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a) \(P=\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}=\dfrac{4x}{\sqrt{x}-3}\)
\(\left(x\ge0;x\ne4;9\right)\)
b)\(P=-1\Leftrightarrow4x+\sqrt{x}-3=0\Leftrightarrow\sqrt{x}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)
c) bpt đưa về dạng \(4mx>x+1\Leftrightarrow\left(4x-1\right)x>1\)
Nếu \(4m-1\le0\) thì tập nghiệm không thể chứa mọi giá trị \(x>9\); Nếu \(4m-1>0\) thì tập nghiệm bpt là \(x>\dfrac{1}{4m-1}\). Do đó bpt tm mọi \(x>9\Leftrightarrow9\ge\dfrac{1}{4m-1}\) và \(4m-1>0\). ta có \(m\ge\dfrac{5}{18}\)
\(P=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(P=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(P=\left(\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(P=\frac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-5x\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{4\sqrt{x}\left(2+5x\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{-4x}{3-\sqrt{x}}\)
\(P=\frac{4x}{\sqrt{x}-3}\)
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:
\(x>9\)
\(\Rightarrow\frac{1}{4m-1}< 9\)
\(\Leftrightarrow1< 9\left(4m-1\right)\)
\(\Leftrightarrow1< 36m-1\)
\(\Leftrightarrow10< 36m\)
\(\Leftrightarrow m< \frac{5}{18}\)
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: \(=\dfrac{4x-8\sqrt{x}+8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}\left(3\sqrt{x}-2\right)}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}=\dfrac{-4x\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
b: \(m\left(\sqrt{x}-3\right)\cdot B>x+1\)
=>\(-4xm\left(3\sqrt{x}-2\right)>\left(\sqrt{x}+2\right)\cdot\left(x+1\right)\)
=>\(-12m\cdot x\sqrt{x}+8xm>x\sqrt{x}+2x+\sqrt{x}+2\)
=>\(x\sqrt{x}\left(-12m-1\right)+x\left(8m-2\right)-\sqrt{x}-2>0\)
Để BPT luôn đúng thì m<-0,3
nhầm rồi, để làm lại
a/ \(P=\left[\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right]\)
\(=\left[\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right]:\left[\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(=\frac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{-\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}\)
\(=\frac{4x}{\sqrt{x}-3}\)
b/ \(P=-1\Rightarrow\frac{4x}{\sqrt{x}-3}=-1\Rightarrow3-\sqrt{x}=4x\Rightarrow4x+\sqrt{x}-3=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=-1\left(l\right)\\\sqrt{x}=\frac{3}{4}\end{cases}\Rightarrow x=\frac{9}{16}}\)
Vậy x = 9/16
ĐKXĐ: x > 0 và \(x\ne4\)
a/ \(P=\left[\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right]\)
\(=\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}\left(\sqrt{x}-1\right)-2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-\sqrt{x}-2}\)
\(=\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{4x}{\left(2+\sqrt{x}\right)\left(\sqrt{x}+1\right)}\)
b/ \(P=-1\Rightarrow\frac{4x}{x+3\sqrt{x}+2}=-1\Rightarrow-x-3\sqrt{x}-2=4x\)
\(\Rightarrow-5x-3\sqrt{x}-2=0\left(1\right)\), vì (1) > 0 => vô nghiệm
Vậy k có giá trị nào của x thỏa P = -1