Bài 1: Tìm x

a) Ta có: \(\dfrac{4}{3}:0.8=\dfrac{2}{3}:\left(0.1\cdot x\right)\)

\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}:\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}\cdot\dfrac{5}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow x\cdot\dfrac{1}{10}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{3}\cdot\dfrac{3}{5}=\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{2}{5}:\dfrac{1}{10}=\dfrac{2}{5}\cdot10=\dfrac{20}{5}=4\)

Vậy: x=4

b) Ta có: \(\left|x\right|=-1.2\)

mà \(\left|x\right|\ge0\forall x\)

nên \(x\in\varnothing\)

Vậy: \(x\in\varnothing\)

Bài 2: Tính

a) Ta có: \(\left(-2.5\right)\cdot\left(-4\right)\cdot\left(-7.9\right)\)

\(=\left(2.5\cdot4\right)\cdot\left(-7.9\right)\)

\(=-7.9\cdot10=-79\)

b) Ta có: \(\left(-0.375\right)\cdot\dfrac{13}{3}\cdot\left(-2\right)^3\)

\(=\dfrac{3}{8}\cdot8\cdot\dfrac{13}{3}\)

\(=3\cdot\dfrac{13}{3}=13\)

\(=\left[\dfrac{1}{4}-\dfrac{6}{5}\right]-\dfrac{4}{5}+\dfrac{3}{4}\)

\(=\dfrac{1}{4}-\dfrac{6}{5}-\dfrac{4}{5}+\dfrac{3}{4}\)

=1/4+3/4-6/5-4/5

=1-2

=-1

\(\left[\left(\dfrac{-1}{2}\right)^2-1,2\right]-\left(0,8-\dfrac{3}{4}\right)\)

\(=\left[\dfrac{1}{4}-\dfrac{6}{5}\right]-\left(\dfrac{4}{5}-\dfrac{3}{4}\right)\)

\(=\dfrac{1}{4}-\dfrac{6}{5}-\dfrac{4}{5}+\dfrac{3}{4}\)

\(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)-\left(\dfrac{6}{5}+\dfrac{4}{5}\right)\)

\(=1-2\)

\(=-1\)

a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)

b: =>x-1/2=1/3

=>x=5/6

c: =>2/3x-1=0 hoặc 3/4x+1/2=0

=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3

d =>4/9:x=10/3:9/4=10/3*4/9=40/27

=>x=4/9:40/27=4/9*27/40=108/360=3/10

$\frac{2}{3}x-1\frac{2}{5}=\frac{1}{2}x+0,8\left(3\right)$

$\frac{2}{3}x-1\frac{2}{5}=\frac{1}{2}x+\mathrm{\backslash (\backslash dfrac\{5\}\{6\}\backslash )}$

$\mathrm{=>\; \backslash (\backslash dfrac\{2\}\{3\}x-\backslash dfrac\{1\}\{2\}x=\backslash dfrac\{5\}\{6\}+\backslash dfrac\{7\}\{5\}\backslash )}$

=> \(\dfrac{1}{6}x=\dfrac{63}{30}\)

=> \(x=\dfrac{63}{30}:\dfrac{1}{6}\)

=> \(x=\dfrac{67}{5}\)

Vậy : x= 67 / 5

\(\dfrac{2}{3}x-1\dfrac{2}{5}=\dfrac{1}{2}x+0,8\left(3\right)\)

\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{2}x=0,8\left(3\right)+1\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{67}{30}\)

\(\Leftrightarrow x=\dfrac{67}{5}\)

Vậy \(x=\dfrac{67}{5}\)

\(\dfrac{x-2}{-1,2}=\dfrac{-5}{2}\Rightarrow x=\dfrac{-5.\left(-1,2\right)}{2}+2=\dfrac{6}{2}+2=3+2=5\\ \dfrac{-6}{x+1}=\dfrac{1,8}{9}\Rightarrow x=\dfrac{-6.9}{1,8}-1=\dfrac{-54}{1,8}-1=-30-1=-31\\ \dfrac{-3}{x}=\dfrac{x}{-12}\Rightarrow x=\sqrt{\left(-12\right).\left(-3\right)}=\sqrt{36}=\sqrt{\left(\pm6\right)^2}=\pm6\)

