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

Tick nha

\(\dfrac{2a+\sqrt{ab}-3b}{2a-5\sqrt{ab}+3b}=\dfrac{2a-2\sqrt{ab}+3\sqrt{ab}-3b}{2a-2\sqrt{ab}-3\sqrt{ab}+3b}\)

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

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

\(=\dfrac{2\sqrt{a}+3\sqrt{b}}{2\sqrt{a}-3\sqrt{b}}\)

\(ĐK:a,b\ge0;a\ne b.\dfrac{2a+\sqrt{ab}-3b}{2a-5\sqrt{ab}+3b}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(2\sqrt{a}+3\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(2\sqrt{a}-3\sqrt{b}\right)}=\dfrac{2\sqrt{a}+3\sqrt{b}}{2\sqrt{a}-3\sqrt{b}}\)

M \(=\) \(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)

đề là gì

a) \(ab+2a-b=7\)

<=> \(a\left(b+2\right)-\left(b+2\right)=5\)

<=> \(\left(a-1\right)\left(b+2\right)=5\)

Vậy có các cặp số nguyên ( a; b ) \(\in\){ ( -4; -3) , ( 0; -7) , ( 2; 3) , ( 6; -1) }

b) \(ab-2a+3b=-5\)

<=> \(\left(ab-2a\right)+\left(3b-6\right)=-5-6\)

<=> \(a\left(b-2\right)+3\left(b-2\right)=-11\)

<=> \(\left(b-2\right)\left(a+3\right)=-11\)

Kẻ bảng rồi làm. Hoặc chia các trường hợp

c) \(2ab-3a+b=10\)

<=> \(4ab-6a+2b=20\)( nhân cả hai vế với 2)

<=> \(2a\left(2b-3\right)+\left(2b-3\right)=20-3\)

<=> \(\left(2a+1\right)\left(2b-3\right)=17\)

Làm tiếp ....