tách nhỏ ra

\(\dfrac{6}{8}-\dfrac{2}{7}=\dfrac{42}{56}-\dfrac{16}{56}=\dfrac{26}{56}=\dfrac{13}{28}\)

b) Ta có: \(x^3-x^2y-xy^2+y^3\)

\(=\left(x^3+y^3\right)-\left(x^2y+xy^2\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)

\(=\left(x+y\right)\left(x-y\right)^2\)

\(\dfrac{x+2}{x-3}+\dfrac{x-2}{x}=\dfrac{x^2+2x+6}{x\left(x-3\right)}\)  đkxđ: x khác 3 , x khác 0

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-3\right)}+\dfrac{\left(x-2\right)\left(x-3\right)}{x\left(x-3\right)}-\dfrac{x^2+2x+6}{x\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+2x}{....}+\dfrac{x^2-3x-2x+6}{.....}-\dfrac{x^2+2x+6}{...}=0\)

\(\Leftrightarrow x^2+2x+x^2-3x-2x+6-x^2-2x-6=0\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

đk : x khác -3 ; 3

\(\Rightarrow\left(x+3\right)^2-4x^2=\left(x-3\right)^2\Leftrightarrow-3x^2+6x+9=x^2-6x+9\)

\(\Leftrightarrow4x^2-12x=0\Leftrightarrow4x\left(x-3\right)=0\Leftrightarrow x=3\left(ktm\right);x=0\)

\(\dfrac{x+3}{x-3}-\dfrac{4x^2}{x^2-9}=\dfrac{x-3}{x+3}\) đkxđ: x khác 3 ; x khác -3

\(\Leftrightarrow\dfrac{x+3}{x-3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+6x+9}{...}-\dfrac{4x^2}{.....}-\dfrac{x^2-6x+9}{....}=0\)

\(\Leftrightarrow x^2+6x+9-4x^2-x^2+6x-9=0\)

\(\Leftrightarrow-4x^2+12x=0\)

\(\Leftrightarrow-4x\left(x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=3\left(kotm\right)\end{matrix}\right.\)

Vậy tập nghiệm của pt có S thuộc \(\left\{0\right\}\)

=>2x=2/3

hay x=1/3

\(\dfrac{9}{7}\times x+\dfrac{5}{7}\times x=\dfrac{2}{3}\)

\(\dfrac{14}{7}\times x=\dfrac{2}{3}\)

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

2x=2/3

hay x=1/3

a: =5/6(1/8+2/3)=5/6*19/24=95/144

b: =3/4(7/5-1/2)=27/40

c: =35/24*2/3=35/36

x+1/6=x+8/3

\(\Leftrightarrow\)x-x=8/3-1/6

\(\Leftrightarrow\)0x=15/6 (vô nghiệm)

\(x+\dfrac{1}{6}=x+\dfrac{8}{3}\)

\(x-x=\dfrac{8}{3}-\dfrac{1}{6}=\dfrac{15}{6}\)(vô nghiệm)

`(x+1)/6=(x+8)/3`

`<=> 3(x+1)=6(x+8)`

`<=> 3x+3=6x+48`

`<=> 3x-6x=48-3`

`<=> -3x=45`

`<=> x=-15`

\(\Leftrightarrow2x^3-3x^2+x+a=\left(x+3\right)\cdot a\left(x\right)\)

Thay \(x=-3\)

\(\Leftrightarrow2\left(-27\right)-3\cdot9-3+a=0\\ \Leftrightarrow-54-27-3+a=0\\ \Leftrightarrow-84+a=0\\ \Leftrightarrow a=84\)

x= 13/21

\(-\left|1,7-x\right|-\dfrac{5}{3}=\dfrac{2}{3}\\ \Rightarrow\left|1,7-x\right|=-\dfrac{5}{3}-\dfrac{2}{3}=-\dfrac{7}{3}\left(l\right)\)

Vậy không có giá trị x thoả mãn