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a) A=\(\dfrac{\sqrt{x}[\left(\sqrt{x}\right)^3-1]}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
A=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\) A=\(\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
A=\(x-\sqrt{x}+1\)
b) A=\(\dfrac{3}{4}\)
=> \(x-\sqrt{x}+1=\dfrac{3}{4}\)
\(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\)
=> \(\sqrt{x}=\dfrac{1}{2}\)
=> \(x=\dfrac{1}{4}\)
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 2: a) Ta có: Q=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) -\(\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\) =\(\dfrac{1}{\sqrt{x}-1}\) -\(\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\left(\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) =
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A = \(\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(\Leftrightarrow A=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}+1}{2\left(x-1\right)}-\dfrac{\sqrt{x}-1}{2\left(x-1\right)}-\dfrac{2\sqrt{x}}{2\left(x-1\right)}\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(x-1\right)}\)
\(\Leftrightarrow A=\dfrac{2\left(1-\sqrt{x}\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{1}{\sqrt{x}+1}\)
b) Khi \(x=\dfrac{4}{9}\) (thảo mãn ĐKXĐ) thì giá trị của A là:
\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{\dfrac{4}{9}}+1}=-\dfrac{3}{5}\)
Vậy .....
c)
+) Khi \(A=-\dfrac{1}{2}\) thì ta có:
\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=1\) (Loại do không thỏa mãn ĐKXĐ)
+) Khi \(A=\dfrac{-1}{4}\) thì ta có:
\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{4}\)
\(\Leftrightarrow x=9\) (thỏa mãn)
Vậy để A = \(-\dfrac{1}{4}\) thì x = 9
a/ ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
= \(\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}+\dfrac{-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{2-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{-1}{\sqrt{x}+1}\)
b/
Thay \(x=\dfrac{4}{9}\) vào A ta được:
\(A=\dfrac{-1}{\sqrt{\dfrac{4}{9}}+1}=\dfrac{-1}{\dfrac{2}{3}+1}=\dfrac{-3}{5}\)
Vậy khi \(x=\dfrac{4}{9}\) thì \(A=\dfrac{-3}{5}\)
c/ Với \(x\ge0,x\ne1\)
* Để \(A=\dfrac{-1}{2}\Leftrightarrow\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{2}\)
\(\Leftrightarrow-2=-\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\) ( ktmđk)-Loại
Vậy không có giá trị nào của x thỏa mãn \(A=\dfrac{-1}{2}\)
* Để \(A=\dfrac{-1}{4}\Leftrightarrow\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{4}\)
\(\Leftrightarrow-4=-\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\) (tmđk)
Vậy để \(A=\dfrac{-1}{4}\) thì \(x=9\)
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)
\(Q=\left(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{x-1-x+4}{x-3\sqrt{x}+2}\right)\)(với x>4)
\(\Leftrightarrow Q=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow Q=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)(với x>4)
b)
\(Q>0\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}>0\)(với x>4)
\(\Leftrightarrow x>4\)
Vậy Q dương khi x>4.
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.
`ĐK: x > 0,x ne 4`
`a)` Thay `x=9` (t/m) vào `P` có:
`P=(1/[9-4]-1/[9+4\sqrt{9}+4).[9+2\sqrt{9}]/\sqrt{9}=4/5`
`b)` Với `x > 0,x ne 4` có:
`P=[\sqrt{x}+2-\sqrt{x}+2]/[(\sqrt{x}-2)(\sqrt{x}+2)^2].[\sqrt{x}(\sqrt{x}+2)]/\sqrt{x}`
`P=4/[x-4]`
`c)` Với `x > 0,x ne 4` có:
`P > -1<=>4/[x-4] > -1`
`<=>4/[x-4]+1 > 0`
`<=>[4+x-4]/[x-4] > 0<=>x/[x-4] > 0<=>[(x > 4),(x < 0):}`
Kết hợp đk
`=>x > 4`.