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câu a) mẫu chung: \(x^3x^2\left(x-1\right)^2\) hơi dài nên mk ko làm ^_^
b)
\(=\frac{xy\left(a-b\right)+\left(x-a\right)\left(y-a\right)b-\left(x-b\right)\left(y-b\right)a}{ab\left(a-b\right)}\)
\(=\frac{xya-xyb+bxy-abx-aby+a^2b-axy+axb+aby-ab^2}{ab\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)}{ab\left(a-b\right)}=1\)
\(b.=\frac{1\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{1c-1a+1a-1b+1b-1c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{2b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Bài 1:
a: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4}{x+1}\)
b: \(=\dfrac{xy\left(x^2+y^2\right)}{x^4y}\cdot\dfrac{1}{x^2+y^2}=\dfrac{x}{x^4}=\dfrac{1}{x^3}\)
c: Đề thiếu rồi bạn
\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(x\left(x+1\right)+x\left(x-3\right)=4x\)
\(x^2+x+x^2-3x=4x\)
\(2x^2-2x=4x\)
\(2x^2-2x-4x=0\)
\(2x\left(x-3\right)=0\)
\(2x=0\Leftrightarrow x=0\)
hoặc
\(x-3=0\Leftrightarrow x=3\)
b) \(ĐKXĐ:x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)
\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)
\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)
\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )
Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)
e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=\dfrac{0}{2\left(x-y\right)\left(x+y\right)}=0\)
\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)
\(=\frac{-2}{x^2}\)
\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)
\(=x\left(x-3\right)\)
\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{x+1}\)
# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha