Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
1) ĐKXĐ: \(x>0;x\ne4;x\ne9\)
(*lười lắm, ko chép lại đề nha :V*)
\(P=\frac{\left(2+\sqrt{x}\right)^2+\sqrt{x}\left(2-\sqrt{x}\right)+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{2\sqrt{x}-\left(\sqrt{x}+3\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}\\ =\frac{4+4\sqrt{x}+x+2\sqrt{x}-x+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\\ =\frac{4x+8\sqrt{x}}{2+\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{4x}{\sqrt{x}-3}\)
2) Để P>0 thì
\(\frac{4x}{\sqrt{x}-3}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>9\\x< 0\left(ktm\right)\end{matrix}\right.\)
Vậy với \(x>9\) thì \(P>0\).
Chúc bạn học tốt nha
.
Bạn giải thêm cho mk câu này đi
c) tìm giá trị của x để P = -1
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\frac{-x+x\sqrt{x}+6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x-\sqrt{x}-x+x\sqrt{x}+6-x-\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x\sqrt{x}-x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{\left(x-4\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\sqrt{x}-2\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(Q=\frac{\left(x+27\right)P}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\frac{\left(x+27\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\frac{x+27}{\sqrt{x}+3}\)
\(Q=\frac{x+27}{\sqrt{x}+3}\ge6\\ \Leftrightarrow\frac{x+27}{\sqrt{x}+3}-6\ge0\\ \Leftrightarrow\frac{x+27-6\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\ge0\\ \Leftrightarrow\frac{x-6\sqrt{x}+45}{\sqrt{x}+3}\ge0\)
Dễ thấy \(x-6\sqrt{x}+45=\left(\sqrt{x}-3\right)^2+36\ge36>0\forall x\ge0\)
\(\sqrt{x}+3\ge3>0\forall x\ge0\)
=> Ko có giá trị nào của x thỏa mãn yêu cầu
P/s: Nếu đề là \(x\sqrt{x}+27\)thì sẽ khác một chút :v
Bạn ơi chỗ kia phải là \(\frac{x-6\sqrt{x}+9}{\sqrt{x}+3}\)
1) ĐKXĐ: \(x\ge0;x\ne9\)
\(P=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\\ =\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\\ =\frac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\frac{1}{\sqrt{x}+1}\\ =\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\frac{-3}{\sqrt{x}+3}\)
2) Ta thấy \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\) (bạn tự biến đổi x cụ thể ra nhé, mà x hoàn toàn thỏa mãn ĐK) nên \(\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\).
Từ đây, thay \(\sqrt{x}=\sqrt{3}-1\) vào P, ta được:
\(P=\frac{-3}{\sqrt{3}-1+3}=\frac{-3}{\sqrt{3}+2}\)
3) Để \(P< \frac{-1}{2}\) thì:
\(\frac{-3}{\sqrt{x}+3}< \frac{-1}{2}=\frac{1}{-2}\\ \Leftrightarrow\sqrt{x}+3>6\\ \Leftrightarrow\sqrt{x}>3\\ \Leftrightarrow x>9\left(t/m\right)\)
Chúc bạn học tốt nha.