\(\dfrac{x-4}{x-1}=\dfrac{3}{5}\\ \Rightarrow5\left(x-4\right)=3\left(x-1\right)\\ \Leftrightarrow5x-20=3x-3\\ \Leftrightarrow5x-3x=-3+20\\ \Leftrightarrow2x=17\\ \Leftrightarrow x=\dfrac{17}{2}\\ ---\\ \dfrac{1,12}{-10}=\dfrac{11,2}{x}\Rightarrow x=\dfrac{11,2.\left(-10\right)}{1,12}=\dfrac{10.1,12.\left(-10\right)}{1,12}=-100\)

\(\left|2\dfrac{1}{5}-x\right|\)\(+\left|x-\dfrac{1}{5}\right|\)\(+8\dfrac{1}{5}\)\(=1,2\)

\(\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=\dfrac{6}{5}-\dfrac{41}{5}\)

\(\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=\dfrac{-36}{5}\) (vô lý vì \(\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|\ge0\))

Vậy: Không tìm được giá trị x thoả mãn.

a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)

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

\(\Leftrightarrow-7x=\dfrac{10}{3}\)

hay \(x=-\dfrac{10}{21}\)

b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)

\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)

\(\Leftrightarrow85x+85=84x+168\)

\(\Leftrightarrow x=83\)

a) 15/11 - (5/7 - 18/11) + 27/7

= 15/11 - 5/7 + 18/11 + 27/7

= (15/11 + 18/11) + (-5/7 + 27/7)

= 3 + 22/7

= 43/7

b) 39/5 + (9/4 - 9/5) - (5/4 + 1,2)

= 39/5 + 9/4 - 9/5 - 5/4 - 6/5

= (39/5 - 9/5 - 6/5) + (9/4 - 5/4)

= 24/5 + 1

= 29/5

c) -1,2 - 0,8 + 0,25 + 5,75 - 2022

= (-1,2 - 0,8) + (0,25 + 5,76) - 2022

= -2 + 6 - 2022

= 4 - 2022

= -2018

d) 0,1 + 16/9 + 5,1 + (-20/9)

= (0,1 + 5,1) + (16/9 - 20/9)

= 5,2 - 4/9

= 419/90

a) \(\dfrac{15}{11}-\left(\dfrac{5}{7}-\dfrac{18}{11}\right)+\dfrac{27}{7}=\dfrac{22}{7}+3=\dfrac{43}{77}\)

b) \(\dfrac{39}{5}+\left(\dfrac{9}{4}-\dfrac{9}{5}\right)-\left(\dfrac{5}{4}+\dfrac{6}{5}\right)=\dfrac{24}{5}+1=\dfrac{29}{5}\)

c) \(-1,2-0,8+0,25+5,75-2022=-2+6-2022=-2018\)

d) \(0,1+\dfrac{16}{9}+5,1+\dfrac{-20}{9}=\dfrac{26}{5}-\dfrac{4}{9}=\dfrac{214}{45}\)

\(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:\left(0,1\right).x\)

\(\Rightarrow\dfrac{4}{3}:\dfrac{4}{5}=\dfrac{2}{3}:\dfrac{1}{10}.x\)

\(\Rightarrow\dfrac{4}{3}.\dfrac{5}{4}=\dfrac{2}{3}.10x\)

\(\Rightarrow\dfrac{5}{3}=\dfrac{2}{3}.10x\)

\(\Rightarrow10x=\dfrac{5}{3}:\dfrac{2}{3}\)

\(\Rightarrow10x=\dfrac{5}{2}\)

\(\Rightarrow x=\dfrac{5}{2}:10=\dfrac{1}{4}\)

Vậy \(x=\dfrac{1}{4}\)

\(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:0,1x\Rightarrow\dfrac{4}{3}:\dfrac{3}{5}=\dfrac{2}{3}:\dfrac{1}{10}x\)

\(\Rightarrow\dfrac{4}{3}.\dfrac{5}{3}=\dfrac{2}{3}.10x\Rightarrow\dfrac{20}{9}=\dfrac{20}{3}x\Rightarrow x=\dfrac{1}{3}\)

Vậy...................................................