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a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
\(M=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{x-9}\)
\(=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
b) Ta có: \(x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{\sqrt{3}-\left|\sqrt{3}-1\right|}\)
\(=\sqrt{\sqrt{3}-\sqrt{3}+1}=\sqrt{1}=1\)( thỏa mãn ĐKXĐ )
Thay \(x=1\)vào M ta được:
\(M=\frac{3\sqrt{1}}{\sqrt{1}-3}=\frac{3}{1-3}=\frac{-3}{2}\)
c) \(M=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\sqrt{x}-9+9}{\sqrt{x}-3}=\frac{3\left(\sqrt{x}-3\right)+9}{\sqrt{x}-3}=3+\frac{9}{\sqrt{x}-3}\)
Vì \(x\inℕ\)\(\Rightarrow\)Để M là số tự nhiên thì \(\frac{9}{\sqrt{x}-3}\inℕ\)
\(\Rightarrow9⋮\left(\sqrt{x}-3\right)\)\(\Rightarrow\sqrt{x}-3\inƯ\left(9\right)\)(1)
Vì \(x\ge0\)\(\Rightarrow\sqrt{x}\ge0\)\(\Rightarrow\sqrt{x}-3\ge-3\)(2)
Từ (1) và (2) \(\Rightarrow\sqrt{x}-3\in\left\{-3;-1;1;3;9\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{0;2;4;6;12\right\}\)\(\Rightarrow x\in\left\{0;4;16;36;144\right\}\)( thỏa mãn ĐKXĐ )
Thử lại với \(x=4\)ta thấy M không là số tự nhiên
Vậy \(x\in\left\{0;16;36;144\right\}\)
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( \(P=\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\)(ĐKXĐ : \(1\le x\ne3\))
\(=\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\left(x-3\right)}=\sqrt{x-1}+\sqrt{2}\)
b) \(x=4\left(2-\sqrt{3}\right)\Rightarrow x-1=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\)
Thay vào P được : \(P=2-\sqrt{3}+\sqrt{2}\)
c) Với mọi \(x\ge1,x\ne3\)ta luôn có \(\sqrt{x-1}\ge0\Rightarrow\) \(P=\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\). Dấu "=" xảy ra khi x = 1
Vậy Min P = \(\sqrt{2}\Leftrightarrow x=1\)
2. a) \(Q=\frac{\sqrt{x+2}-1}{x+1}\)(ĐKXĐ: \(-2\le x\ne-1\))
\(=\frac{\left(\sqrt{x+2}-1\right)\left(\sqrt{x+2}+1\right)}{\left(x+1\right)\left(\sqrt{x+2}+1\right)}=\frac{x+2-1}{\left(x+1\right)\left(\sqrt{x+2}+1\right)}=\frac{x+1}{\left(x+1\right)\left(\sqrt{x+2}+1\right)}=\frac{1}{\sqrt{x+2}+1}\)b) \(x=40,25=\frac{161}{4}\Rightarrow x+2=\frac{169}{4}\Rightarrow Q=\frac{1}{\sqrt{\frac{169}{4}}+1}=\frac{1}{\frac{13}{2}+1}=\frac{2}{15}\)
c) Ta có : \(Max_Q\Leftrightarrow Min_{\left(\sqrt{x+2}+1\right)}\)
Mà : \(\sqrt{x+2}+1\ge1\) với mọi \(-2\le x\ne-1\)
Do đó Max Q = 1 \(\Leftrightarrow x=-2\)
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
A=\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right)\):\(\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)Đk x>0 x#0 x#1
=\(\frac{x-1}{\sqrt{x}\left(\sqrt{x-1}\right)}\):\(\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{\sqrt{x}+1}{\left(\sqrt{x-1}\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}}.\sqrt{x}-1\)
=\(\frac{x-1}{\sqrt{x}}\)
Ta có 3+\(2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)(thay và A ta dc
=>\(\frac{3+2\sqrt{2}-1}{\sqrt{2}+1}\)
= \(\frac{2\sqrt{2}+2}{\sqrt{2}+1}\)
=2
