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\(\frac{2a+5}{a+2}+\frac{4a+6}{a+2}-\frac{3a}{a+2}=\frac{2a+5+4a+6-3a}{a+2}\)
\(=\frac{3a+11}{a+2}=\frac{3\left(a+2\right)+5}{a+2}=\frac{3\left(a+2\right)}{a+2}+\frac{5}{a+2}=3+\frac{5}{a+2}\in Z\)
\(\Rightarrow5⋮a+2\)
\(\Rightarrow a+2\inƯ\left(5\right)=\left\{1;5;-1;-5\right\}\)
\(\Rightarrow a=3\) (a nguyên dương)
\(A=\frac{2a+5}{a+2}+\frac{4a+6}{a+2}-\frac{3a}{a+2}\)\(=\frac{2a+5+4a+6-3a}{a+2}\)
\(=\frac{3a+1}{a+2}=\frac{3\left(a+2\right)+5}{a+2}=\frac{3\left(a+2\right)}{a+2}+\frac{5}{a+2}=3+\frac{5}{a+2}\in Z\)
\(\Rightarrow5⋮a+2\)
\(\Rightarrow a+2\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(\Rightarrow a=3\) (a nguyên dương)
a/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(\dfrac{2a+7b}{3a-4b}=\dfrac{2bk+7b}{3bk-4b}=\dfrac{b\left(2k+7\right)}{b\left(3k-4\right)}=\dfrac{2k+7}{3k-4}\left(1\right)\)
\(\dfrac{2c+7d}{3c-4d}=\dfrac{2dk+7d}{3dk-4d}=\dfrac{d\left(2k+7\right)}{d\left(3k-4\right)}=\dfrac{2k+7}{3k-4}\)\(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
b/ tương tự
Bài 1:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
link đây tham khảo nhé:
https://hoc24.vn/hoi-dap/question/207558.html
đặt M = \(\dfrac{2a+5}{a+2}+\dfrac{4a+6}{a+2}-\dfrac{3a}{a+2}\)
ta có:
\(M=\dfrac{2a+5}{a+2}+\dfrac{4a+6}{a+2}-\dfrac{3a}{a+2}\)
\(M=\dfrac{2a+5+4a+6-3a}{a+2}\\ M=\dfrac{6a+11-3a}{a+2}\\ M=\dfrac{3a+11}{a+2}\\ M=\dfrac{\left(3a+6\right)+5}{a+2}\\ M=\dfrac{5}{a+2}+3\)
Để M nguyên => \(\dfrac{5}{a+2}\) là số nguyên
\(\Rightarrow5⋮a+2\)
=> \(a+2\inƯ_{\left(5\right)}=\left\{\mp1,\mp5\right\}\)
ta có bảng sau:
vậy a = {-1; -3; 3; -7 } mà a là số nguyên dương nên a = 3
vậy a = 3