\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)

\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)

Vậy \(x\in\left\{22;23;24;...\right\}\)

\(\dfrac{????????}{????????????}\)

a) \(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

   \(\dfrac{11}{55}+\dfrac{10}{55}< \dfrac{x}{55}< \dfrac{22}{55}+\dfrac{1}{55}\)

   \(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{23}{55}\)

\(\Rightarrow\) \(x=22\)

b) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)

  \(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}< x\le\dfrac{26}{8}+\dfrac{14}{8}\)

  \(1< x\le5\)

  \(\Rightarrow\) \(x\in\) {\(2;3;4;5\)}

c) \(\dfrac{1}{3}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\)

 Ko biết làm

d) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)

   \(\dfrac{79}{15}+\dfrac{21}{15}+\dfrac{-40}{15}\le x\le\dfrac{40}{12}+\dfrac{45}{12}+\dfrac{23}{12}\)

   \(4\le x\le9\)

   \(\Rightarrow\) \(x\in\) {\(4;5;6;7;8;9\)}

ai giúp mik ik T_T

a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)

= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))

= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))

= \(\dfrac{1}{13}\) \(\times\)  \(\dfrac{317}{73}\)

= \(\dfrac{317}{949}\)

b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)

=   \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)

= \(\dfrac{199}{45}\)

c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)

=   \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)

= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)

= \(\dfrac{20696}{315}\)

d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)

= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)

= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))

= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)

= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)

e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))

= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)

=  -1 - \(\dfrac{2}{11}\)  - \(\dfrac{8}{33}\)

= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) -  \(\dfrac{8}{33}\)

= - \(\dfrac{47}{33}\)

a: =>x-3=9

=>x=12

b: =>10-x=-26

=>x=36

c: =>x:4-1=2

=>x:4=3

=>x=12

d: =>x^2=4

=>x=2 hoặc x=-2

e: =>(x-2)^2=100

=>x-2=10 hoặc x-2=-10

=>x=12 hoặc x=-8

câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)

b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)

⇒ \(6^{x+1}-6^x=6^{14}-6^{13}\)

⇒ \(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)

⇒ \(x=13\)

c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)

⇒ \(2^{x+3}-2^x=2^{13}-2^{10}\)

⇒ \(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)

⇒ \(x=10\)

d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)

⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)

⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)

⇒ \(x=13\)

câu d chưa có đóng ngoặc kìa bn

Hoàng Tử Hà

Y

Dương Bá Gia Bảo

Nguyễn Thị Diễm Quỳnh

lê thị hương giang

Vũ Huy Hoàng

Nguyễn Thị Linh Chi

Luân Đào

DƯƠNG PHAN KHÁNH DƯƠNG

Nguyễn Ngân Hòa

Nguyễn Huy Tú

Lightning Farron

m.n làm giúp mình với ak

Lời giải:
$\frac{x}{3}-\frac{1}{4}=\frac{-5}{6}$

$\frac{x}{3}=\frac{1}{4}-\frac{5}{6}=\frac{-7}{12}$

$x=\frac{-7}{12}.3=\frac{-7}{4}$

-------------------

$\frac{2x}{3}=\frac{6}{x}$ ($x\neq 0$)

$\Rightarrow 2x^2=18$

$x^2=18:2=9=(-3)^2=3^2$

$\Rightarrow x=\pm 3$

---------------------

$\frac{x-1}{14}=\frac{4}{x}$ ($x\neq 0$)

$\Rightarrow x(x-1)=14.4=56$

$x^2-x-56=0$

$(x+7)(x-8)=0$

$\Rightarrow x+7=0$ hoặc $x-8=0$

$\Leftrightarrow x=-7$ hoặc $x=8$

---------------------------

$\frac{-x}{8}=\frac{-50}{x}$ ($x\neq 0$)

$\Rightarrow -x^2=8(-50)$

$x^2=400=20^2=(-20)^2$

$\Rightarrow x=\pm 20$

a)=2

b)=1/3

a,2

b,1/3

\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{14}{30}.\dfrac{15}{32}=\dfrac{1}{2^x}\)

\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot14\cdot15}{4\cdot6\cdot8\cdot10\cdot...\cdot30\cdot32}=\dfrac{1}{2^x}\)

\(\Rightarrow\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot14\cdot15}{2\cdot4\cdot6\cdot8\cdot10\cdot...\cdot30\cdot32}=\dfrac{1}{2^{x+1}}\)

\(\Rightarrow\dfrac{1}{2^{15}\cdot32}=\dfrac{1}{2^{x+1}}\)

\(\Rightarrow2^{15}.2^5=2^{x+1}\)

\(\Rightarrow2^{20}=2^{x+1}\)

\(\Rightarrow x+1=20\Rightarrow x=19\)

Vậy x = 19.

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

\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)

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

b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)

\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)

c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=64\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)

d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)

\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)

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

ko biết =))

Bài 2:

\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)

\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)

\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)