\(\left(\frac{5}{12}+\frac{1}{2}.\frac{1}{3}\right):\left(2.\frac{1}{3}-1.\frac{5}{12}\right)\)

\(=\left(\frac{5}{12}+\frac{1}{6}\right):\left(\frac{2}{3}-\frac{5}{12}\right)\)

\(=\frac{7}{12}:\frac{1}{4}\)

\(=\frac{7}{3}\)

_Chúc bạn học tốt_

Đề bài 

= ( 5/12 +    1/6 ) : (    2/3 - 1/12 ) 

= ( 5/12 + 2/12 ) :   ( 8/12 - 1/12 ) 

= 7/12 : 7/12 

= 1 

K Trắng nha .

=1 nha

\(A=\frac{5}{2\times1}+\frac{4}{1\times11}+\frac{3}{11\times2}+\frac{1}{2\times15}+\frac{13}{15\times4}\)

      \(=\frac{5}{2}+\frac{4}{11}+\frac{3}{22}+\frac{1}{30}+\frac{13}{60}\)

        \(=\frac{13}{4}\)hoặc 3,25

Vậy A = 13/4 hoặc 3,25

\(\left(\frac{5}{12}+\frac{1}{2}\times\frac{1}{3}\right):\left(2\times\frac{1}{3}-1\times\frac{1}{12}\right)\)

\(=\left(\frac{5}{12}+\frac{1}{6}\right):\left(\frac{2}{3}-\frac{1}{12}\right)\)

\(=\frac{7}{12}:\frac{7}{12}\)

\(=1\)

k bk tính nhanh!

1234567890

Tìm x
a) 3(1-4x) (x-1)+a (3x-2) (x+3)= -27
b) 5(2x+3) (x+2)- 2(5x-4) (x-1) = 75

a,=(1/3+3/5+1/15)+(3/4+-1/36)+(1/72-2/9)=1+26/36-15/72=1+(52-15)/72=1+37/72=109/72

b,=1/100-(1/1x2+1/2x3+...+1/97x98+1/98x99+1/99x100)

   =1/100-(1/1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)

   =1/100-(1/1-1/100)=1/100-99/100=-98/100=-49/50

chỉ có mk mk giải thôi đó l-i-k-e đi

mình cũng đang bí bài này

Câu a là 1676250 nhé

\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\frac{99}{100}\)

\(C=-\frac{98}{100}=-\frac{49}{50}\)

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

   \(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

    \(=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)

      \(=\frac{1}{100}-\left(\frac{1}{100}-1\right)\)

      \(=1\)

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{-49}{50}\)