Câu 5: 

\(\Leftrightarrow-x^2+7x-9+2x-9=0\)

\(\Leftrightarrow x^2-9x+18=0\)

=>x=3

=>Chọn A

bài nào zậy bạn

Câu 3 và caau4 bài giải phương trình nhé

Bài 3. Đặt ẩn phụ là 

\(a=2x-\frac{5}{x}\\\)

\(b=x-\frac{1}{x}\)

pt <=> \(b-a=\sqrt{a}-\sqrt{b}\\ \)

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

\(\left(\sqrt{b}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{a}+1\right)=0\)

tới đây xét 2 TH bạn tự giải nhé

\(b,B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\left(x\ge0;x\ne4;x\ne9\right)\\ B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

\(c,B< A\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}< \dfrac{\sqrt{x}+1}{\sqrt{x}-2}\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 0\\ \Leftrightarrow\dfrac{-5}{\sqrt{x}-2}< 0\Leftrightarrow\sqrt{x}-2>0\left(-5< 0\right)\\ \Leftrightarrow x>4\\ d,P=\dfrac{B}{A}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}:\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}-4}{\sqrt{x}+1}=1-\dfrac{5}{\sqrt{x}+1}\in Z\\ \Leftrightarrow5⋮\sqrt{x}+1\Leftrightarrow\sqrt{x}+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-2;0;4\right\}\\ \Leftrightarrow x\in\left\{0;16\right\}\left(\sqrt{x}\ge0\right)\)

\(e,P=1-\dfrac{5}{\sqrt{x}+1}\)

Ta có \(\sqrt{x}+1\ge1,\forall x\Leftrightarrow\dfrac{5}{\sqrt{x}+1}\ge5\Leftrightarrow1-\dfrac{5}{\sqrt{x}+1}\le-4\)

\(P_{max}=-4\Leftrightarrow x=0\)