\(\frac{2}{3}\) .\(\frac{3}{4}\)\(\le\)\(\frac{x}{18}\) \(\le\)\(\frac{7}{3}\).\(\frac{1}{3}\)

\(\frac{1}{2}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{9}{18}\le\frac{x}{18}\le\frac{14}{18}\)

\(\Rightarrow x\in\){9:10;11;12;13;14}

\(\frac{2}{3}.\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\)

\(\frac{2}{3}.\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\frac{1}{3}\)

\(\frac{2}{3}.\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{11}{18}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)

Vậy \(x\in\left\{11;12;13\right\}\)

\(\frac{2}{3}\cdot\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}\cdot\left(\frac{1}{2}-\frac{1}{6}\right)\)

\(\frac{2}{3}\cdot\left(\frac{2}{4}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}\cdot\left(\frac{3}{6}-\frac{1}{6}\right)\)

\(\frac{2}{3}\cdot\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}\cdot\frac{1}{3}\)

\(\frac{2}{3}\cdot\left(\frac{15}{12}-\frac{4}{12}\right)\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{2}{3}\cdot\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)

Để \(x\)phải nhỏ hơn hoặc bằng thì x lần lượt bằng \(\left\{11;12;13;14\right\}\)

1)

\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)       

=  \(\frac{7.2^8.2-5.2^8.2^2}{16^2}\)

=  \(\frac{2^8.\left(2.7-5.2^2\right)}{2^8}\)

=   \(\frac{2^8.\left(-6\right)}{2^8}\)

=   \(-6\)

2)

b)\(\frac{-28}{4}\le x\le\frac{-21}{7}\)

\(\Rightarrow-7\le x\le-3\)

\(\Rightarrow x=\left\{-7;-6;-5;-4;-3\right\}\)

2)

a)\(\frac{108}{12}\le x\le\frac{91}{7}\)

\(\Rightarrow9\le x\le13\)

\(\Rightarrow x=\left\{9;10;11;12;13\right\}\)

=> \(-\frac{70}{7}\le\frac{7x}{7}\le-\frac{11}{7}\)

=> \(-70\le7x\le-11\)

=> 7x \(\in\) {-70; -69; -68; ...;-11}

Để x nguyên thì 7x \(\in\) B(7)

=> 7x \(\in\) {-70; -63; -56; -49;-42;-35;-28;-21;-14}

=> x \(\in\) {-10; -9; -8; -7; -6;-5;-4; -3;-2}

\(\frac{-5}{3}< \frac{7}{6}-\left(x-1\right)\le\frac{11}{12}\)

\(\Leftrightarrow\frac{-20}{12}< \frac{14}{12}-\frac{12\left(x-1\right)}{12}\le\frac{11}{12}\)

\(\Leftrightarrow-20< 14-12\left(x-1\right)\le11\)

\(\Leftrightarrow-20< 14-12x+12\le11\)

\(\Leftrightarrow-20< 26-12x\le11\)

\(\Leftrightarrow26-46< 26-12x\le26-15\)

\(\Leftrightarrow-46< 12x\le-15\)

\(\Leftrightarrow x\in\left\{2;3\right\}\)

Vậy x = 2 hoặc x = 3

ta có:\(\frac{-5}{3}\)<\(\frac{7}{6}\)-(x-1)<=\(\frac{11}{12}\)

=>\(\frac{-20}{12}\)<\(\frac{14}{12}\)-(x-1)<=\(\frac{11}{12}\)

=>\(\frac{-20}{12}\)<\(\frac{14}{12}\)-x+1<=\(\frac{11}{12}\)

=>\(\frac{-20}{12}\)<\(\frac{26}{12}\)-x<=\(\frac{11}{12}\)

=>\(\frac{26}{12}\)-x={-19/12,-18.12,-17/12......,11/12}

=>x={45/12,44/12,43/12........15/12}

=>x={45/12,22/6,43/12......5/4}

Vậy.............................

mik nghĩ zậy!!!!!

x-1 ko phải x + 1 đâu bạn

\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)

\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)

2 câu sau tự làm nha

\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)

\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)

=> \(x\in\left\{-2;-1;0;1;2\right\}\)

thank nha