các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha

a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)

              -\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)

              -6 < \(x\) < -4

             vì \(x\) \(\in\) Z nên \(x\) = -5

- \(\dfrac{19}{6}+\dfrac{-15}{2}+\dfrac{11}{3}\) < \(x\) < - \(\dfrac{5}{4}\) + \(\dfrac{19}{12}\) - \(\dfrac{10}{3}\)

- \(\dfrac{32}{3}+\dfrac{11}{3}< x< \) \(\dfrac{1}{3}\) - \(\dfrac{10}{3}\)

-7 < \(x\) < - 3

Vì \(x\in\) Z nên \(x\in\) {-6;-5;-4}

\(-\frac{1}{10}< =x< =\frac{3}{5}\) 

\(\frac{-4}{9}< x< =\frac{2}{3}\)

Đoàn Thị Thanh Ngân viết lời giải hộ thanks

bn tính từng vế thì thành như thế mà

đây bạn

a) \(x\cdot3\dfrac{1}{4}+\left(-\dfrac{7}{6}\right)\cdot x-1\dfrac{2}{3}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{3}{4}x-\dfrac{7}{6}x-\dfrac{2}{3}=\dfrac{5}{12}\)

\(\Leftrightarrow9x-14x-8=5\)

\(\Leftrightarrow-5x-8=5\)

\(\Leftrightarrow-5x=5+8\)

\(\Leftrightarrow-5x=13\)

\(\Rightarrow x=-\dfrac{13}{5}\)

Vậy \(x=-\dfrac{13}{5}\)

b) \(5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Rightarrow5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\left(đk:x\ne0\right)\)

\(\Leftrightarrow\dfrac{93}{17}\cdot\dfrac{1}{x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{93}{17x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{93}{17x}+2x-\dfrac{3}{4}=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}\ge0\right)\\\dfrac{93}{17x}-\left(2x-\dfrac{3}{4}\right)=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}< 0\right)\end{matrix}\right.\)

đến đây bạn giải tiếp nhé

c) \(\left(x+\dfrac{1}{2}\right)\cdot\left(\dfrac{2}{3}-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0-\dfrac{1}{2}\\2x=0+\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}:2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{1}{2};x_2=\dfrac{1}{3}\)

Bài 1:

a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)

b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)

c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)

d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)

e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)

hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)

dad

A = ( -4/5 + 4/3 ) + (-5/4 + 14/5) - 7/3

  =   8/15 + 31/20 - 7/3

  =  25/12 - 7/3

  = -1/4

B = 8/3 x 2/5 x 3/8 x 10x 19/92

   = 16/15 x 15/4 x 19/92

   = 4x19/92

   = 19/23

C = - \(\dfrac{5}{7}\) x \(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\) x \(\dfrac{9}{14}\) + \(\dfrac{1}{57}\)

   = - \(\dfrac{10}{77}\) - \(\dfrac{45}{98}\) + \(\dfrac{1}{57}\)

  = - \(\dfrac{635}{1078}\) + \(\dfrac{1}{57}\)

 = - \(\dfrac{36195}{61446}\) + \(\dfrac{1078}{61446}\)

 = - \(\dfrac{35117}{61446}\)

D = \(\dfrac{12}{19}\) x \(\dfrac{7}{15}\) x \(\dfrac{-13}{17}\) x \(\dfrac{19}{12}\)  x \(\dfrac{17}{13}\)

D =  (\(\dfrac{12}{19}\) x \(\dfrac{19}{12}\)) x (- \(\dfrac{13}{17}\) x \(\dfrac{17}{13}\)) x \(\dfrac{7}{15}\)

D =   1  x  (- 1) x \(\dfrac{7}{15}\)

D = -1 x \(\dfrac{7}{15}\)

D = - \(\dfrac{7}{15}\)

Bài 46:

11: Ta có: \(-4\left|x-2\right|=-8\)

\(\Leftrightarrow\left|x-2\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: x∈{0;4}

12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)

\(\Leftrightarrow5\left|x+2\right|=20\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy: x∈{-6;2}

13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)

\(\Leftrightarrow6\left|x-2\right|=-6\)(1)

Ta có: \(\left|x-2\right|\ge0\forall x\)

\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)

Ta có: -6<0(3)

Từ (1), (2) và (3) suy ra x∈∅

Vậy: x∈∅

14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)

\(\Leftrightarrow-7\left|x+4\right|=-7\)

\(\Leftrightarrow\left|x+4\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: x∈{-5;-3}

15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)

\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)

\(\Leftrightarrow4\left|x+1\right|=24\)

\(\Leftrightarrow\left|x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Vậy: x∈{-7;5}

16: Ta có: \(3\left|x+5\right|=-9\)(4)

Ta có: |x+5|≥0∀x

⇒3|x+5|≥0∀x(5)

Ta có: -9<0(6)

Từ (4), (5) và (6) suy ra x∈∅

Vậy: x∈∅

17: Ta có: \(-8\left|x-3\right|=24-16:2\)

\(\Leftrightarrow-8\left|x-3\right|=16\)

\(\Leftrightarrow\left|x-3\right|=-2\)

mà |x-3|≥0>-2∀x

nên x∈∅

Vậy: x∈∅

18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)

\(\Leftrightarrow-3\left|x+6\right|=3\)

\(\Leftrightarrow\left|x+6\right|=-1\)

mà |x+6|≥0>-1∀x

nên x∈∅

Vậy: x∈∅

19: Ta có: \(5-\left|x+7\right|=4\)

\(\Leftrightarrow\left|x+7\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)

Vậy: x∈{-8;-6}

20: Ta có: \(12-\left|x+8\right|=10\)

\(\Leftrightarrow\left|x+8\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)

Vậy: x∈{-10;-6}

dap an bang 170

dap an bang 170

(17 x 12 + 17 x 20 - 17 x 22 ) x ( \(\frac{3}{2}-\frac{4}{6}+\frac{1}{6}\) )

= 17 x ( 12 + 20 - 22 ) x ( \(\frac{9}{6}-\frac{4}{6}+\frac{1}{6}\))

= 17 x 10 x 1

= 170