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Phần 3:
Ta đã rút gọn được \(P=\frac{4x}{\sqrt{x}-3}\)
Ta có: \(m(\sqrt{x}-3)P> x+1\) với mọi \(x>4\)
\(\Leftrightarrow m(\sqrt{x}-3).\frac{4x}{\sqrt{x}-3}> x+1\) với mọi \(x>4\)
\(\Leftrightarrow 4mx> x+1\) với mọi \(x>4\)
\(\Leftrightarrow m> \frac{x+1}{4x}\) với mọi \(x>4\)
Điều này xảy ra khi mà \(m> max \left(\frac{x+1}{4x}\right)\)
Ta có: \(\frac{x+1}{4x}=\frac{1}{4}+\frac{1}{4x}<\frac{1}{4}+\frac{1}{4.4}\Leftrightarrow \frac{x+1}{4x}< \frac{5}{16}\) (do \(x>4\) )
\(\Rightarrow max\left(\frac{x+1}{4x}\right)< \frac{5}{16}\)
Do đó \(m\geq \frac{5}{16}\) thỏa mãn điều kiện đã cho.
Lời giải:
ĐK: \(x>0; x\neq 4\)
Có: \(K=\left(\frac{4\sqrt{x}(2-\sqrt{x})}{(2+\sqrt{x})(2-\sqrt{x})}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-2)}-\frac{2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\right)\)
\(=\frac{8\sqrt{x}-4x+8x}{(2+\sqrt{x})(2-\sqrt{x})}: \frac{\sqrt{x}-1-2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\)
\(=\frac{8\sqrt{x}+4x}{(2+\sqrt{x})(2-\sqrt{x})}.\frac{\sqrt{x}(\sqrt{x}-2)}{-\sqrt{x}+3}\)
\(=\frac{4\sqrt{x}(2+\sqrt{x})}{2+\sqrt{x}}. \frac{-\sqrt{x}}{3-\sqrt{x}}=\frac{-4\sqrt{x}.\sqrt{x}}{3-\sqrt{x}}=\frac{4x}{\sqrt{x}-3}\)
b)
\(K=-1\Leftrightarrow \frac{4x}{\sqrt{x}-3}=-1\Rightarrow 4x=-(\sqrt{x}-3)\)
\(\Leftrightarrow 4x+\sqrt{x}-3=0\)
\(\Leftrightarrow (4\sqrt{x}-3)(\sqrt{x}+1)=0\)
Vì \(\sqrt{x}+1>0\Rightarrow 4\sqrt{x}-3=0\Rightarrow x=\frac{9}{16}\)
c) \(m(\sqrt{x}-3)K>x+1\)
\(\Leftrightarrow m. (\sqrt{x}-3).\frac{4x}{\sqrt{x}-3}>x+1\)
\(\Leftrightarrow m> \frac{x+1}{4x}\)
\(\Leftrightarrow m> max(\frac{4x}{x+1}), \forall x< 9\)
Với đk đã cho thì ta thấy \(\frac{4x}{x+1}\) có min thôi.
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
\(P=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(P=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(P=\left(\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(P=\frac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-5x\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{4\sqrt{x}\left(2+5x\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{-4x}{3-\sqrt{x}}\)
\(P=\frac{4x}{\sqrt{x}-3}\)
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:
\(x>9\)
\(\Rightarrow\frac{1}{4m-1}< 9\)
\(\Leftrightarrow1< 9\left(4m-1\right)\)
\(\Leftrightarrow1< 36m-1\)
\(\Leftrightarrow10< 36m\)
\(\Leftrightarrow m< \frac{5}{18}\)
Rút gọn P:
\(P=\dfrac{4x}{\sqrt{x}-3}\)
\(\Rightarrow m\left(\sqrt{x}-3\right)P>x+1\)
\(\Leftrightarrow4mx>x+1\)
\(\Leftrightarrow\left(4m-1\right)x>1\)(1)
Xét \(4m-1=0\)
\(\Leftrightarrow m=\dfrac{1}{4}\)(loại vì (1) sai)
\(\Leftrightarrow x>\dfrac{1}{4m-1}\)
Xét \(\dfrac{1}{4m-1}< 0\)
\(\Rightarrow\left\{{}\begin{matrix}x< 0\\x>9\end{matrix}\right.\)(loại)
Xét \(\Leftrightarrow\dfrac{1}{4m-1}>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}4m-1>0\\\dfrac{1}{4m-1}\le9\end{matrix}\right.\)
\(\Leftrightarrow m\ge\dfrac{5}{18}\)
Có mấy cái dâu tương đương. Copy từ cái hàng trên xuống dưới quên xóa nha. Mà chắc đọc hiểu thôi ah.