mk nhầm....\(\frac{x-1}{\sqrt{x}}>0\)=> \(x-1>0\Rightarrow x>1\)
mk làm r nhé
\(A=\left(\frac{1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\frac{1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right).\sqrt{3}\)
\(=\left(\frac{1+\sqrt{3}-1+\sqrt{3}}{-2}\right).\sqrt{3}=-3\)
\(B=\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
Để \(A=\frac{B}{6}\Leftrightarrow B=6A\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}}=-18\)
\(\Rightarrow\sqrt{x}-1=-18\sqrt{x}\Rightarrow\sqrt{x}=\frac{1}{19}\Rightarrow x=\frac{1}{361}\)
\( 1)P = \left( {\dfrac{{2x + 1}}{{\sqrt {{x^3}} - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\left( {1 - \dfrac{{x + 4}}{{x + \sqrt x + 1}}} \right)\\ = \left( {\dfrac{{2x + 1}}{{x\sqrt x - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{x + \sqrt x + 1 - \left( {x + 4} \right)}}{{x + \sqrt x + 1}}\\ = \left[ {\dfrac{{2x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{1}{{\sqrt x - 1}}} \right]:\dfrac{{\sqrt x - 3}}{{x + \sqrt x + 1}}\\ = \dfrac{{2x + 1 - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x - 3}}\\ = \dfrac{{x - \sqrt x }}{{\sqrt x - 1}}.\dfrac{1}{{\sqrt x - 3}}\\ = \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}.\dfrac{1}{{\sqrt x - 3}}\\ = \dfrac{{\sqrt x }}{{\sqrt x - 3}} \)
1) Để ý rằng : \(x\sqrt{x}-1=\sqrt{x^3}-\sqrt{1^3}=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
\(P=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\)
\(P=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
2) \(x=28-6\sqrt{3}=\left(3\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=3\sqrt{3}-1\)
Thay vào P ta được :
\(P=\frac{3\sqrt{3}-1}{28-6\sqrt{3}+3\sqrt{3}-1+1}\)
\(P=\frac{3\sqrt{3}-1}{28-3\sqrt{3}}\)
3) \(P=\frac{\sqrt{x}}{x+\sqrt{x}+1}< \frac{1}{3}\)
\(\Leftrightarrow x+\sqrt{x}+1>3\sqrt{x}\)
\(\Leftrightarrow x-2\sqrt{x}+1>0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)
BĐT cuối luôn đúng \(\forall x>1\)
Ta có đpcm
4) \(P=\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{2}{7}\)
\(\Leftrightarrow2x+2\sqrt{x}+2=7\sqrt{x}\)
\(\Leftrightarrow2x-5\sqrt{x}+2=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy...
5) \(P=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(\Leftrightarrow Px+P\sqrt{x}+P=\sqrt{x}\)
\(\Leftrightarrow x\cdot P+\sqrt{x}\left(P-1\right)+P=0\)
Phương trình trên có nghiệm khi \(\Delta\ge0\)
\(\Leftrightarrow\left(P-1\right)^2-4P^2\ge0\)
\(\Leftrightarrow P^2-2P+1-4P^2\ge0\)
\(\Leftrightarrow-3P^2-2P+1\ge0\)
\(\Leftrightarrow-3\left(P^2+\frac{2}{3}P-\frac{1}{3}\right)\ge0\)
\(\Leftrightarrow P^2+\frac{2}{3}P-\frac{1}{3}\le0\)
\(\Leftrightarrow P^2+2\cdot P\cdot\frac{1}{3}+\frac{1}{9}-\frac{4}{9}\le0\)
\(\Leftrightarrow\left(P+\frac{1}{3}\right)^2\le\left(\frac{2}{3}\right)^2\)
\(\Leftrightarrow P+\frac{1}{3}\le\frac{2}{3}\)
\(\Leftrightarrow P\le\frac{1}{3}\)
Vậy \(maxP=\frac{1}{3}\Leftrightarrow x=1\)??
Đoạn này sai sai ta ?
Akai Haruma câu 5 sai sai ha chị ?