a. \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x-2}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b. Ta có \(x=11-6\sqrt{2}=\left(3-\sqrt{2}\right)^2\)
Thay \(x=\left(3-\sqrt{2}\right)^2\) vào \(M\), ta được
\(M=\dfrac{\sqrt{\left(3-\sqrt{2}\right)^2}+1}{\sqrt{\left(3-\sqrt{2}\right)^2}-3}=\dfrac{3-\sqrt{2}+1}{3-\sqrt{2}-3}=\dfrac{4+\sqrt{2}}{-\sqrt{2}}=\dfrac{\sqrt{2}\left(2\sqrt{2}+1\right)}{-\sqrt{2}}=1-2\sqrt{2}\)c. Ta có \(M=2\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=2\Leftrightarrow\sqrt{x}+1=2\sqrt{x}-6\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\)
d. \(M< 1\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}< 1\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\)
Vì \(4>0\) nên để \(\dfrac{4}{\sqrt{x}-3}< 0\) thì \(\sqrt{x}-3< 0\Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)
e. Ta có \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
Để \(M\) nguyên thì \(4⋮\left(\sqrt{x}-3\right)\Rightarrow\left(\sqrt{x}-3\right)\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng giá trị
| \(\sqrt{x}-3\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(4\) | \(-4\) |
| \(x\) | \(16\) | \(4\) | \(25\) | \(1\) | \(49\) | x |
Kết hợp với đk, ta có tất cả cá x thỏa ycbt là x=16;x=25,x=1 hoặc x=49
a: Ta có: \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b: Thay \(x=11-6\sqrt{2}\) vào M, ta được
\(M=\dfrac{3-\sqrt{2}+1}{3-\sqrt{2}-3}=\dfrac{4-\sqrt{2}}{-\sqrt{2}}=-2\sqrt{2}+1\)
c: Ta có: M=2
nên \(\sqrt{x}+1=2\sqrt{x}-6\)
hay x=49
