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![](https://rs.olm.vn/images/avt/0.png?1311)
e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)
= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)
= \(\dfrac{2x-6}{2x\left(x+3\right)}\)
= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)
c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
![](https://rs.olm.vn/images/avt/0.png?1311)
2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}-\dfrac{3a+1}{1-a^2}\right):\dfrac{2a+1}{a^2-1}\left(dk:a\ne1,a\ne-1\right)\)
\(=-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a+1}{a^2-1}\right):\dfrac{2a+1}{\left(a-1\right)\left(a+1\right)}\\ =-\left(\dfrac{\left(a-1\right)^2-a\left(a+1\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\dfrac{a^2-2a+1-a^2-a+3a+1}{\left(a-1\right)\left(a+1\right)}.\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\)
\(=-\dfrac{2}{2a+1}\)
a: \(=\dfrac{a}{9\left(a-2\right)}-\dfrac{b-1}{6b}-\dfrac{ab-3a+6}{9b\left(a-2\right)}\)
\(=\dfrac{2ab}{18b\left(a-2\right)}-\dfrac{3\left(b-1\right)\left(a-2\right)}{18b\left(a-2\right)}-\dfrac{2ab-6a+12}{18b\left(a-2\right)}\)
\(=\dfrac{2ab-3\left(ba-2b-a+2\right)-2ab+6a-12}{18b\left(a-2\right)}\)
\(=\dfrac{6a-12-3ab+6b+3a-6}{18b\left(a-2\right)}\)
\(=\dfrac{3a+12b-3ab-18}{18b\left(a-2\right)}\)
\(=\dfrac{a+4b-ab-6}{6b\left(a-2\right)}\)
b: \(=\dfrac{xa-2x+ax+a-x\left(a-2\right)}{a\left(a-2\right)}\)
\(=\dfrac{2ax-2x+a-xa+2x}{a\left(a-2\right)}=\dfrac{xa+a}{a\left(a-2\right)}=\dfrac{x+1}{a-2}\)