\(=27x^3-9x+27x^3+8=8-9x\)

Bài làm

a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)

\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)

\(\Leftrightarrow6x+4=0\)

\(\Leftrightarrow x=-\frac{4}{6}\)

\(\Leftrightarrow x=-\frac{2}{3}\)

Vậy x = -2/3 là nghiệm.

@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x²-4

Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)

ĐKXĐ: \(x\ne\left\{-\dfrac{1}{3};\dfrac{1}{3};0;-\dfrac{4}{3}\right\}\)

\(M=\left(\dfrac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right):\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

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

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

a: \(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)

\(=x^4-16-x^4+9=-7\)

b: \(=27x^3-8-27x^3+6=-2\)

c: \(=\left(3x+5+2-3x\right)^2=7^2=49\)

Giúp mình với 

a.

ĐKXĐ: \(x\ge-\dfrac{5}{3}\)

\(9x^2-3x-\left(3x+5\right)-\sqrt{3x+5}=0\)

Đặt \(\sqrt{3x+5}=t\ge0\)

\(\Rightarrow9x^2-3x-t^2-t=0\)

\(\Delta=9+36\left(t^2+t\right)=\left(6t+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+6t+3}{18}=\dfrac{t+1}{3}\\x=\dfrac{3-6t-3}{18}=-\dfrac{t}{3}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+5}=3x-1\left(x\ge\dfrac{1}{3}\right)\\\sqrt{3x+5}=-3x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+5=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\\3x+5=9x^2\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

c.

ĐKXĐ: \(x\ge-5\)

\(x^2-3x+2-x-5-\sqrt{x+5}=0\)

Đặt \(\sqrt{x+5}=t\ge0\)

\(\Rightarrow-t^2-t+x^2-3x+2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1-x\left(x\le1\right)\\\sqrt{x+5}=x-2\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2-2x+1\left(x\le1\right)\\x+5=x^2-4x+4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

Lời giải:
c. 

$(x-3)(x^2+3x+9)-x^3=x^3-3^3-x^3=-27$ không phụ thuộc vào giá trị của biến

Ta có đpcm

d. 

$(3x+2)(9x^2-6x+4)-9x(3x^2+1)+9x$

$=(3x)^3+2^3-27x^3-9x+9x$

$=27x^3+8-27x^3=8$ không phụ thuộc vào giá trị của biến 

Ta có đpcm

c) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x^3\)

\(=x^3-27-x^3\)

=-27

d) Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-9x\left(3x^2+1\right)+9x\)

\(=27x^3+8-27x^3-9x+9x\)

=8

a) Sửa đề: \(A=\left(3x-2\right)\left(9x^2+6x+4\right)-3x\left(9x^2-2\right)\)

\(=27x^3-8-27x^3+6=-2\)

b: Ta có: \(B=\left(3x+5\right)^2+\left(6x+10\right)\left(2-3x\right)+\left(2-3x\right)^2\)

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

=49