1.\(\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
= \(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
= \(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
= \(\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
= \(\frac{-3}{\sqrt{x}+3}\)
2. x = 4 - \(2\sqrt{3}\)
= \(\left(\sqrt{3}-1\right)^2\)
=> \(\sqrt{x}=\sqrt{3}-1\)
Thay vào P, ta có:
P = \(\frac{-3}{\sqrt{3}+2}\)
3. Để P < -1/2
=> \(\frac{-3}{\sqrt{x}+3}< \frac{-1}{2}\)
<=> \(\frac{3}{\sqrt{x}+3}>\frac{1}{2}\)
<=> \(\sqrt{x}+3< 6\)
<=> \(\sqrt{x}< 3\)
<=> x < 9
Mà x \(\ge0\)
=> \(0\le x< 9\) thì P < - 1/2
\(Q=\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(1-\frac{1}{x-1}\right)\)
\(=\frac{\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}.\frac{x}{x-1}\)
\(=\frac{\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}}{\sqrt{\left(x-2\right)^2}}.\frac{x}{x-1}\)
\(=\frac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}\)
Nếu \(x\ge2\) thì
\(Q=\frac{\sqrt{x-1}-1+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}=\frac{2x\sqrt{x-1}}{\left(x-2\right)\left(x-1\right)}=\frac{2x}{\left(x-2\right)\left(\sqrt{x-1}\right)}\)
Nếu \(x< 2\) thì \(Q=\frac{1-\sqrt{x-1}+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}=\frac{2x}{\left(x-2\right)\left(x-1\right)}\)
Cảm ơn bạn nhiều nhưng mình thấy \(1-\frac{1}{x-1}=\frac{x-2}{x-1}\) mà bạn sao lại bằng \(\frac{x}{x-1}\)được
c/ \(C'=\frac{1}{\frac{1}{3-2\sqrt{x}}}.\frac{1}{\frac{1}{\sqrt{3-2\sqrt{x}}}+1}=\frac{\sqrt{\left(3-2\sqrt{x}\right)^3}}{1+\sqrt{\left(3-2\sqrt{x}\right)}}\)
Đặt \(\sqrt{\left(3-2\sqrt{x}\right)}=a\)
\(\Rightarrow C'=\frac{a^3}{a+1}=a^2-a+1-\frac{1}{a+1}\)
Đế C' nguyên thì a + 1 là ước của 1
\(\Rightarrow a=0\)
\(\Rightarrow\sqrt{\left(3-2\sqrt{x}\right)}=0\)
\(\Rightarrow x=\frac{9}{4}\left(l\right)\)
Vậy không có x.
Không biết có nhầm chỗ nào không nữa. Lam biếng kiểm tra lại quá. You kiểm tra lại hộ nhé. Thanks
a/ \(C=\left(\frac{2\sqrt{x}}{2x-5\sqrt{x}+3}-\frac{5}{2\sqrt{x}-3}\right):\left(3+\frac{2}{1-\sqrt{x}}\right)\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\frac{5}{2\sqrt{x}-3}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-1}\right)\)
\(=\left(\frac{2\sqrt{x}-5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-1}\right)\)
\(=\frac{5-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}.\frac{\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=\frac{1}{3-2\sqrt{x}}\)
Câu b, c tự làm nhé
ĐKXĐ : \(x\ge0\)
\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2}{\left[1+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2\right]\left[1+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2\right]}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2-2\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)}{\left[1+\frac{\left(2\sqrt{x}+1\right)^2}{3}\right]\left[1+\frac{\left(2\sqrt{x}-1\right)^2}{3}\right]}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{2+\left(\frac{4\sqrt{x}}{\sqrt{3}}\right)^2-\frac{2\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{3}}{\left(\frac{4x+4\sqrt{x}+4}{3}\right)\left(\frac{4x-4\sqrt{x}+4}{3}\right)}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{2+\frac{16x}{3}-\frac{2\left(4x-1\right)}{3}}{\frac{16\left(x+1+\sqrt{x}\right)\left(x+1-\sqrt{x}\right)}{9}}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{\frac{6+16x-8x+2}{3}}{\frac{16\left(x+1\right)^2-16x}{9}}.\frac{2010}{x+1}\)
\(A=\frac{x+1}{x^2+x+1}.\frac{2010}{x+1}=\frac{2010}{x^2+x+1}\le2010\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=0\)
...
1) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\left(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}< 0\)
Dễ thấy \(\sqrt{x}+2\ge2>0\forall x\ge0\)
Nên để \(P< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
Vậy với \(0\le x< 1\)thì P<0
1) ĐKXĐ \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\\ =\left(\frac{\sqrt{x}-5\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\frac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\\ =\frac{-4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\frac{-4\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{-4}\\ =\sqrt{x}+1\)
2)
\(P=\sqrt{x}+1=\sqrt{\frac{3-\sqrt{5}}{2}}+1\\ \sqrt{\frac{6-2\sqrt{5}}{4}}+1\\ =\sqrt{\frac{5-2\cdot\sqrt{5}\cdot1+1}{4}}+1\\ =\sqrt{\frac{\left(\sqrt{5}-1\right)^2}{4}}+1\\ =\frac{\sqrt{5}-1}{2}+1\\ \frac{\sqrt{5}-1+2}{2}\\ =\frac{\sqrt{5}+1}{2}\)