Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐK \(\hept{\begin{cases}x\ge0\\x\ne4;x\ne9\end{cases}}\)
a. P=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right):\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+2+\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
b. Với \(x=4-2\sqrt{3}\Rightarrow P=\frac{\sqrt{4-2\sqrt{3}}+1}{4-2\sqrt{3}-4}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+1}{-2\sqrt{3}}\)
\(=\frac{\sqrt{3}-1+1}{-2\sqrt{3}}=-\frac{1}{2}\)
c. Để \(\frac{1}{P}\le\frac{-5}{2}\Leftrightarrow\frac{x-4}{\sqrt{x}+1}+\frac{5}{2}\le0\Leftrightarrow\frac{2x-8+5\sqrt{x}+5}{2\left(\sqrt{x}+1\right)}\le0\)
\(\Leftrightarrow\frac{2x+5\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}\le0\Leftrightarrow2x+5\sqrt{x}-3\le0\)vì \(2\left(\sqrt{x}+1\right)\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\le0\Leftrightarrow2\sqrt{x}-1\le0\Leftrightarrow0\le x\le\frac{1}{4}\left(tm\right)\)
Vậy với \(0\le x\le\frac{1}{4}\)thì \(\frac{1}{P}\le-\frac{5}{2}\)
d. Ta có \(B=P\left(\sqrt{x}-2\right)=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+2}=1-\frac{1}{\sqrt{x}+2}\)
Gỉa sử \(B\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(1\right)\Leftrightarrow\sqrt{x}+2\in\left\{-1;1\right\}\Leftrightarrow x\in\left\{\phi\right\}\)
Vậy B không nhận giá trị nguyên với mọi x để P có nghĩa
Lời giải:
a)
Để $M$ có nghĩa thì \(x-3\geq 0\Leftrightarrow x\geq 3\)
b)
\(M=\sqrt{x-2-2\sqrt{x-3}}-\sqrt{x+1-4\sqrt{x-3}}\)
\(=\sqrt{(x-3)-2\sqrt{x-3}+1}-\sqrt{(x-3)-4\sqrt{x-3}+4}\)
\(=\sqrt{(\sqrt{x-3}-1)^2}-\sqrt{(\sqrt{x-3}-2)^2}\)
\(=|\sqrt{x-3}-1|-|\sqrt{x-3}-2|\)
Với \(3\leq x\leq 4\Rightarrow 0\leq \sqrt{x-3}\leq 1\)
\(\Rightarrow \sqrt{x-3}-1\leq 0; \sqrt{x-3}-2< 0\)
\(\Rightarrow M=|\sqrt{x-3}-1|-|\sqrt{x-3}-2|=(1-\sqrt{x-3})-(2-\sqrt{x-3})=-1\)
Lời giải:
a)
Để $M$ có nghĩa thì \(x-3\geq 0\Leftrightarrow x\geq 3\)
b)
\(M=\sqrt{x-2-2\sqrt{x-3}}-\sqrt{x+1-4\sqrt{x-3}}\)
\(=\sqrt{(x-3)-2\sqrt{x-3}+1}-\sqrt{(x-3)-4\sqrt{x-3}+4}\)
\(=\sqrt{(\sqrt{x-3}-1)^2}-\sqrt{(\sqrt{x-3}-2)^2}\)
\(=|\sqrt{x-3}-1|-|\sqrt{x-3}-2|\)
Với \(3\leq x\leq 4\Rightarrow 0\leq \sqrt{x-3}\leq 1\)
\(\Rightarrow \sqrt{x-3}-1\leq 0; \sqrt{x-3}-2< 0\)
\(\Rightarrow M=|\sqrt{x-3}-1|-|\sqrt{x-3}-2|=(1-\sqrt{x-3})-(2-\sqrt{x-3})=-1\)
a) ĐKXĐ: \(\hept{\begin{cases}x-9\ne0\\\sqrt{x}\ge0\\\sqrt{x}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ge0\\x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne9\\x>0\end{cases}}}\)
\(A=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{x+\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}+3}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{x-9}\)
b) \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow x=\sqrt{4+4\sqrt{2}+2}-\sqrt{2+2\sqrt{2}+1}\)
\(\Leftrightarrow x=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(\Leftrightarrow x=\left|2+\sqrt{2}\right|-\left|\sqrt{2}+1\right|\)
\(\Leftrightarrow x=2+\sqrt{2}-\sqrt{2}-1=1\left(TM\right)\)
Vậy với x= 1 thì giá trị của biểu thức \(A=\frac{\left(1+1\right)\left(1-3\right)}{1-9}=\frac{2.\left(-2\right)}{-8}=\frac{-4}{-8}=\frac{1}{2}\)
c)
Ta có :
\(\frac{x-9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)
+) \(\frac{1}{A}\)nguyên
\(\Leftrightarrow1+\frac{2}{\sqrt{x}+1}\)nguyên
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)
\(\Leftrightarrow x=1\)
Vậy ..............
a)Dễ thấy: \(M=\sqrt{\left(\sqrt{x-3}-1\right)^2}+\sqrt{\left(\sqrt{x-3}-2\right)^2}\)
\(\Rightarrow M\)có nghĩa\(\Leftrightarrow x-3\ge0\Leftrightarrow x\ge3\)
b) với \(3\le x\le4\)M xác định
\(3\le x\le4\Rightarrow\sqrt{x-3}\le1\)
\(\Rightarrow M=\left|\sqrt{x-3}-1\right|+\left|\sqrt{x-3}-2\right|=1-\sqrt{x-3}+2-\sqrt{x-3}=3-2\sqrt{x-3}\)