Cho P=\(\frac{5}{\sqrt{x}+1}\)
Tìm x để P > 2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/
\(=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{x-1-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{1}\right)\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b/ Biểu thức nhận giá trị dương khi
\(\sqrt{x}-1>=0\)
\(x>=1\)
Vậy với x>=1 thì biểu thức dương
c/ biểu thức nhận giá trị âm khi
\(\sqrt{x}-1
a)
\(M=\frac{-(\sqrt{x}+1)\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-2\right)\left(x+2\right)}+\frac{-2\sqrt{x}\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)\left(x+2\right)}+\frac{2+5\sqrt{x}}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{-x-3\sqrt{x}-2-2x+4\sqrt{x}+2+5\sqrt{x}}{4-x}\)
\(=\frac{-3x+6\sqrt{x}}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-3\sqrt{x}\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-3\sqrt{x}}{-\sqrt{x}-2}\)
a) \(ĐKXĐ:x\ne4;x\ne9\)
b) \(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{-\sqrt{x}+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
c) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\) (ĐK: x thuộc Z)
\(\sqrt{x}-3\) | 1 | -1 | 2 | -2 | 4 | -4 |
\(\sqrt{x}\) | 4 | 2 | 5 | 1 | 7 | -1 |
x | 2 | \(\sqrt{2}\) | \(\sqrt{5}\) | \(\sqrt{1}\) | \(\sqrt{7}\) | \(\varnothing\) |
Vậy để A thuộc Z khi x = {2;\(\sqrt{2};\sqrt{5};\sqrt{1};\sqrt{7}\) }
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
Hay \(A=\frac{3\sqrt{x}+6-6}{\sqrt{x}+2}\)
\(=3-\frac{6}{\sqrt{x}+2}\)
Để \(A\) nguyên \(\Leftrightarrow6\) chia hết cho \(\sqrt{x}+2\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy ..............................
Bổ sung xíu :))
\(Đk:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(\frac{5}{\sqrt{x}+1}>2\\ \frac{5}{\sqrt{x}+1}>\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ 5>2.\left(\sqrt{x}+1\right)\\ 5>2\sqrt{x}+2\\ 3>2\sqrt{x}\\ \sqrt{x}=\frac{3}{2}\)
x= 9/4
đỗ c3 nhá :))
\(\frac{5}{\sqrt{x}+1}>2\)
\(\Leftrightarrow\sqrt{x}+1< \frac{5}{2}\)
\(\Leftrightarrow\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow x< \frac{9}{4}\)