\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\1-\frac{1}{x+3}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\pm3\\x\ne-2\end{cases}}}\)

a ) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right):\left(1-\frac{1}{x+3}\right)\)

\(=\frac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+3-1}{x+3}\)

\(=\frac{21+x^2-x-12-\left(x^2-4x+3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+2}{x+3}\)

\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)

\(=\frac{3.\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}\)

\(=\frac{3}{x-3}\) 

b ) \(B=-\frac{3}{5}\Leftrightarrow\frac{3}{x-3}=-\frac{3}{5}\)

\(\Leftrightarrow x-3=-5\Leftrightarrow x=-2\) ( do \(x\ne\pm3;x\ne-2\) ) 

c ) \(B< 0\Leftrightarrow\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow\) \(\hept{\begin{cases}x< 3\\x\ne-2\\x\ne-3\end{cases}}\)

không bạn nha

x²+2>0 r

x(x²+2)=0

=> x=0

hai pt trên không tương đương

a) \(\frac{3}{7}x-1=\frac{1}{7}x\left(3x-7\right)\)

<=> \(3x-7=x\left(3x-7\right)\)

<=> \(\left(3x-7\right)-x\left(3x-7\right)=0\)

<=> \(\left(3x-7\right)\left(1-x\right)=0\)

<=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x=1\end{cases}}\)

Vậy S = { 7/3; 1}

b) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

<=> \(\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

<=> \(\left(3x-1\right)\left(x^2-7x+12\right)=0\)

<=> \(\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

<=> \(\left(3x-1\right)\left(x\left(x-3\right)-4\left(x-3\right)\right)=0\)

<=> \(\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

<=> x = 1/3 hoặc x = 3 hoặc x = 4.

Vậy S = { 1/3; 3; 4}

\(a,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+1\right)\left(x-3\right)-\left(x-1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(7-5x\right)=0\)

\(\Leftrightarrow x-1=0;x+2=0\)hoặc \(7-5x=0\)

\(\Leftrightarrow x=1;x=-2\)hoặc \(x=\frac{7}{5}\)

KL....

\(b,\left(5x^2-2x+10\right)^2=\left(x^2+10x-8\right)^2\)

\(\Leftrightarrow\left(5x^2-2x+10\right)^2-\left(x^2+10x-8\right)^2=0\)

\(\Leftrightarrow\left(5x^2-2x+10-x^2-10x+8\right)\left(5x^2-2x+10+x^2+10x-8\right)=0\)

\(\Leftrightarrow\left(4x^2-12x+18\right)\left(6x^2+8x+2\right)=0\)

\(\Leftrightarrow\left(x^2-3x+\frac{9}{2}\right)\left(6x^2+6x+2x+2\right)=0\)

\(\Leftrightarrow\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{9}{4}\right)\left(6x+2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[\left(x-\frac{3}{2}\right)^2+\frac{9}{4}\right]\left(3x+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=-1\end{cases}}\)Vì \(\left(x-\frac{3}{2}\right)^2+\frac{9}{4}>0\forall x\)

Vậy ..