Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, ĐKXĐ : \(x> 0 ; x \neq 1 \)
P = \(\dfrac{3x+3\sqrt{x} - 3}{\sqrt{x^2} +2\sqrt{x} - \sqrt{x} - 2}\) \(- \dfrac{\sqrt{x}+1}{\sqrt{x}+2} + \dfrac{\sqrt{x}-2}{\sqrt{x}} . \dfrac{1-( 1 -\sqrt{x})}{1-\sqrt{x}}\)
= \(\dfrac{3x+3\sqrt{x} - 3 }{\sqrt{x}(\sqrt{x}+2)-(\sqrt{x} - 2)}\) \(- \dfrac{\sqrt{x}+1}{\sqrt{x}+2} + \dfrac{\sqrt{x}-2}{\sqrt{x}}. \dfrac{ 1-1+\sqrt{x}}{1-\sqrt{x}}\)
= \(\dfrac{3x+3\sqrt{x} - 3 }{(\sqrt{x}+2)(\sqrt{x}-1)}\) \(- \dfrac{\sqrt{x}+1}{\sqrt{x}+2} + \dfrac{\sqrt{x}-2}{(\sqrt{x}-1)} \)
= \(\dfrac{3x+3\sqrt{x}-3-(\sqrt{x}-1)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
= \(\dfrac{3x+3\sqrt{x}-3-(\sqrt{x^2}- 1^2) - (\sqrt {x^2}-2^2)}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
= \(\dfrac{3x+3\sqrt{x} - 3 - x+1-x+4}{(\sqrt{x}+2)(\sqrt{x}-1)} \)
= \(\dfrac{x+3\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x} - 1)}\)
= \(\dfrac{\sqrt{x^2}+2\sqrt{x} +\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x} - 1)} \)
= \(\dfrac{\sqrt{x}(\sqrt{x}+2)+(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x} - 1)} \)
= \(\dfrac{(\sqrt{x}+2)(\sqrt{x}+1)}{(\sqrt{x}+2)(\sqrt{x} - 1)} \)
= \(\dfrac{\sqrt{x}+1}{\sqrt{x} - 1} \)
c, Để P = \(\sqrt{x}\) \(\Leftrightarrow\) \(\dfrac{\sqrt{x}+1}{\sqrt{x} - 1} \) = \(\sqrt{x} \)
\(\Rightarrow\) \(\sqrt{x}+1= \sqrt{x}(\sqrt{x}-1)\)
\(\Leftrightarrow\) \(\sqrt{x}+1 = \sqrt{x^2} - \sqrt{x}\)
\(\Leftrightarrow\) \( \sqrt{x^2} -\sqrt{x} - \sqrt{x} - 1 = 0\)
\(\Leftrightarrow\) \(\sqrt{x^2} - 2\sqrt{x} +1-1-1=0\)
\(\Leftrightarrow\) \((\sqrt{x}-1)^2 - (\sqrt{2})^2 \) = 0
\(\Leftrightarrow\) \((\sqrt{x} - 1 - \sqrt{2})(\sqrt{x} - 1+\sqrt{2})\)
\(\Leftrightarrow\) \(\begin{cases} \sqrt{x} - 1 - \sqrt{2}=0 \\ \sqrt{x} - 1 +\sqrt{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} \sqrt{x} = 1 +\sqrt{2} \\ \sqrt{x} = 1 - \sqrt{2} \end{cases} \) \(\Leftrightarrow\)\(\begin{cases} x = 1+\sqrt{2} = 3+2\sqrt{2} \\ \sqrt{x} = 1-\sqrt{2} < 0 ( LOẠI ) \end{cases} \)
P/s : mk không biết làm phần b
Mình bị nhầm
b) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Để P\(\in Z\) thì \(\sqrt{x}-1\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)
Vì \(\sqrt{x}-1\ge-1\)
Vậy \(\sqrt{x}-1\in\left\{\pm1;2\right\}\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x}-1=-1\\\sqrt{x}-1=2\\\sqrt{x}-1=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=0\left(tm\right)\\x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\)
Vậy x=0, x=4,x=9 thì P\(\in Z\)
a)
\(=\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với \(x\ge0;x\ne1\)
b)
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Vì 1 \(\in Z\) nên
Để P \(\in\) Z thì \(2⋮\sqrt{x}-1=>\sqrt{x}-1\in\) Ư(2) = { -2;-1;1;2 }
=> \(\sqrt{x}\) = { -1;0;2;3 }
=> x ={0;4;9} thỏa mãn đkxđ
Vậy, ...............
Đặt: \(\sqrt{x}=t\)( \(t\ge0;t\ne1\)) => \(A\ne0\)
Ta có: \(A=\frac{t-1}{t^2+t+1}\)
<=> \(At^2+At+A=t-1\)
<=> \(At^2+\left(A-1\right)t+\left(A+1\right)=0\) (1)
(1) có nghiệm <=> \(\Delta\ge0\)<=> \(-3A^2-6A+1\ge0\)<=> \(-1-\frac{2}{\sqrt{3}}\le A\le-1+\frac{2}{\sqrt{3}}\)
Theo đề ra A thuộc Z ; A khác 0
=> A \(\in\){ - 2; -1 }
+) Với A = - 2 thế vào (1) ta có: \(-2t^2-3t-1=0\) <=> \(\orbr{\begin{cases}t=-1\left(loai\right)\\t=-\frac{1}{2}\left(loai\right)\end{cases}}\)
+) Với A = -1 thế vào (1) ta có: \(-t^2-2t=0\)<=> \(\orbr{\begin{cases}t=0\left(tm\right)\\t=-2\left(loai\right)\end{cases}}\)
Với t = 0 ta có: \(\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
Vậy x = 0 ; A = -1
E cảm ơn cô