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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 \(m\ge\dfrac{5}{18}\)
a) P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}\) với x > 0 và x≠4
=\(\left(\dfrac{\sqrt{x}.\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{x-4}\right).\dfrac{x-4}{2\sqrt{x}}\)
=\(\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\right)\).\(\dfrac{x-4}{2\sqrt{x}}\)
=\(\dfrac{2x}{x-4}.\dfrac{x-4}{2\sqrt{x}}\)
=\(\dfrac{x}{\sqrt{x}}\)
b) \(\dfrac{x}{\sqrt{x}}\) >3
<=> x> \(3\sqrt{x}\)
<=> x>9
a: \(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
b: Để P>3 thì \(\sqrt{x}>3\)
hay x>9
a: \(B=\dfrac{x+4\sqrt{x}+4-\left(x-4\sqrt{x}+4\right)+4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(4-x\right)}{\left(3-3\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\dfrac{5x+4\sqrt{x}+4-x+4\sqrt{x}-4}{1}\cdot\dfrac{-\sqrt{x}}{\left(3-3\sqrt{x}\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4x+8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{3\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}}{3\sqrt{x}-3}\)
b: Để B>0 thì \(3\sqrt{x}-3>0\)
hay x>1
c: Để B là số nguyên thì \(4\sqrt{x}⋮3\sqrt{x}-3\)
\(\Leftrightarrow12\sqrt{x}-12+12⋮3\sqrt{x}-3\)
\(\Leftrightarrow3\sqrt{x}-3\in\left\{1;-1;2;-2;3;-3;4;6;12\right\}\)
hay \(x\in\left\{\dfrac{16}{9};\dfrac{4}{9};\dfrac{25}{9};\dfrac{1}{9};\dfrac{49}{9};9;25\right\}\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
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: \(A=\dfrac{\sqrt{3}+1}{\sqrt{3}+1}+\sqrt{5}+3-3-\sqrt{5}=1\)
b: \(B=\dfrac{-\sqrt{x}-3+x-3\sqrt{x}-x-9}{x-9}=\dfrac{-4\sqrt{x}-12}{x-9}=\dfrac{-4}{\sqrt{x}-3}\)
Để B>1 thì \(\dfrac{-4-\sqrt{x}+3}{\sqrt{x}-3}>0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay 0<x<9
Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
a/ đkxđ: x > 0; x≠1
b/ \(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)