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\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
Lời giải:
a.
\(B=\frac{2\sqrt{x}(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-2x}{(\sqrt{x}+3)(\sqrt{x}-3)}=\frac{x-3\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}=\frac{\sqrt{x}}{\sqrt{x}+3}\)
b.
\(P=AB=\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}+3}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
Để $P<0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+3}<0$
Mà $\sqrt{x}+3>0$ nên $\sqrt{x}-2<0$
$\Leftrightarrow 0< x< 4$
Kết hợp với ĐKXĐ suy ra $0< x< 4$
Mà $x$ nguyên nên $x\in left\{1; 2; 3\right\}$
a: \(A=\dfrac{\sqrt{3}+1}{\sqrt{3}+1}+\sqrt{5}+3-3-\sqrt{5}=1\)
b: \(B=\dfrac{-\sqrt{x}-3+x-3\sqrt{x}-x-9}{x-9}=\dfrac{-4\sqrt{x}-12}{x-9}=\dfrac{-4}{\sqrt{x}-3}\)
Để B>1 thì \(\dfrac{-4-\sqrt{x}+3}{\sqrt{x}-3}>0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay 0<x<9
a: \(=\dfrac{x-3\sqrt{x}-x-9}{x-9}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
b: Để A>-1 thì A+1>0
=>\(-3\sqrt{x}+2\sqrt{x}+4>0\)
=>-căn x>-4
=>0<x<16
a, Rút gọn P
\(\dfrac{3}{\sqrt{x}+3}-\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2-\sqrt{x}}{\sqrt{x}+3}\right)\)
\(\Leftrightarrow\left(1-\dfrac{\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{x+3\sqrt{x}-2\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{-\left(\sqrt{x}-2\right)\sqrt{x}+3}\right)\)
\(\Leftrightarrow\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+3}\right):\left(\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2-\sqrt{x}}{\sqrt{x}+3}\right)\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2-\sqrt{x}}{\sqrt{x}+3}\right)\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{\left(\sqrt{x}+3\right).\left(3-\sqrt{x}\right).\left(x+\sqrt{3}\right).\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right).\left(2-\sqrt{x}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)+x-9-\left(2\sqrt{x}-x-4+2\sqrt{x}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{9-x+x-9-\left(4\sqrt{x}-x-4\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{-4\sqrt{x}+x+4}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{\left(\sqrt[]{x}-2\right)^2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}.\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(\Leftrightarrow3.\dfrac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow\)\(\dfrac{3}{\sqrt{x}-2}\)
Ta có :
a , \(M=2\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\left[\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(M=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}-\dfrac{2\left(x+9\right)}{x-9}\right]:\left[\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(M=\left(\dfrac{2x-6\sqrt{x}-2x-18}{x-9}\right).\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\right]\)
\(M=\dfrac{-6\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(2\sqrt{x}+4\right)}\)
\(M=\dfrac{-6\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(M=-\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b , mik ko chắc chắn nên mik chưa làm nhé !