a) Ta có :\(20< 25\Rightarrow\sqrt{20}< \sqrt{25}\Leftrightarrow2\sqrt{5}< 5\)

b) Ta có : \(\dfrac{16}{9}< 12\Rightarrow\sqrt{\dfrac{16}{9}}< \sqrt{12}\Leftrightarrow\dfrac{1}{3}\cdot\sqrt{16}< \sqrt{12}\)

\(M=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}=\dfrac{2\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

ĐKXĐ : x >= 0 , x khác 9

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{x+5\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

hơi xấu nhỉ:)

\(=\left[\dfrac{1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]\cdot\dfrac{1-\sqrt{a}}{\sqrt{a}}\)

\(=\dfrac{2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\dfrac{1-\sqrt{a}}{\sqrt{a}}=\dfrac{2}{\sqrt{a}+1}\)

Với \(a=3-2\sqrt{2}tmđk\)thay vào M ta được :

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

Ta có : \(18M=18\cdot\dfrac{2}{\sqrt{a}+1}=\dfrac{36}{\sqrt{a}+1}\)

Đặt \(\dfrac{36}{\sqrt{a}+1}=x^2\left(x\in N\cdot\right)\Rightarrow x^2\left(\sqrt{a}+1\right)=36\)

Ta lại có a².b² = (a.b)² => \(\left\{{}\begin{matrix}x^2\\\sqrt{a}+1\end{matrix}\right.\)phải là bình phương của các số tự nhiên

mà \(x^2\left(\sqrt{a}+1\right)=36\)=> Ta có các trường hợp sau :

\(\left\{{}\begin{matrix}x^2=1\\\sqrt{a}+1=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\a=1225\end{matrix}\right.\)(tm)

\(\left\{{}\begin{matrix}x^2=36\\\sqrt{a}+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\a=0\end{matrix}\right.\)(ktm)

\(\left\{{}\begin{matrix}x^2=4\\\sqrt{a}+1=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\a=64\end{matrix}\right.\)(tm)

\(\left\{{}\begin{matrix}x^2=9\\\sqrt{a}+1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\a=9\end{matrix}\right.\)(tm)

mình làm ở bài kia rồi bạn xem lại nhé 

\(a,=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

Với \(x=28-6\sqrt{3}tmđk\)thay vào P ta có :

\(P=\dfrac{\sqrt{28-6\sqrt{3}}}{28-6\sqrt{3}+\sqrt{28-6\sqrt{3}}+1}=\dfrac{\sqrt{\left(3\sqrt{3}-1\right)^2}}{29-6\sqrt{3}+\sqrt{\left(3\sqrt{3}-1\right)^2}}=\dfrac{3\sqrt{3}-1}{29-6\sqrt{3}+3\sqrt{3}-1}=\dfrac{3\sqrt{3}-1}{28-3\sqrt{3}}=\dfrac{\left(3\sqrt{3}-1\right)\left(28+3\sqrt{3}\right)}{784-27}=\dfrac{81\sqrt{3}-1}{757}\)

Hoành độ giao điểm của d₁ và d₂ là nghiệm của phương trình :

2x = -x - 3 <=> 3x = -3 <=> x = -1

Thế x = -1 vào d₁ => y = -2

=> d₁ và d₂ đồng quy tại điểm ( -1 ; -2 )

Để d₁ , d₂ , d₃ đồng quy thì d₃ phải đi qua điểm ( -1 ; -2 )

tức -2 = -m + 5 <=> m = 7