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`(a^2-b^2)/(a^2b + ab^2) = ((a-b)(a+b))/(ab(a+b)) = (a-b)/(ab)`.
`(a-b)/(ab) = ((a-b)(a+b))/(ab(a+b)) = (a^2-b^2)/(ab(a+b))`
a) \(\dfrac{3ac}{a^3b}=\dfrac{3c}{a^2b}\)
\(\dfrac{6c}{2a^2b}=\dfrac{3c}{a^2b}\)
\(\Rightarrow\dfrac{3ac}{a^3b}=\dfrac{6c}{2a^2b}\)
b) \(\dfrac{3ab-3b^2}{6b^2}=\dfrac{3b\left(a-b\right)}{6b^2}=\dfrac{a-b}{2b}\left(dpcm\right)\)
`a, (3ac)/(a^3b) = (3c)/(a^2b)`
`(6c)/(2a^2b) = (3c)/(a^2b)`
Vậy hai phân thức `=` nhau
`b, (3ab-3b^2)/(6b^2) = (3b(a-b))/(6b^2) = (a-b)/(2b)`
Vậy hai phân thức `=` nhau
\(+\) Rút gọn \(A,B\)
\(A=\left(\dfrac{1}{3}a-\dfrac{1}{3}b\right)-\left(a-2b\right)\)
\(=\dfrac{1}{3}a-\dfrac{1}{3}b-a+2b\)
\(=\left(\dfrac{1}{3}a-a\right)+\left(2b-\dfrac{1}{3}b\right)\)
\(=-\dfrac{2}{3}a+\dfrac{5}{3}b\)
\(B=\dfrac{1}{3}a-\dfrac{1}{3}b-\left(a-b\right)\)
\(=\dfrac{1}{3}a-\dfrac{1}{3}b-a+b\)
\(=\left(\dfrac{1}{3}a-a\right)+\left(b-\dfrac{1}{3}b\right)\)
\(=-\dfrac{2}{3}a+\dfrac{2}{3}b\)
\(+\) Tính \(A+B\)
\(A+B=-\dfrac{2}{3}a+\dfrac{5}{3}b-\dfrac{2}{3}a+\dfrac{2}{3}b\)
\(=\left(-\dfrac{2}{3}a-\dfrac{2}{3}a\right)+\left(\dfrac{5}{3}b+\dfrac{2}{3}b\right)\)
\(=-\dfrac{4}{3}a+\dfrac{7}{3}b\)
\(+\) Tính \(A-B\)
\(A-B=\left(-\dfrac{2}{3}a+\dfrac{5}{3}b\right)-\left(-\dfrac{2}{3}a+\dfrac{2}{3}b\right)\)
\(=-\dfrac{2}{3}a+\dfrac{5}{3}b+\dfrac{2}{3}a-\dfrac{2}{3}b\)
\(=\left(-\dfrac{2}{3}a+\dfrac{2}{3}a\right)+\left(\dfrac{5}{3}b-\dfrac{2}{3}b\right)\)
\(=0+\dfrac{3}{3}b\)
\(=b\)
\(A=\left(\dfrac{1}{3}a-\dfrac{1}{3}b\right)-\left(a-2b\right)\\ A=\dfrac{1}{3}a-\dfrac{1}{3}b-a+2b\\ A=-\dfrac{2}{3}a+\dfrac{5}{3}b\\ \\ \)
\(B=\dfrac{1}{3}a-\dfrac{1}{3}b-\left(a-b\right)\\ B=\dfrac{1}{3}a-\dfrac{1}{3}b-a+b\\ B=-\dfrac{2}{3}a+\dfrac{2}{3}b\\ B=\dfrac{2}{3}\left(-a+b\right)
\)
\(A+B=-\dfrac{2}{3}a+\dfrac{5}{3}b+\left(-\dfrac{2}{3}a+\dfrac{2}{3}b\right)\\ A+B=-\dfrac{2}{3}a+\dfrac{5}{3}b-\dfrac{2}{3}a+\dfrac{2}{3}b\\ A+B=-\dfrac{4}{3}a+\dfrac{7}{3}b\)
\(A-B=-\dfrac{2}{3}a+\dfrac{5}{3}b-\left(-\dfrac{2}{3}a+\dfrac{2}{3}b\right)\\
A-B=-\dfrac{2}{3}a+\dfrac{5}{3}b+\dfrac{2}{3}a-\dfrac{2}{3}b\\
A-B=\dfrac{5}{3}b-\dfrac{2}{3}b\\
A-B=b\)
Bài 3:
\(C=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)
Lời giải:
\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)
Lời giải:
\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)
a) \(\dfrac{a-1}{a+1}+\dfrac{3-a}{a+1}\)
\(=\dfrac{a-1+3-a}{a+1}\)
\(=\dfrac{2}{a+1}\)
b) \(\dfrac{b}{a-b}+\dfrac{a}{b-a}\)
\(=\dfrac{b}{a-b}+\dfrac{-a}{a-b}\)
\(=\dfrac{b-a}{a-b}\)
\(=-1\)
c) \(\dfrac{\left(a+b\right)^2}{ab}-\dfrac{\left(a-b\right)^2}{ab}\)
\(=\dfrac{\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)+\left(a-b\right)\right]}{ab}\)
\(=\dfrac{4ab}{ab}\)
\(=4\)
a: \(\dfrac{a+b}{ab}=\dfrac{a\left(a+b\right)}{a^2b}=\dfrac{a^2+ab}{a^2b}\)
\(\dfrac{a-b}{a^2}=\dfrac{ab-b^2}{a^2b}\)
b: \(A+B=\dfrac{a^2+ab+ab-b^2}{a^2b}=\dfrac{a^2+2ab-b^2}{a^2b}\)
\(A-B=\dfrac{a^2+ab-ab+b^2}{a^2b}=\dfrac{a^2+b^2}{a^2b}\)