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a) Ta có A = 2 x + 1 x : 2 x − 1 x = 2 x + 1 2 x − 1
b) Ta có B = a + 2 a − 2 : a 2 + 4 a + 4 a 2 + 2 a + 4 = a + 2 a − 2 . a 2 + 2 a + 4 ( a + 2 ) 2 = a 2 + 2 a + 4 a 2 − 4
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\(B=\frac{1+\frac{2}{x-1}}{1+\frac{2x}{x^2+1}}\)
\(B=\left(1+\frac{2}{x-1}\right):\left(1+\frac{2x}{x^2+1}\right)\)
\(=\left(\frac{x-1}{x-1}+\frac{2}{x-1}\right):\left(\frac{x^2+1}{x^2+1}+\frac{2x}{x^2+1}\right)\)
\(=\frac{x-1+2}{x-1}:\frac{x^2+1+2x}{x^2+1}\)
\(=\frac{x+1}{x-1}:\frac{\left(x+1\right)^2}{x^2+1}\)
\(=\frac{x+1}{x-1}.\frac{x^2+1}{\left(x+1\right)^2}\)
\(=\frac{x^2+1}{\left(x-1\right)\left(x+1\right)}\)
Chúc bạn học tốt !!!
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1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
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Ta có:
\(\dfrac{x^2-4}{x+1}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{x+1}\)
Và:
\(\dfrac{x+2}{2x}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{2x\left(x-2\right)}\)
Vậy ta đã biến đổi hai phân thức đó để chúng bằng phân thức cũ và có tủ bằng nhau
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\(A=\dfrac{\dfrac{x}{x-1}-\dfrac{x+1}{x}}{\dfrac{x}{x+1}-\dfrac{x-1}{x}}=\dfrac{\dfrac{x^2-\left(x^2-1\right)}{x\left(x-1\right)}}{\dfrac{x^2-\left(x^2-1\right)}{x\left(x+1\right)}}=\dfrac{\dfrac{1}{x\left(x-1\right)}}{\dfrac{1}{x\left(x+1\right)}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{0;\pm1\right\}\\A=\dfrac{x+1}{x-1}\end{matrix}\right.\)