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\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\sqrt{x}-3\in\left(1;4;-1;-4\right)\)
\(\Rightarrow\sqrt{x}\in\left(4;7;2;-1\right)\)
\(\Rightarrow\sqrt{x}=4\Leftrightarrow x=2\)
\(4,A=x+\sqrt{x}+1\)
\(A=\left(\sqrt{x}\right)^2+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(A=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}.\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)
Dấu "=" xảy ra khi :
\(\sqrt{x}+\frac{1}{2}=0\Leftrightarrow\sqrt{x}=-\frac{1}{2}\)
Vậy Min A = 3/4 khi căn x = -1/2
\(a,B=\left(\frac{15-\sqrt{x}}{x-25}+\frac{2}{\sqrt{x}+5}\right):\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
\(B=\left(\frac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\frac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(B=\frac{5+\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\frac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(B=\frac{1}{\sqrt{x}+1}\)
\(b,P=A.B=\frac{4\left(\sqrt{x}+1\right)}{25-x}.\frac{1}{\sqrt{x}+1}\)
\(P=\frac{4}{25-x}\)
bổ sung điều kiện cho câu b là x nguyên
\(TH1:x>25< =>P< 0\left(KTM\right)\)
\(TH2:x< 25< =>P>0\)mà x nguyên
\(\frac{4}{25-x}\le4\)
dấu "=" xảy ra khi \(x=24\)
\(< =>MAX:P=4\)
a) đk: \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
Ta có:
\(M=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(M=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\sqrt{x}}=\frac{2}{x-1}\)
b) Để M nguyên => \(\left(x-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x\in\left\{-1;0;2;3\right\}\) mà theo đk nên \(x\in\left\{2;3\right\}\)