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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 ...........................
Bài 1:
a) Rút gọn:
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{3-x}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right).\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{\left(3-x\right).\left(x+3\right)^2}{\left(x+3\right).\left(x-3\right).\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-\left(x-3\right).\left(x+3\right)^2}{\left(x+3\right)^2.\left(x-3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(-1+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-1.\left(x+3\right)}{x+3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-x-3}{x+3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-x-3+x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}:\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}.\frac{x+3}{3x^2}\)
\(A=\frac{-3.\left(x+3\right)}{\left(x+3\right).3x^2}\)
\(A=\frac{-1}{x^2}.\)
b) Ta có:
\(\left|x\right|=-\frac{1}{2}\)
Vì \(\left|x\right|\ge0\) \(\forall x.\)
\(\Rightarrow\left|x\right|>-\frac{1}{2}\)
\(\Rightarrow\left|x\right|\ne-\frac{1}{2}\)
\(\Rightarrow x\in\varnothing.\)
Vậy biểu thức A không có giá trị tại \(\left|x\right|=-\frac{1}{2}.\)
Chúc bạn học tốt!
Bài 1: Cho biểu thức: \(A=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
a) Rút gọn biểu thức \(A\)
\(A=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-\left(x-3\right)\left(x+3\right)\left(x+3\right)}{\left(x+3\right)\left(x+3\right)\left(x-3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(-1+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-x-3}{x+3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}:\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}\cdot\frac{x+3}{3x^2}\)
\(A=\frac{-3\left(x+3\right)}{\left(x+3\right)3x^2}\)
\(A=\frac{-1}{x^2}\)
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