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a: ĐKXĐ: x<>1; x<>-1
\(A=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x-1}\)
\(=\dfrac{x+1}{x-1}-\dfrac{1}{x-1}=\dfrac{x}{x-1}\)
b: x^2+3x+2=0
=>x=-1(loại) hoặc x=-2(nhận)
Khi x=-2 thì A=-2/(-3)=2/3
a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
a) ĐKXĐ: \(x\ne-10;x\ne0;x\ne-5\)
b) \(P=\dfrac{x^2+2x}{2x+20}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+2x}{2\left(x+10\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+2x\right)\left(x+5\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+10\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{\left(50-5x\right)\left(x+10\right)}{2x\left(x+5\right)\left(x+10\right)}\)
\(=\dfrac{x^4+7x^3+10x^2+2x^2+10x-100+500-5x^2}{2x\left(x+10\right)\left(x+5\right)}\)
\(=\dfrac{x^4+7x^3+7x^2+10x+400}{2x\left(x+10\right)\left(x+5\right)}\)
c) \(P=0\Rightarrow x^4+7x^3+7x^2+10x+400=0\Leftrightarrow...\)
Số xấu thì câu c, d làm cũng như không. Bạn xem lại đề.
\(M=\dfrac{\left|x+1\right|+2x}{3x^2-2x-1}\)
a) ĐKXĐ : \(x\ne-\dfrac{1}{3};1\).
Rút gọn :
+) Khi \(x\ge-1\) :
\(M=\dfrac{x+1+2x}{\left(x+\dfrac{1}{3}\right)\cdot\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{3x+1}{\left(x+\dfrac{1}{3}\right)\cdot\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{3\cdot\left(x+\dfrac{1}{3}\right)}{\left(x+\dfrac{1}{3}\right)\cdot\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{3}{x-1}\).
+) Khi \(x< -1\) :
\(M=\dfrac{-1-x+2x}{\left(x+\dfrac{1}{3}\right)\cdot\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{x-1}{\left(x+\dfrac{1}{3}\right)\cdot\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{1}{x+\dfrac{1}{3}}\)
b) pt \(4x^2+5x-9=0\) có hai nghiệm \(\left[{}\begin{matrix}x=1\left(KTMĐK\right)\\x=-\dfrac{9}{4}\left(TMĐK\right)\end{matrix}\right.\)
Mà \(-\dfrac{9}{4}< -1\).
=> Giá trị của M : \(M=\dfrac{1}{-\dfrac{9}{4}+\dfrac{1}{3}}=-\dfrac{12}{23}\)