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đkxd: \(x\ne\left\{\pm3\right\}\)
a) B= \(\frac{21+\left(x-4\right)\left(x+3\right)-\left(x+1\right)\left(x-3\right)}{x^2-9}:\left(\frac{x+3-1}{x+3}\right)\)
=\(\frac{21+x^2-x-12-x^2+2x+3}{x^2-9}.\frac{x+3}{x+2}\)
=\(\frac{x+12}{x-3}\)
b)|2x+1|=5
<=> \(\left[\begin{array}{nghiempt}2x+1=-5\\2x+1=5\end{array}\right.\)<=> x=-3 hoặc x=2
với x=-3 thì B=\(\frac{-3}{2}\)
với x=2 thì B=-14
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
Đề sai sửa luôn !
\(a,M=\left(\frac{21}{x^2-9}+\frac{4-x}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21-\left(4-x\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+3-1}{x+3}\right)\)
\(=\frac{21-4x-12+x^2+3x-x^2+3x+x-3}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)
\(=\frac{3x+6}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{3}{x-3}\)
\(b,x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Kết hợp ĐKXĐ => x = 2
Thay vào \(M=\frac{3}{2-3}=\frac{3}{-1}=-3\)
Vậy ...........................
biết đề ghê vậy :D ?!