Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(P=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(P=\left(\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(P=\frac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-5x\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{4\sqrt{x}\left(2+5x\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{-4x}{3-\sqrt{x}}\)
\(P=\frac{4x}{\sqrt{x}-3}\)
Có:
\(m\left(\sqrt{x}-3\right)P>x+1\)
\(\Leftrightarrow m\left(\sqrt{x}-3\right).\frac{4x}{\sqrt{x}-3}>x+1\)
\(\Leftrightarrow4mx>x+1\)
\(\Leftrightarrow4mx-x>1\)
\(\Leftrightarrow\left(4m-1\right)x>1\)
\(\Leftrightarrow x>\frac{1}{4m-1}\)
Lại có:
\(x>9\)
\(\Rightarrow\frac{1}{4m-1}< 9\)
\(\Leftrightarrow1< 9\left(4m-1\right)\)
\(\Leftrightarrow1< 36m-1\)
\(\Leftrightarrow10< 36m\)
\(\Leftrightarrow m< \frac{5}{18}\)
a)\(\frac{\left(x-1\right)}{\sqrt{x}}\)
b) để P>0\(\Rightarrow\)\(\frac{\left(x-1\right)}{\sqrt{x}}>0\)
do \(\sqrt{x}>0\Rightarrow x-1>0\)
\(\Leftrightarrow x>1\)
c)P=\(\frac{8}{3}\)
Ta có: \(M=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{9-x}:\frac{\sqrt{x}-2-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{x+3\sqrt{x}}{9-x}:\frac{4-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{9-x}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{4-\sqrt{x}}=\frac{x}{\sqrt{x}-4}\)
Khi x > 16 thì \(\sqrt{x}-4>0\), như vậy \(M>y\Leftrightarrow x>m-3x+1\Leftrightarrow4x-1>m\) với mọi x > 16. Vậy m < 15 thì \(M>y\) với mọi x > 16.
Chúc em học tốt ^^
a) DK de P xác dinh : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
b) \(P=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{1-x}+\frac{\left(\sqrt{x}-2\right)^2+3\sqrt{x}-x}{1-\sqrt{x}}\)
\(=\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{-\sqrt{x}+4}{1-\sqrt{x}}\)
\(=\frac{4}{1-\sqrt{x}}\)
c) de P > o thì \(1-\sqrt{x}>0\Rightarrow\sqrt{x}< 1\Rightarrow0< x< 1\)
a/ \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{x-4}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
=> \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
=> \(B=\frac{\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{1}{\sqrt{x}-2}\)
=> \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
b/ B>2 <=> \(\frac{\sqrt{x}+5}{\sqrt{x}+2}>2\) <=> \(\sqrt{x}+5>2\sqrt{x}+4\)
<=> \(1>\sqrt{x}\)=> \(-1\le x\le1\)
c/ \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}=\frac{\sqrt{x}+2+3}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}\)
Để Bmax thì \(\sqrt{x}+2\) đạt giá trị nhỏ nhất . Do \(\sqrt{x}+2\ge2\)=> Đạt nhỏ nhất khi x=0
Khí đó giá trị lớn nhất của B là: \(1+\frac{3}{2}=\frac{5}{2}\)Đạt được khi x=0