Ta có :\(\frac{x}{6}-\frac{7}{y}=\frac{1}{12}\)(y khác 0)

=> \(\frac{xy-42}{6y}=\frac{1}{12}\)

=> 12(xy - 42) = 6y

=> 12xy - 504 = 6y

=> 12xy - 6y = 504

=> 2xy - y = 84

=> y(2x - 1) = 84

Ta có 84 = 1.84 = (-1).(-84) = 42.2 = (-42).(-2) = 21.4 = (-21).(-4) = 7.12 = (-7).(-12) = (-3).(-28) = 28.3 = 14.6 = (-14).(-6)

Lập bảng xét 24 trường hợp

Vậy các cặp (y;x) thỏa mãn là : (84;1) ; (4 ; 11) ; (12 ; 4) ; (28 ; 2) ; (-4 ; - 10) ; (-12 ; -3) ; (-28 ; -1) ; (-84 ; 0)

so sánh ba số sau:12/13;13/14 và 0

Ta so sánh 3 số với 1 ta có :

1 - 12/13 = 1/13                 1 - 13/14 = 1/14               1 - 0 = 1

Vậy phần thừa của phân số lớn hơn thì phân số đó nhỏ hơn.

Vậy 0 < 12/13 < 13/14.

BPTT là gì vậy?

biện pháp tu từ

\(\frac{12}{10}\cdot\frac{25}{12}-\left(\frac{1}{2}+\frac{1}{3}\right):\left(-\frac{5}{6}\right)\)

\(=\frac{5}{2}-\frac{5}{6}:\left(-\frac{5}{6}\right)\)

\(=\frac{5}{2}+1\)

\(=\frac{7}{2}\)

kết quả : 3/2

học tốt .....

a)\(\frac{5}{3}-\frac{2}{3}\times x=1\)

=>\(\frac{2}{3}\times x=\frac{5}{3}-1\)

=>\(\frac{2}{3}\times x=\frac{2}{3}\)

=>\(x=\frac{2}{3}:\frac{2}{3}\)

=>\(x=1\)

b)\(\frac{1}{2}+\frac{5}{7}:x=\frac{1}{6}\)

=>\(\frac{5}{7}:x=\frac{1}{6}-\frac{1}{2}\)

=>\(\frac{5}{7}:x=-\frac{1}{3}\)

=>\(x=-\frac{1}{3}\times\frac{5}{7}\)

=>\(x=-\frac{5}{21}\)

thank you!

c) \(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{15}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}.\frac{27}{8}\)

\(2+\frac{3}{4}x=\frac{21}{8}\)

\(\frac{3}{4}x=\frac{21}{8}-2\)

\(\frac{3}{4}x=\frac{21}{8}-\frac{16}{8}\)

\(\frac{3}{4}x=\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{3}{4}\)

\(x=\frac{5}{8}.\frac{4}{3}\)

\(x=\frac{5}{6}\)

Vậy \(x=\frac{5}{6}\).

d) \(\left|x-\frac{1}{3}\right|-\frac{3}{4}=\frac{5}{3}\)

\(\left|x-\frac{1}{3}\right|=\frac{5}{3}+\frac{3}{4}\)

\(\left|x-\frac{1}{3}\right|=\frac{20}{12}+\frac{9}{12}\)

\(\left|x-\frac{1}{3}\right|=\frac{29}{12}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{29}{12}\\x-\frac{1}{3}=-\frac{29}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=-\frac{25}{12}\end{cases}}\)

Vậy \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).

a, \(\frac{2}{3}x+\frac{5}{6}x+\frac{1}{2}=\frac{-3}{4}\)

\(\Leftrightarrow\frac{3}{2}x=\frac{-5}{4}\)

\(\Leftrightarrow x=\frac{-5}{6}\)

b, \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}\)

\(\Leftrightarrow\frac{3}{5}.\left(3x-\frac{37}{10}\right)=\frac{-57}{10}\)

\(\Leftrightarrow3x-\frac{37}{10}=\frac{-19}{2}\)

\(\Leftrightarrow3x=\frac{-29}{5}\)

\(\Leftrightarrow x=\frac{-29}{15}\)

a) \(\frac{2}{3}x+\frac{5}{6}x+\frac{1}{2}=-\frac{3}{4}\)

\(x\left(\frac{2}{3}+\frac{5}{6}\right)+\frac{1}{2}=-\frac{3}{4}\)

\(x\cdot\frac{3}{2}=\frac{-5}{4}\)

\(x=-\frac{5}{6}\)

\(\frac{2}{5}+\frac{3}{5}\left(3x-3,7\right)=-\frac{53}{10}\)

\(\frac{3}{5}\left(3x-\frac{37}{10}\right)=-\frac{57}{10}\)

\(3x-\frac{37}{10}=-\frac{19}{2}\)

\(3x=\frac{-29}{5}\)

\(x=\frac{-29}{15}\)

c) \(M=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{1}{2}.\frac{4}{4}.\frac{6}{6}...\frac{100}{100}=\frac{1}{2}\)

a) M . N = \(\left(\frac{1}{2.}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)=\frac{1.2.3.4....100}{2.3.4.5...101}=\frac{1}{101}\)