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Đề khá hay đấy! Nhưng lần sau đừng viết sai đề nx!
a) ĐK: \(x>4\)
b) \(P=\dfrac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{1-\dfrac{8}{x}+\dfrac{16}{x^2}}}\)
= \(\dfrac{\sqrt{\left(x-4\right)+4\sqrt{x-4}+4}+\sqrt{\left(x-4\right)-4\sqrt{x-4}+4}}{\sqrt{1-2.\dfrac{4}{x}+\left(\dfrac{4}{x}\right)^2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}}{\sqrt{\left(1-\dfrac{4}{x}\right)^2}}\)
= \(\dfrac{\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|}{\left|1-\dfrac{4}{x}\right|}\)
= \(\dfrac{\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|}{1-\dfrac{4}{x}}\) = \(\left[{}\begin{matrix}\dfrac{2x\sqrt{x-4}}{x-4}khix\ge8\\\dfrac{4x}{x-4}khi4< x< 8\end{matrix}\right.\)
Xét \(P=\dfrac{2x}{\sqrt{x-4}}\left(x\ge8\right)\) thì:
Để \(P\in Z\) khi \(\dfrac{2x-8+8}{\sqrt{x-4}}\in Z\)
<=> \(2.\left(\sqrt{x-4}\right)+\dfrac{8}{\sqrt{x-4}}\in Z\)
<=> \(\left\{{}\begin{matrix}\sqrt{x-4}\in Z^+\\\sqrt{x-4}\inƯ\left(8\right)\end{matrix}\right.\)
Mà \(x\ge8\) => \(\left[{}\begin{matrix}\sqrt{x-4}=2\\\sqrt{x-4}=4\\\sqrt{x-4}=8\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=8\\x=20\\x=68\end{matrix}\right.\)
Xét \(P=\dfrac{4x}{x-4}\left(4< x< 8\right)\) thì:
Để \(P\in Z\) khi \(\dfrac{4x-16+16}{x-4}\in Z\) <=> \(4+\dfrac{16}{x-4}\in Z\)
=> \(x-4\inƯ\left(16\right)\) mà \(0< x-4< 4\)
=> \(x-4=2\) => \(x=6\)
Vậy \(x\in\left\{6;8;20;68\right\}\) thì \(P\in Z\)
P/s: Vì bài này dài nên mk lm khá tắt, ko hiểu cứ hỏi!
1) +) ta có : \(C-\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}=\dfrac{3\sqrt{x}-x+\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-\left(x-4\sqrt{x}+4\right)+3}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-\left(\sqrt{x}-2\right)^2+3}{3\left(x+\sqrt{x}+1\right)}\)
không thể cm được đâu bn --> xem lại đề
2) +) ta có : \(D=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)
--> để \(D\in Z\Leftrightarrow\sqrt{x}+2\) là ước của 3 \(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x=1\) vậy \(x=1\)
3) +) tương tự 2)
4) a) +) điều kiện xác định : \(x>0;x\ne4\)
ta có : \(A=\left(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-2}{x+3\sqrt{x}}\)
\(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
b) ta có : \(A=3\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=3\Leftrightarrow\sqrt{x}-3=3\sqrt{x}-6\)
\(\Leftrightarrow2\sqrt{x}=3\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\) vậy \(x=\dfrac{9}{4}\)
c) ta có : \(B=A.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{x-9}{x-4}=1-\dfrac{5}{x-4}\)
tương tự 2 )
\(\)
ĐKXĐ:\(x>0,x\ne4\)
\(M=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right)\)
\(M=\left(\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(M=\dfrac{4\sqrt{x}}{\left(2-\sqrt{x}\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(M=\dfrac{4x}{\sqrt{x}-3}\)
a) ĐKXĐ: \(x\ge0;x\ne9\) . Rút gọn: \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{x-2\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{x+\sqrt{x}-3\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x+4\sqrt{x}-7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-3\sqrt{x}-2\sqrt{x}+6+x+\sqrt{x}+3\sqrt{x}+3-x+4\sqrt{x}-7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
A>-1\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)>-1\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+1>0\Leftrightarrow\dfrac{\sqrt{x}+2+\sqrt{x}-3}{\sqrt{x}-3}>0\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\sqrt{x}-1>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}2\sqrt{x}-1< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0,5\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0,5\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0,25\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0,25\\x< 9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow}}\left[{}\begin{matrix}x>9\\0\le x< 0,25\end{matrix}\right.\)
a: Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b: Thay \(x=\dfrac{1}{4}\) vào P, ta được:
\(P=\left(\dfrac{1}{2}-1\right):\left(\dfrac{1}{2}+1\right)=\dfrac{-1}{2}:\dfrac{3}{2}=-\dfrac{1}{3}\)
c: Ta có: \(P< \dfrac{1}{2}\)
\(\Leftrightarrow P-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
điều kiện \(x\ge0\)và x khác 1/4
Q= \(\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x+7\sqrt{x}-4}=\frac{3x+14\sqrt{x}+8+2x-3\sqrt{x}+1-x+6\sqrt{x}-5}{2x+7\sqrt{x}-4}\)
=\(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}\)
đề Q>1/2 thì \(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}>\frac{1}{2}\)
<=> \(8x+34\sqrt{x}+8>2x+7\sqrt{x}-4\)<=> \(6x+27\sqrt{x}+12>0\) với mọi x>=0
vậy Q>1/2 khi x>=0 và x khác 1/4
\(A=\dfrac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{1-\dfrac{8}{x}+\dfrac{16}{x^2}}}\)
\(=\dfrac{\sqrt{\left(x-4\right)+4\sqrt{x-4}+4}+\sqrt{\left(x-4\right)-4\sqrt{x-4}+4}}{\sqrt{\left(\dfrac{4}{x}-1\right)^2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{x-4}+2\right)}^2+\sqrt{\left(\sqrt{x-4}-2\right)}^2}{\left|\dfrac{4}{x}-1\right|}\)
\(=\dfrac{\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|}{\left|\dfrac{4}{x}-1\right|}\)
\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}=0\left(x\ne1\right)\\ \Leftrightarrow x=0\)
\(d,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\left(\dfrac{2}{\sqrt{x}+1}>0\right)\\ e,P=1-\dfrac{2}{\sqrt{x}+1}\\ \sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-\dfrac{2}{1}=-2\\ \Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-\left(-2\right)=3\)
Dấu \("="\Leftrightarrow x=0\)
a) ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp đk:
\(\Leftrightarrow x\in\left\{0\right\}\)
d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)
e) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Do \(\sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-2\)
\(\Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)
\(minP=-1\Leftrightarrow x=0\)
a) \(B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\left(đk:x\ge0,x\ne4\right)\)
\(=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}.\dfrac{x-4}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}\)
c) \(C=A\left(B-2\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\dfrac{-2}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{1;-1;2-2\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0\right\}\)
\(\Rightarrow x\in\left\{0;1;9;16\right\}\)
Đc rồi
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