a) \(6x^2-5x+3=2x-3x\left(3-2x\right)\)

\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)

\(\Leftrightarrow2x=-3\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(\left(3x-1\right)\left(4x+3\right)=2\left(3x-1\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(4x+1\right)=0\)

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

a)  Ta có:   \(A=\dfrac{x}{x+2}-\dfrac{2x}{x-2}+\dfrac{x^2+12}{x^2-4}\left(x\ne\pm2\right)\)

\(A=\dfrac{x\left(x-2\right)-2x\left(x+2\right)+x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(A=\dfrac{x^2-2x-2x^2-4x+x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(A=\dfrac{-6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(A=\dfrac{-6}{x+2}\)

b) Để A có giá trị nguyên thì \(x+2\inƯ\left(6\right)\)

Mà \(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Từ đó, ta có:

\(x+1=1\Leftrightarrow x=0\) ( nhận )

\(x+1=-1\Leftrightarrow x=-2\)  ( loại )

\(x+1=2\Rightarrow x=1\) ( nhận )

\(x+1=-2\Rightarrow x=-3\) ( nhận )

\(x+1=3\Rightarrow x=2\) ( loại )

\(x+1=-3\Rightarrow x=-4\) ( nhận )

\(x+1=6\Rightarrow x=5\) ( nhận )

\(x+1=-6\Rightarrow x=-7\) ( nhận )

Vậy để A nhận giá trị nguyên thì \(x\in\left\{-7;-4;-3;0;1;5\right\}\)

Ta có: (a+b+c)2=a2+b2+c2

<=>a2+b2+c2+2ab+2bc+2ca=a2+b2+c2

<=>ab+bc+ca=0

<=>ab+bc+ca/abc = 0

<=> 1/a+1/b+1/c =0

<=>1/a+1/b =-1/c (1)

<=> (1/a+1/b)³ =(-1/c)³

<=> 1/a³+3/a²b+3/ab²+1/b³ =-1/c³

<=>1/a³+3/ab(1/a+1/b)+1/b³ = -1/c³ (2)

thay (1) vào (2) ta được :

1/a³+3/abc+1/b³ = -1/c³

<=> 1/a³+1/c³+1/b³ =3/abc

tha

a. \(x\ne5\) là ĐKXĐ của biểu thức P

b. P =\(\dfrac{\left(x-5\right)^2}{x-5}\)=\(x-5\)

c. P = -1 <=> x-5 =-1 <=> x=4

\(\dfrac{x}{x-5}+\dfrac{4x}{x+5}+\dfrac{x\left(x-15\right)}{x^2-25}\)

  = \(\dfrac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{4x\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x\left(x-15\right)}{\left(x+5\right)\left(x-5\right)}\)

  = \(\dfrac{x^2+5x+4x^2-20x+x^2-15x}{\left(x-5\right)\left(x+5\right)}\)

  = \(\dfrac{6x^2-30x}{\left(x-5\right)\left(x+5\right)}\)

  = \(\dfrac{6x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\)

  = \(\dfrac{6x}{x+5}\)

\(đk:x\ne1\\ \dfrac{x^2+5x}{3x^2-6x+3}:\dfrac{7x+35}{6x-6}\\ =\dfrac{x\left(x+5\right)}{3\left(x^2-2x+1\right)}:\dfrac{7\left(x+5\right)}{6\left(x-1\right)}\\ =\dfrac{x\left(x+5\right)}{3\left(x-1\right)^2}\times\dfrac{6\left(x-1\right)}{7\left(x+5\right)}\\ =\dfrac{2x}{7\left(x-1\right)}\)

\(đk:x\ne1\)

\(\dfrac{x^2+5}{3x^2-6x+3}.\dfrac{7x+35}{6x-6}\\ =\dfrac{x^2+5}{3\left(x^2-2x+1\right)}.\dfrac{7\left(x+5\right)}{6\left(x-1\right)}\\ =\dfrac{x^2+5}{3\left(x-1\right)^2}.\dfrac{7\left(x+5\right)}{6\left(x-1\right)}\\ =\dfrac{7\left(x^2+5\right)\left(x+5\right)}{18.\left(x-1\right)^3}\)