Chứng tỏ hai phân thức \(\dfrac{{{a^2} - {b^2}}}{{{a^2}b + a{b^2}}}\) và \(\dfrac{{a - b}}{{ab}}\) bằng nhau theo hai cách khác nhau.
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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
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: \(A+B=\dfrac{a^2+ab+ab-b^2}{a^2b}=\dfrac{a^2+2ab-b^2}{a^2b}\)
\(A-B=\dfrac{a^2+ab-ab+b^2}{a^2b}=\dfrac{a^2+b^2}{a^2b}\)
d: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{A}\)
hay A=x-2
a) a − b = a . ( − 1 ) − b . ( − 1 ) = − a b
b) − a − b = − a . ( − 1 ) − b . ( − 1 ) = a b
Ta có: \(\left(a^2-a+1\right)\left(a+2\right)\)
\(=a^3+2a^2-a^2-2a+a+2\)
\(=a^3+a^2-a+2\)(1)
Ta có: \(\left(a^2+3a-1\right)\left(a-2\right)\)
\(=a^3-2a^2+3a^2-6a-a+2\)
\(=a^3+a^2-7a+2\)(2)
Từ (1) và (2) suy ra \(\dfrac{a^2-a+1}{a-2}\ne\dfrac{a^2+3a-1}{a+2}\)
\(A=\sqrt{\dfrac{b^2\left(a-b\right)^2+a^2\left(a-b\right)^2+a^2b^2}{a^2b^2\left(a-b\right)^2}}\)
\(=\sqrt{\dfrac{b^2\left(a^2-2ab+b^2\right)+a^2\left(a^2-2ab+b^2\right)+a^2b^2}{a^2b^2\left(a-b\right)^2}}\)
\(=\sqrt{\dfrac{b^4+a^4-2ab^3-2a^3b+3a^2b^2}{a^2b^2\left(a-b\right)^2}}=\sqrt{\dfrac{\left(b^2+a^2\right)^2-2ab\left(a^2+b^2\right)+a^2b^2}{a^2b^2\left(a-b\right)^2}}\)
\(=\sqrt{\dfrac{\left(b^2+a^2-ab\right)}{a^2b^2\left(a-b\right)^2}}=\left|\dfrac{a^2+b^2-ab}{ab\left(a-b\right)}\right|\)
Do a,b là số hữu tỉ\(\Rightarrow\)\(\left|\dfrac{a^2+b^2-ab}{ab\left(a-b\right)}\right|\) là số hữu tỉ hay A là số hữu tỉ
Ta có: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
Ta có: \(P=\dfrac{ab^2}{a^2+b^2-c^2}+\dfrac{bc^2}{b^2+c^2-a^2}+\dfrac{ca^2}{c^2+a^2-b^2}\)
\(=\dfrac{ab^2}{\left(a+b\right)^2-c^2-2ab}+\dfrac{bc^2}{\left(b+c\right)^2-a^2-2bc}+\dfrac{ca^2}{\left(c+a\right)^2-b^2-2ac}\)
\(=\dfrac{ab^2}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+\dfrac{bc^2}{\left(b+c+a\right)\left(b+c-a\right)-2bc}+\dfrac{ca^2}{\left(c+a+b\right)\left(c+a-b\right)-2ac}\)
\(=\dfrac{ab^2}{-2ab}+\dfrac{bc^2}{-2bc}+\dfrac{ca^2}{-2ac}\)
\(=\dfrac{-ab\cdot b}{2ab}+\dfrac{-bc^2}{2bc}+\dfrac{-ca^2}{2ac}\)
\(=\dfrac{-b}{2}+\dfrac{-c}{2}+\dfrac{-a}{2}=\dfrac{-\left(a+b+c\right)}{2}=\dfrac{0}{2}=0\)
`(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))`