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a) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=1+\dfrac{4}{\sqrt{x}-2}\)
Để A nguyên thì 4 ⋮ √x - 2
\(\Rightarrow\sqrt{x}-2\inƯ\left(4\right)\)
\(\Rightarrow\sqrt{x}-2\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0;6;-2\right\}\)
Mà x \(\sqrt{x}\ge0\)
=> x thuộc {9; 1; 16; 0; 36}
b)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Với \(x=3\)( thỏa mãn ĐKXĐ ) ta có \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)
c) A ở đâu ???? '-'
\(A=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}ĐKXĐ:x\ge0\)
\(\Rightarrow A=\frac{2+\sqrt{2}+1}{\sqrt{2}+1}=\frac{3+\sqrt{2}}{1+\sqrt{2}}=2\sqrt{2}-1\)
\(B=\frac{1}{\sqrt{x}-1}-\frac{x+2}{x\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+1-x-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\left(1-\sqrt{x}\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{-\sqrt{x}}{x+\sqrt{x}+1}\)
\(C=-A.B=-\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\cdot\frac{-\sqrt{x}}{x+\sqrt{x}+1}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1-1}{\sqrt{x}+1}=1-\frac{1}{\sqrt{x}+1}\)
\(ĐểC\in Z\Rightarrow\frac{1}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\sqrt{x}+1\in\left\{1\right\}=\left\{1;-1\right\}\)
\(\Leftrightarrow x\in\left\{0\right\}\)
Kết hợp ĐKXĐ =>...