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P = A.B = \(\dfrac{x-7}{\sqrt{x}+2}=\dfrac{\left(x-4\right)-3}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)-3}{\sqrt{x}+2}\)
\(=\sqrt{x}-2-\dfrac{3}{\sqrt{x}+2}\)
\(P\inℤ\) <=> x là số chính phương và \(\dfrac{3}{\sqrt{x}+2}\inℤ\)
mà \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\inℤ\Leftrightarrow\sqrt{x}+2=3\)
\(\Leftrightarrow x=1\) (thỏa)
Vậy x = 1 thì P \(\inℤ\)
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-x}-\frac{\sqrt{x}+2}{1-x}\)
\(\Leftrightarrow M=\frac{1}{1-\sqrt{x}}-\frac{\sqrt{x}+2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(\Leftrightarrow M=\frac{1+\sqrt{x}-\sqrt{x}-2}{1-x}\)
\(\Leftrightarrow M=\frac{-1}{1-x}\)
\(\Leftrightarrow M=\frac{1}{x-1}\)
b) Để M nhận giá trị nguyên
\(\Leftrightarrow\frac{1}{x-1}\inℤ\)
\(\Leftrightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{0;2\right\}\)
Mà \(x>0\)
Vậy để M nguyên \(\Leftrightarrow x=2\)
\(P=\frac{4\sqrt{x}+3}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(P=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}\inℤ\Leftrightarrow x+4\sqrt{x}+3⋮\sqrt{x}\)
Giải tiếp nhé sau đó thử chọn :V
\(p=\frac{4\sqrt{x}+3}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+3}{\sqrt{x}}=1+\frac{3}{\sqrt{x}}\)
Để \(x\in Z\Rightarrow P\in Z\)
\(\Rightarrow\sqrt{x}\inƯ\left(3\right)= \left\{-3;3\right\}\)
\(\Leftrightarrow x=9\left(t.mĐKXĐ\right)\)
a) Đkxđ: \(x\ne4\)
Thay x=9 vào A ta được:
\(\frac{9+3}{\sqrt{9}-2}=\frac{12}{3-2}=12\)
b)Ta có \(B=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(\Rightarrow B=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c) TA có \(\frac{4B}{A}=\frac{4\sqrt{x}}{\sqrt{x}-2}:\frac{x+3}{\sqrt{x}-2}=\frac{\left(4\sqrt{x}\right).\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(x+3\right)}\)
\(=\frac{4\sqrt{x}}{x+3}\)
Để \(\frac{4B}{A}=\frac{4\sqrt{x}}{x+3}\in Z\)thì \(x+3\inƯ\left(4\right);x=a^2\left(a\in Z\right)\)
Với \(x+3\inƯ\left(4\right)\Rightarrow x\in\left\{-5;-4;-2;\pm1;7\right\}\)mà \(x=a^2\Rightarrow x=1\left(TM\right)\)
Vậy x=1
Hok tốt!
Mình nghĩ nên bổ sung x nguyên
\(A=\frac{3\sqrt{x}-2}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)-5}{\sqrt{x}+1}=3-\frac{5}{\sqrt{x}+1}\)
Để A nguyên thì \(\frac{5}{\sqrt{x}+1}\in Z\Leftrightarrow\sqrt{x}+1\in\left\{1;5\right\}\Leftrightarrow x=0\)
Thay x = 0 vào thì ta có A âm vậy ............