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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^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))`
\(a,đk\left(B\right):x\ne\pm3\\ B=\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\\ =\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}\\ =\dfrac{3\left(x+3\right)+6x+x\left(x-3\right)}{x^2-9}\\ =\dfrac{3x+9+6x+x^2-3x}{x^2-9}\\ =\dfrac{x^2+6x+9}{x^2-9}\\ =\dfrac{\left(x+3\right)^2}{x^2-9}\\ =\dfrac{x+3}{x-3}\)
\(b,P=A.B\\ =\dfrac{x+1}{x+3}\times\dfrac{x+3}{x-3}\\ =\dfrac{x+1}{x-3}\)
\(c,\) Để P nguyên
\(\dfrac{x+1}{x-3}=1+\dfrac{4}{x-3}\)
=> \(x-3\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{-1;1;2;-2;4;-4\right\}\)
\(=>x=\left\{2;4;5;1;7;-1\right\}\)
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:
\(C=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)
a) Điều phải chứng minh tương đương với:
\(a^3+b^3-a^2b-b^2a\ge0\\ \Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0\\ \Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\\ \Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\left(luon.dung\right)\)
Dấu = xảy ra khi a=b
b) Áp dụng bất đẳng thức ở phần a ta có:
\(\dfrac{1}{a^3+b^3+1}\le\dfrac{1}{a^2b+b^2a+abc}=\dfrac{1}{ab\left(a+b+c\right)}\\ =\dfrac{abc}{ab\left(a+b+c\right)}=\dfrac{c}{a+b+c}\left(do.abc=1\right)\)
Tương tự : \(\dfrac{1}{b^3+c^3+1}\le\dfrac{a}{a+b+c};\dfrac{1}{c^3+a^3+1}\le\dfrac{b}{a+b+c}\)
\(\Rightarrow P\le\dfrac{a+b+c}{a+b+c}=1\)
Dấu = xảy ra <=> a=b=c=1
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: \(A+B=\dfrac{a^2+ab+ab-b^2}{a^2b}=\dfrac{a^2+2ab-b^2}{a^2b}\)
\(A-B=\dfrac{a^2+ab-ab+b^2}{a^2b}=\dfrac{a^2+b^2}{a^2b}\)