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

\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{7}{3}-\dfrac{3}{4}=2\\x\cdot\dfrac{7}{3}-\dfrac{3}{4}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{33}{28}\\x=-\dfrac{15}{28}\end{matrix}\right.\)

b: \(\Leftrightarrow\left|x\cdot\dfrac{2}{3}-\dfrac{1}{3}\right|=\dfrac{6}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{6}{5}\\x\cdot\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-13}{10}\\x=\dfrac{23}{10}\end{matrix}\right.\)

a)\(=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)

b)\(=\dfrac{5}{9}\times\dfrac{3}{2}=\dfrac{15}{18}=\dfrac{5}{6}\)

d)\(=\left(\dfrac{12}{8}-\dfrac{3}{8}\right)\times2=\dfrac{9}{8}\times2=\dfrac{18}{8}=\dfrac{9}{4}\)

c)\(=\dfrac{4}{3}-\dfrac{5}{6}=\dfrac{8}{6}-\dfrac{5}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

a) 1 + 4/3 = 7/3

b) 5/9 : 2/3 = 5/6

c ) 4/3 -1/3 x 5/2

= 1 x 5/2

= 5/2

d) ( 3/2 - 3/8) : 1/2

= 9/8 : 1/2

= 9/4

e) 15/16 : 3/8 x 3/4 

= 5/2 x 3/4

= 15/8

f) 7/19 x 1/3 x 7/19 x 2/3

= 7/19 x (1/3 x 2/3)

= 7/19 x 2/9

= 14/171

g) 3/5 x 8/27 x 25/3

= 3/5 x 25/3 x 8/27 

= 5 x 8/27

= 40/27

h) 1/5 + 4/11 + 4/5 + 7/11

= (1/5 + 4/5) + (4/11 + 7/11)

= 1 + 1

= 2

e)\(\dfrac{15}{16}\times\dfrac{8}{3}\times\dfrac{3}{4}=\dfrac{360}{192}=\dfrac{15}{8}\)

f) \(=\dfrac{7}{19}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{7}{19}\times\dfrac{3}{3}=\dfrac{7}{19}\)

g)\(=\dfrac{600}{120}=\dfrac{600:120}{120:120}=\dfrac{5}{1}=5\)

h)\(=\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\left(\dfrac{4}{11}+\dfrac{7}{11}\right)=\dfrac{5}{5}+\dfrac{11}{11}=1+1=2\)

Yêu cầu đề bài là gì bạn nên ghi đầy đủ ra nhé.

Yêu cầu đề bài là gì bạn nên ghi đầy đủ ra nhé. Và nên viết đề bằng công thức toán để mọi người hiểu đề của bạn hơn.

a: ĐKXĐ: \(a\notin\left\{3;-3\right\}\)

\(\dfrac{a}{a-3}-\dfrac{3}{a+3}\)

\(=\dfrac{a\left(a+3\right)-3\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}\)

\(=\dfrac{a^2+3a-3a+9}{\left(a-3\right)\left(a+3\right)}=\dfrac{a^2+9}{a^2-9}\)

b: ĐKXĐ: x<>0

\(\dfrac{1}{2x}+\dfrac{2}{x^2}\)

\(=\dfrac{x+2\cdot2}{2x^2}=\dfrac{x+4}{2x^2}\)

c: ĐKXĐ: \(x\notin\left\{1;-1\right\};b< >0\)

\(\dfrac{4}{x^2-1}-\dfrac{2}{3b}\)

\(=\dfrac{4\cdot3b-2\left(x^2-1\right)}{3b\left(x^2-1\right)}\)

\(=\dfrac{12b-2x^2+2}{3b\left(x^2-1\right)}\)

có cần giải chi tiết k bạn

a: \(\dfrac{7}{x}-\dfrac{2}{6}-2>\dfrac{2\left(x+1\right)}{3}\)

=>\(\dfrac{7}{x}>\dfrac{2x+2}{3}+\dfrac{1}{3}+2=\dfrac{2x+2+1+6}{3}=\dfrac{2x+9}{3}\)

=>\(\dfrac{7}{x}-\dfrac{2x+9}{3}>0\)

=>\(\dfrac{21-2x^2-9x}{3x}>0\)

=>\(\dfrac{2x^2+9x-21}{x}< 0\)

TH1: x>0 và 2x^2+9x-21<0

=>x>0 và \(\dfrac{-9-\sqrt{249}}{4}< x< \dfrac{-9+\sqrt{249}}{4}\)

=>\(0< x< \dfrac{-9+\sqrt{249}}{4}\)

TH2: x<0 và 2x^2+9x-21>0

=>x<0 và (x<-9-căn 249)/4 hoặc (x>(-9+căn 249)/4))

=>x<(-9-căn 249)/4

b: =>x-4=-4

hay x=0

`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`

`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`