Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)
b) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=-\frac{3}{2\left(\sqrt{x}-3\right)}\)c) Để P nguyên thì \(2\left(\sqrt{x}-3\right)\in\left\{-3;-1;1;3\right\}\)=> x thuộc rỗng.
1) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(P=\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\)
\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4x+8\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4\sqrt{x}}{2-\sqrt{x}}\)
2) Để \(P=2\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=2\)
\(\Leftrightarrow4\sqrt{x}=4-2\sqrt{x}\)
\(\Leftrightarrow6\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{4}{9}\)
Vậy để \(P=2\Leftrightarrow x=\frac{4}{9}\)
3) Khi \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\2\sqrt{x}-1==0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\left(ktm\right)\\x=\frac{1}{4}\left(tm\right)\end{cases}}\)
Thay \(x=\frac{1}{4}\)vào P, ta được :
\(\Leftrightarrow P=\frac{4\sqrt{\frac{1}{4}}}{2-\sqrt{\frac{1}{4}}}=\frac{4\cdot\frac{1}{2}}{2-\frac{1}{2}}=\frac{2}{\frac{3}{2}}=\frac{4}{3}\)
4) Để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow8x-4\sqrt{x}=-x-\sqrt{x}+6\)
\(\Leftrightarrow9x-3\sqrt{x}-6=0\)
\(\Leftrightarrow3x-\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=3x-2\)
\(\Leftrightarrow x=9x^2-12x+4\)
\(\Leftrightarrow9x^2-13x+4=0\)
\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}9x-4=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{9}\\x=1\end{cases}}\)
Thử lại ta được kết quá : \(x=\frac{4}{9}\left(ktm\right)\); \(x=1\left(tm\right)\)
Vậy để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\Leftrightarrow x=1\)
5) Để biểu thức nhận giá trị nguyên
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}\inℤ\)
\(\Leftrightarrow4\sqrt{x}⋮2-\sqrt{x}\)
\(\Leftrightarrow-4\left(2-\sqrt{x}\right)+8⋮2-\sqrt{x}\)
\(\Leftrightarrow8⋮2-\sqrt{x}\)
\(\Leftrightarrow2-\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)
Ta loại các giá trị < 0
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;6;10\right\}\)
\(\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
\(\)
\(a,x>0;x\ne4,9\)
\(b,Q=\left(\frac{1}{\sqrt{x}-3}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\)
\(Q=\left(\frac{\sqrt{x}-\sqrt{x}+3}{x-3\sqrt{x}}\right):\left(\frac{x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(Q=\frac{3}{x-3\sqrt{x}}:\frac{-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(Q=\frac{3}{\sqrt{x}\left(\sqrt{x}-3\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{-5}\)
\(Q=\frac{3\sqrt{x}-6}{-5\sqrt{x}}\)
\(c,Q< 0< =>\frac{3\sqrt{x}-6}{-5\sqrt{x}}\)
\(-5\sqrt{x}< 0\)
\(< =>3\sqrt{x}-6>0\)
\(\sqrt{x}>2\)
\(x>4\)
\(A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\) ĐKXĐ : x > 0 , x khác 9
\(A=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(A=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(A=\frac{-3\sqrt{x}}{\sqrt{x}+3}.\frac{1}{\sqrt{x}+1}\)
\(A=\frac{-3\sqrt{x}}{x+4\sqrt{x}+4}\)
\(A=\frac{-3\sqrt{x}}{\left(\sqrt{x}+2\right)^2}\)
a) ĐKXĐ : x>hoặc = 0 ; x khác 9
Còn câu b,c,d để vài bữa mình làm tiếp cho bây giờ mình đi ngủ đã buồn ngủ quá !
----------------- -Học tốt-----------------
a) ĐK: \(x-9\ne0\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\ne0\)
Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3>0\)
Nên \(\sqrt{x}-3\ne0\Leftrightarrow x\ne9\)
b) \(P=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left(\frac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right)\)
\(=\left[\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\left[\frac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\left(\frac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
\(=\left(\frac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\right)\left(\frac{1}{\sqrt{x}+1}\right)\)
\(=\frac{-3}{\sqrt{x}+3}\)
c) Ta có: \(\sqrt{x}+3\ge3\)
\(\Rightarrow\frac{3}{\sqrt{x}+3}\le\frac{3}{3}=1\)
\(\Rightarrow\frac{-3}{\sqrt{x}+3}\ge-1\)
Dấu "=" xảy ra khi \(x=0\)
Vậy \(P_{min}=-1\) khi \(x=0\)
d) \(\frac{-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow-\left(\sqrt{x}+3\right)< -9\)
\(\Leftrightarrow-\sqrt{x}< -6\)
\(\Leftrightarrow\sqrt{x}>6\)
\(\Leftrightarrow x>36\)
e) Thế \(x=3-2\sqrt{2}\) vào P ta được:
\(\frac{-3}{\sqrt{3-2\sqrt{2}}+3}=\frac{-3}{\sqrt{2}-1+3}=\frac{-3}{\sqrt{2}+2}=\frac{-3\left(\sqrt{2}-2\right)}{\left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right)}=\frac{6-3\sqrt{2}}{-2}=\frac{3\sqrt{2}-6}{2}\)
f) \(P=\frac{-3}{\sqrt{x}+3}=-2\Leftrightarrow\sqrt{x}+3=6\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)