A=\(\dfrac{5}{9}\)

\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{2a-5b}{-14}=\dfrac{a-3b}{-9}=\dfrac{4a+b}{16}=\dfrac{8a-2b}{16}\\ \Leftrightarrow A=\dfrac{-14}{-9}-\dfrac{16}{16}=\dfrac{14}{9}-1=\dfrac{5}{9}\)

A=5/9

Với \(a,b\in\mathbb{Z};a,b\ne0;a\ne3b;a\ne-5b\), ta có:

\(E=\dfrac{b\left(2a^2+10ab+a+5b\right)}{a-3b}:\dfrac{a^2b+5ab^2}{a^2-3ab}\)

\(=\dfrac{b\left[2a\left(a+5b\right)+\left(a+5b\right)\right]}{a-3b}:\dfrac{ab\left(a+5b\right)}{a\left(a-3b\right)}\)

\(=\dfrac{b\left(2a+1\right)\left(a+5b\right)}{a-3b}:\dfrac{b\left(a+5b\right)}{a-3b}\)

\(=\dfrac{b\left(2a+1\right)\left(a+5b\right)}{a-3b}\cdot\dfrac{a-3b}{b\left(a+5b\right)}\)

\(=2a+1\)

Vì \(2a+1\) là số nguyên lẻ với mọi a nguyên

nên \(E\) là số nguyên lẻ.

\(\text{#}Toru\)

Lời giải:

a)\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow4a=3b\)

Và \(4a.5=3b.5\Leftrightarrow20a=15b\Leftrightarrow\dfrac{20a}{3}=5b\)

Khi đó:

\(A=\dfrac{2a-5b}{a-3b}=\dfrac{2a-\dfrac{20}{3}a}{a-4a}=\dfrac{-\dfrac{14}{3}a}{-3a}=\dfrac{-14}{\dfrac{3}{-3}}=14\)

b) Ta có:

\(a-b=7\Leftrightarrow b=a-7\)

\(B=\dfrac{3a-b}{2a+7}+\dfrac{3b-a}{2b-7}=\dfrac{3a-\left(a-7\right)}{2a+7}+\dfrac{3\left(a-7\right)-a}{2\left(a-7\right)-7}\)

\(B=\dfrac{3a-a+7}{2a+7}+\dfrac{3a-21-a}{2a-14-7}\)

\(B=\dfrac{2a+7}{2a+7}+\dfrac{2a-21}{2a-21}=1+1=2\)

`a/b=3/5=>a=3/5b`

Thay `a=3/5b` vào `[2a-5b]/[a-3b]` có:

     `[2. 3/5b-5b]/[3/5b-3b]`

`=[6/5b-5b]/[3/5b-3b]`

`=[-19/5b]/[-12/5b]`

`=[-19/5]/[-12/5]=19/12`

\(\dfrac{2a-5b}{a-3b}=\dfrac{2\left(\dfrac{a}{b}\right)-5}{\left(\dfrac{a}{b}\right)-3}=\dfrac{2.\dfrac{3}{4}-5}{\dfrac{3}{4}-3}=\dfrac{14}{9}\)

\(a-b=11\)

\(P=\dfrac{5a-b}{4a+11}+\dfrac{5b-a}{4b-11}=\dfrac{5a-b}{4a+a-b}+\dfrac{5b-a}{4b-\left(a-b\right)}\)

\(=\dfrac{5a-b}{5a-b}+\dfrac{5b-a}{5b-a}\)

\(=2\)

Vậy...

Giải

a, 2A+3B=0 <=> \(\dfrac{10}{2m+1}+\dfrac{12}{2m-1}=0\)

<=>10(2m-1)+ 12(2m+1) =0

<=> 44m +2 =0 

<=> m=-1/22

b, AB= A+B <=> \(\dfrac{20}{\left(2m-1\right)\left(2m+1\right)}=\dfrac{5}{2m+1}+\dfrac{4}{2m-1}\)

<=> 20 = 5(2m -1) + 4(2m+1) 

<=> 20 = 18m - 1

<=> m=7/6

\(Q=\dfrac{2002}{a}+\dfrac{2017}{b}+2996a-5501b=\left(\dfrac{2002}{a}+8008a\right)+\left(\dfrac{2017}{b}+2017b\right)-\left(5012a+7518b\right)\)

\(=\left(\dfrac{2002}{a}+8008a\right)+\left(\dfrac{2017}{b}+2017b\right)-2506\left(2a+3b\right)\)

Áp dụng bất đẳng thức Cosi cho 2 số dương:

\(\left\{{}\begin{matrix}\dfrac{2002}{a}+8008\ge2\sqrt{\dfrac{2002}{a}.8008}=8008\\\dfrac{2017}{b}+2017b\ge2\sqrt{\dfrac{2017}{b}.2017b}=4034\end{matrix}\right.\)

Ta có: \(2a+3b=4\Rightarrow-\left(2a+3b\right)=-4\Leftrightarrow-2506\left(2a+3b\right)=-10024\)

\(\Rightarrow Q\ge8008+4034-10024=2018\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\)

Áp dụng bđt \(\dfrac{9}{a+b+c}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

Khi đó \(\dfrac{9.ab}{a+3b+2c}=ab.\dfrac{9}{\left(a+c\right)+\left(c+b\right)+2b}\le\dfrac{ab}{a+c}+\dfrac{ab}{c+b}+\dfrac{a}{2}\)

Tương tự và cộng theo vế suy ra \(9A\le\dfrac{3\left(a+b+c\right)}{2}=9< =>A\le1\)

Dấu "=" xảy ra khi và chỉ khi a = b = c = 2