\(2xy-3x+2y=6\)

\(\Leftrightarrow x\left(2y-3\right)+2y-3=3\)

\(\Leftrightarrow\left(x+1\right)\left(2y-3\right)=3\)

Ta có bảng giá trị: 

A=\(\frac{-1}{9}:\left[\frac{-3}{2}.\left(1-\frac{4}{5}\right)-\frac{2}{3}.\left(\frac{-4}{5}-\frac{1}{-2}\right)\right]\)

=\(\frac{-1}{9}:\left[\frac{-3}{2}.\frac{1}{5}-\frac{2}{3}.\frac{-3}{10}\right]\)

=\(\frac{-1}{9}:\left(\frac{-3}{10}-\frac{-1}{5}\right)\)

=\(\frac{-1}{9}:\frac{-1}{10}=\frac{-1}{9}.\left(-10\right)=\frac{10}{9}\)

B=\(\frac{9}{2}:\left[\frac{1}{2}.\left(-1-\frac{3}{4}\right)+\frac{2}{3}.\left(\frac{-3}{4}-\frac{-1}{2}\right)\right]\)

B=\(\frac{9}{2}:\left[\frac{1}{2}.\frac{-7}{4}+\frac{2}{3}.\frac{-1}{4}\right]=\frac{9}{2}:\left(\frac{-7}{8}+\frac{-1}{6}\right)=\frac{9}{2}:\frac{-25}{24}=\frac{-108}{25}\)

C=\(\frac{5}{4}:\left[\frac{-5}{4}.\left(4-\frac{3}{5}\right)-\frac{3}{2}.\left(\frac{-3}{4}-\frac{3}{-5}\right)\right]\)

C=\(\frac{5}{4}:\left[\frac{-5}{4}.\frac{17}{5}-\frac{3}{2}.\frac{-3}{20}\right]=\frac{5}{4}:\left(\frac{-17}{4}-\frac{-9}{40}\right)=\frac{5}{4}:\frac{-161}{40}=\frac{-50}{161}\)

D=\(\frac{4}{7}.\left[\frac{-3}{4}:\left(\frac{5}{-4}-\frac{-4}{5}\right)+\frac{2}{3}:\left(\frac{-4}{5}-\frac{1}{-2}\right)\right]\)

=\(\frac{4}{7}.\left[\frac{-3}{4}:\frac{-9}{20}+\frac{2}{3}:\frac{-3}{10}\right]=\frac{4}{7}.\left(\frac{5}{3}+\frac{-20}{9}\right)=\frac{4}{7}.\frac{-5}{9}=\frac{-20}{63}\)

E=\(\frac{2}{-5}.\left[\frac{3}{2}:\left(\frac{-3}{4}+\frac{3}{5}\right)+\frac{2}{3}:\left(\frac{-7}{5}+\frac{4}{3}\right)\right]\)

E=\(\frac{2}{-5}.\left[\frac{3}{2}:\frac{-3}{20}+\frac{2}{3}:\frac{-1}{15}\right]=\frac{-2}{5}.\left[\left(-10\right)+\left(-10\right)\right]=\frac{-2}{5}.\left(-20\right)=8\)

F=)tương tự mấy câu trên)

\(3x-2=3x-6+4=3\left(x-2\right)+4⋮\left(x-2\right)\Leftrightarrow4⋮\left(x-2\right)\)

\(\Leftrightarrow x-2\inƯ\left(4\right)=\left\{-4,-2,-1,1,2,4\right\}\)

\(\Leftrightarrow x\in\left\{-2,0,1,3,4,6\right\}\).

Hình...đâu bạn..?

Tất cả các bộ ba điểm thẳng hàng là: A, B, C;  A, B, D;  A, C, D và  B, C, D.

D=\(\frac{-21}{9}-\frac{4}{11}.\frac{-7}{2}+\frac{2}{3}.\left(\frac{-5}{2}\right)\)

=\(\left(-3\right)-\frac{-14}{11}+\frac{-5}{3}=\frac{-99}{33}-\frac{-42}{33}+\frac{-55}{33}=\frac{-112}{33}\)

E=\(\frac{5}{36}:\frac{-1}{6}+\frac{7}{12}.\frac{5}{-2}-\frac{11}{3}.\frac{-1}{4}\)

=\(\frac{-5}{6}+\frac{-35}{24}-\frac{-11}{12}=\frac{-20}{24}+\frac{-35}{24}-\frac{-22}{24}=\frac{-33}{24}=\frac{-11}{8}\)

G=\(\frac{-3}{4}+\frac{-1}{3}.\left(\frac{1}{2}-\frac{5}{18}\right)\)

=\(\frac{-3}{4}+\frac{-1}{3}.\frac{2}{9}=\frac{-3}{4}+\frac{-2}{27}=\frac{-89}{108}\)

I=\(\frac{-19}{9}.\left(2-\frac{4}{-11}-\frac{-2}{3}\right)\)

=\(\frac{-19}{9}.\left(\frac{26}{11}-\frac{-2}{3}\right)=\frac{-19}{9}.\frac{100}{33}=\frac{-1900}{297}\)

J=\(\frac{-5}{6}:\left(\frac{1}{2}.\frac{3}{5}+\frac{-7}{12}-\frac{1}{3}\right)\)

=\(\frac{-5}{6}:\left(\frac{3}{10}+\frac{-11}{12}\right)=\frac{-5}{6}:\frac{-37}{60}=\frac{-5}{6}.\frac{60}{-37}=\frac{50}{37}\)

K=\(\frac{-3}{4}:\left(\frac{11}{3}.\frac{-9}{2}-\frac{5}{18}.\frac{-9}{3}\right)\)

=\(-\frac{3}{4}:\left(-\frac{33}{2}-\frac{-5}{6}\right)=\frac{-3}{4}:\frac{-47}{3}=\frac{-3}{4}.\frac{3}{-47}=\frac{9}{188}\)

1152-(374+1152)+(-65+374)

=1152-374-1152-65+374

=(1152-1152)+(374-374)-65

=0+0-65

=-65

thank you!