Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 2:
a: \(\dfrac{25}{5-2\sqrt{3}}=\dfrac{125+10\sqrt{3}}{13}\)
b: \(\dfrac{8}{\sqrt{5}+2}=8\sqrt{5}-32\)
c: \(\dfrac{6}{2\sqrt{3}-\sqrt{7}}=\dfrac{12\sqrt{3}+6\sqrt{7}}{5}\)
d: \(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}=\dfrac{\sqrt{6}}{2}\)
1) \(x\ge\frac{1}{6}\)
2.\(x\le0\)
3.\(4-5x\ge0\Leftrightarrow x\le\frac{4}{5}\)
4.mọi x
1)
a) \(6=\sqrt{36}< \sqrt{40}\)
b) \(3=\sqrt{9}< \sqrt{10}\)
c) \(2\sqrt{3}< 2\sqrt{4}=4\)
d) \(3\sqrt{2}=\sqrt{18}< \sqrt{36}=6\)
e) \(7=\sqrt{49}< \sqrt{50}\)
2)
a) \(x\ge0\)
b) \(-2x+1\ge0\Leftrightarrow-2x\ge-1\Leftrightarrow x\le\dfrac{1}{2}\)
c) \(5-a\ge0\Leftrightarrow a\le5\)
d) \(2x-3>0\Leftrightarrow2x>3\Leftrightarrow x>\dfrac{3}{2}\)
e) \(-3< x< 1\)
f) \(-3x\ge-4\Leftrightarrow x\le\dfrac{4}{3}\)
g) \(x^2-2x-3\ge0\Leftrightarrow\left(x+1\right)\left(x-3\right)\ge0\Leftrightarrow-1\le x\le3\)
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
1)\(\sqrt{3-2\sqrt{2}}-\sqrt{2}=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{2}=\sqrt{2}-1-\sqrt{2}=-1\left(đpcm\right)\)
2) \(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}-1=-2\)
3) \(ĐK:\)\(\left\{{}\begin{matrix}\dfrac{x-1}{x+3}\ge0\\x+3\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\\x\ne-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x< -3\end{matrix}\right.\)
4) \(ĐK:\left\{{}\begin{matrix}7-x\ge0\\a\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le7\\a\ge0\end{matrix}\right.\)