\(\dfrac{1}{2.2}.\dfrac{2}{2.3}.....\dfrac{31}{64}=2^x\\ =>\dfrac{1}{2.2.2.....2.64}=2^x\\ \dfrac{1}{2^{30}.26}=2^x\\ =>\dfrac{1}{2^{36}}=2^x\\ =>2^{-36}=2^x\\ =>x=-36\)

\(n=-36\)

Ta có:  \(2^n=\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}....\dfrac{30}{62}.\dfrac{31}{64}\)

⇔ \(2^n=\dfrac{1.2.3.4....31}{2.\left(2.3.4.....1\right).64}=\)

⇔ \(2^n=\dfrac{1}{2}.\dfrac{1}{64}=\dfrac{1}{128}\)  \(\Leftrightarrow\)   \(2^n=\dfrac{1}{2^6}\)

⇔ \(2^{x+6}=1\)

⇔ \(x+6=0\)

⇒  \(\left\{{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Trả lời

Trả lời:

a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)

hay \(x=-\dfrac{17}{21}\)

Vậy: \(x=-\dfrac{17}{21}\)

b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)

\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)

Vậy: \(x=\dfrac{4}{5}\)

c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)

\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)

\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)

hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)

Vậy: \(x=-\dfrac{5}{7}\)

f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)

\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)

\(\Leftrightarrow-x-\dfrac{9}{60}=0\)

\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)

hay \(x=-\dfrac{3}{20}\)

Vậy: \(x=-\dfrac{3}{20}\)

g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)

a) \(x=0,05\)

b) \(x=1,125\)

c) \(x=0,96\)

d) \(x=0,025\)

Bạn tự làm đi dễ mà . Cố mag vận động đầu óc đừng copy làm bài nữa khó lắm mới hỏi thôi

Chuẩn đấy

a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)

Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)

Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)

Suy ra \(-15\le x\le-2\), x ϵ Z

b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)

Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)

Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)

Suy ra \(-1\le x\le0\), x ϵ Z

b, -4\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) < \(x\) < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))

   - \(\dfrac{13}{3}\).\(\dfrac{1}{3}\) < \(x\) < - \(\dfrac{2}{3}\).(-\(\dfrac{11}{12}\))

    - \(\dfrac{13}{9}\) < \(x\) < \(\dfrac{11}{18}\)

     \(x\) \(\in\) { -1; 0; 1}

a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)

  - \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) <   \(x\)   < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)

   -  \(\dfrac{46}{3}\)     <  \(x\) < - \(\dfrac{13}{7}\) 

          \(x\) \(\in\) {-15; -14;-13;..; -2}

\(\dfrac{1}{2}+\dfrac{-1}{3}+\dfrac{-2}{3}\le x< \dfrac{-3}{5}+\dfrac{1}{6}+\dfrac{-2}{5}+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{1}{2}+\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)\le x< \left(\dfrac{-3}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{1}{6}+\dfrac{3}{2}\right)\)

\(\Leftrightarrow\dfrac{1}{2}+\left(-1\right)\le x< -1+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{-1}{2}\le x< \dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{-3}{6}\le x< \dfrac{4}{6}\)

\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3\right\}\)

⇔x∈{−3;−2;−1;0;1;2;3}

`[-5]/6+8/3+[-29]/6 <= x <= [-1]/2+2+5/2`

`[-5]/6+16/6+[-29]/6 <= [6x]/6 <= [-3]/6+12/6+15/6`

`-5+16-29 <= 6x <= -3+12+15`

`-18 <= 6x <= 24`

`-18:6 <= 6x:6 <= 24:6`

`-3 <= x <= 4`

Vậy `-3 <= x <= 4`

\(=>-3\le x\le4\)

\(=>x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)

`=>` Ta có STN:

`<=> -3 <= x <= 4`

`<=> x = {-3; -2; -1; 0; 1; 2; 3; 4}`

`<=> x in Z`

1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)

=\(0\)

mink chịu bài này nó rất khó