả cũng sai mà ghi như tht v?

Diện tích thửa ruộng :

\(\dfrac{1}{2}.\left(15+20\right).5=87,5\) \(\left(m^2\right)\)

Số kg thóc thu hoạch :

\(87,5.0,6=52,5\) \(\left(kg\right)\)

Diện tích mảnh ruộng là:

   1/2 . 5 ( 15 + 20 ) = 87,5(m² )

Số kg thóc mạnh ruộng cho sản là:

    0,6 . 87,5 = 52,5 ( kg )

  Vậy ...

Gọi 3 số tự nhiên liên tiếp là x; x+1 ; x+2

Ta có:

x.(x+1) + 50 = (x+1).(x+2)

=> (x+1).(x+2) - x.(x+1) = 26

=> (x+1).(x+2-x) = 26

=> (x+1).2 = 26

=> x+1 = 26 : 2 = 13

tick nha

giúp với

\(-\dfrac{3}{5}:\left(-\dfrac{3}{4}\right)=\dfrac{3}{5}.\dfrac{4}{3}=\dfrac{4}{5}\)

\(-\dfrac{5}{2}:\left(-\dfrac{3}{4}\right)=\dfrac{5}{2}.\dfrac{4}{3}=\dfrac{10}{3}\)

\(x\left(1-3y\right)+1-3y-1=-4\)

\(\left(1-3y\right)\left(x+1\right)=-3\)

⇒ (x+1) và (1-3y) ϵ {-1;1;-3;3}

\(\Rightarrow\left(x;y\right)\in\left\{\left(-2;-\dfrac{2}{3}\right);\left(0;\dfrac{4}{3}\right);\left(-4;0\right);\left(2;\dfrac{2}{3}\right)\right\}\)