\(\dfrac{1}{4}-\dfrac{1}{3}+\dfrac{5}{12}\\ =\dfrac{3}{12}-\dfrac{4}{12}+\dfrac{5}{12}\\ =\dfrac{3-4+5}{12}\\ =\dfrac{4}{12}\\ =\dfrac{1}{3}\)

a,3/7 +1/2-[-3]/70

=13/14 -[-3]/70

=65/70 -[-3]/70

=34/35

b,3/5+-1/25 -35/100

=14/25 - 35/100

=56 /100 -35/100

= 21/100

c ,5/12 -3/-16+3/4

=29/48 +3/4

= 29/48 +36/48

=65/48

d, 5/15 +4/-12 +1/7 -1/-6

= 1/3 +1/-3 +1/7-1/-6

=0+1/7-1/-6

= 13/42

A.\(\dfrac{3}{7}+\dfrac{1}{2}+\dfrac{\left(-3\right)}{70}\)

=\([\dfrac{30}{70}+\dfrac{\left(-3\right)}{70}]\)+\(\dfrac{1}{2}\)

= \(\dfrac{27}{70}+\dfrac{35}{70}\)

=\(\dfrac{62}{70}=\dfrac{31}{35}\)

B.\(\dfrac{3}{5}+\dfrac{-1}{25}-\dfrac{35}{100}\)

=\(\dfrac{60}{100}+\dfrac{-4}{100}-\dfrac{35}{100}\)

=\(\dfrac{60+\left(-4\right)-35}{100}\)

=\(\dfrac{21}{100}\)

C.\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}\)

=\(\left(\dfrac{5}{12}+\dfrac{9}{12}\right)-\dfrac{-3}{16}\)

=\(\dfrac{7}{6}-\dfrac{-3}{16}\)

=\(\dfrac{56}{48}-\dfrac{-9}{48}\)=\(\dfrac{65}{48}\)

D.\(\dfrac{5}{15}+\dfrac{4}{-12}+\dfrac{1}{7}-\dfrac{1}{-6}\)

=\(\dfrac{1}{3}+\dfrac{1}{7}+(\dfrac{-2}{6}-\dfrac{-1}{6})\)

=\(\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{-1}{6}\)

=\((\dfrac{1}{3}+\dfrac{-1}{6})+\dfrac{1}{7}\)

=\(\dfrac{1}{6}+\dfrac{1}{7}\)=\(\dfrac{13}{42}\)

\(\frac{-1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)

\(=\frac{-1}{12}-\left(\frac{21}{8}-\frac{1}{3}\right)\)

\(=\frac{-1}{12}-\left(\frac{63}{24}-\frac{8}{24}\right)\)

\(=\frac{-1}{12}-\frac{55}{24}\)

\(=\frac{-2}{24}-\frac{55}{24}\)

\(=\frac{-19}{8}\)

\(A=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(A=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

\(A=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(a=x^2+8x+11\)

\(\Rightarrow A=\left(a-4\right)\left(a+4\right)+15\)

\(\Leftrightarrow A=a^2-16+15\)

\(\Leftrightarrow A=a^2-1\)

Thay a vào A ( :v ) ta có :

\(A=\left(x^2+8x+11\right)^2-1\)

\(A=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)\)

\(A=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(A=\left(x^2+2x+6x+12\right)\left(x^2+8x+10\right)\)

\(A=\left[x\left(x+2\right)+6\left(x+2\right)\right]\left(x^2+8x+10\right)\)

\(A=\left(x+6\right)\left(x+2\right)\left(x^2+8x+10\right)⋮x+6\left(đpcm\right)\)

giai ro rang

 (-1)+2-3+4-....-99+100 = 50

5/11+1/2+2/5+6/11+3/4+16/25+5/16

=1441/400

19/4+37/100+1/8+132/25+5/2+37/12

=1933/120

Ban lam sao duooc nhu the giai ra cho minh voi

B = ( \(\dfrac{5}{7}.0,6:3\dfrac{1}{2}\)) . ( 40% - 1,4) . \(\left(-2\right)^3\)

= (\(\dfrac{5}{7}.\dfrac{3}{5}\): \(\dfrac{7}{2}\)) . ( \(\dfrac{2}{5}-\dfrac{7}{5}\)) . (-8)

= (\(\dfrac{15}{35}:\dfrac{7}{2}\)) . \(\dfrac{-5}{5}\) . ( -8)

= (\(\dfrac{3}{7}.\dfrac{2}{7}\)) . (-1) . (-8)

= \(\dfrac{6}{49}\) . (-1) . (-8)

= \(\dfrac{-6}{49}\) . (-8)

= \(\dfrac{48}{49}\)

Vậy: B =\(\dfrac{48}{49}\)

Nhớ tick nha

B=(\(\dfrac{3}{7}.\dfrac{2}{7}\)).(\(\dfrac{4}{10}-\dfrac{14}{10}\)).(-8)

B=\(\dfrac{3}{7}.\left(-1\right)\left(-8\right)\)

B=\(\dfrac{24}{7}\)

=> B = (1 + 5 + 9 + ... + 2017) - (3 + 7 + 11 + ... + 2019)

Đặt C = 1 + 5 + 9 + ... + 2017

Ta thấy các số hạng liền kề của C hơn kém nhau 4 đơn vị

=> Số số hạng của C là

(2017 - 1)/4 + 1 = 505 (số)

=> C = (1 + 2017) * 505 / 2 = 509545 (1)

Đặt D = 3 + 7 + 11 + ... + 2019

Bạn làm tương tự tính được 

D = (3 + 2019) * 505 / 2 = 510555 (2)

Từ (1) và (2)

=> B = C - D = 509545 - 510555 = -1010

Ai tra loi dung toi se t cho

Ta có 

B = 1 - 3 + 5 - 7 + 9 - 11 + ... + 2017 - 2019

   = (1 - 3) + (5 - 7) + (9 - 11) + ... + (2017 - 2019)

   = -2 + (-2) + (-2) + ....+ (-2)

   Số cặp số hạng của dãy là : 

((2019 - 1 ) : 2 ) :2 +1 = 505.5 (cặp)

= > B = (-2) x 505.5 = -1011 

Vậy B = 1011