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a , thu gọn
\(A=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3x+3}{x-9}\right]:\left[\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right]\)
\(A=\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(A=-\dfrac{3}{\sqrt{x}+3}\)
b , tự làm
\(a\text{) Để biểu thức xác định }\\ \text{thì }\Rightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3\ne0\\x-9\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
\(\text{b) }A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\\ =\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right)\\ =\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ =\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\dfrac{-3}{\sqrt{x}+3}\)
\(c\text{) Để }A\le-\dfrac{1}{3}\\ \text{thì }\Rightarrow\dfrac{-3}{\sqrt{x}+3}\le-\dfrac{1}{3}\\ \Rightarrow\dfrac{3}{\sqrt{x}+3}\ge\dfrac{1}{3}\\ \Rightarrow\dfrac{3}{\sqrt{x}+3}-\dfrac{1}{3}\ge0\\ \Rightarrow\dfrac{9}{3\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}\ge0\\ \Rightarrow\dfrac{9-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}\ge0\\ \Rightarrow\dfrac{\sqrt{x}-6}{\sqrt{x}+3}\le0\\ \Leftrightarrow\sqrt{x}-6\ge0\left(\text{Vì }\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}\ge6\\ \Leftrightarrow x\ge36\)
\(d\text{) Do }\sqrt{x}\ge0\\ \Rightarrow\sqrt{x}+3\ge3\\ \Rightarrow\dfrac{-3}{\sqrt{x}+3}\ge-1\\ \text{Dấu }"="\text{ }xảy\text{ }ra\text{ }khi:\text{ }x=0\)
Vậy..............
a) Ta có:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{x-9}\right):\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\)
b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)
.....Chưa nghĩ ra....
c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)
Vậy Min P = 0 khi x =9.
k - kb với tớ nhia mn!
Bài 1:
a: \(A=\dfrac{\sqrt{x}+2}{2\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4+x-4\sqrt{x}+4}{2\left(x-4\right)}\)
\(=\dfrac{2x+8}{2\left(x-4\right)}=\dfrac{x+4}{x-4}\)
b: Để A=8 thì x+4=8(x-4)
=>x+4=8x-32
=>-7x=-36
hay x=36/7(nhận)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{2\sqrt{x}}{x-4}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)
\(=\sqrt{x}\)
b) Để P>4 thì \(\sqrt{x}>4\)
hay x>16
Kết hợp ĐKXĐ, ta được: x>16
Vậy: Khi x>16 thì P>4
Ta có A=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) với x≥ 9, x ∈ R
Để A > 0 \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) > 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}< -1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}>-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x>1\end{matrix}\right.\end{matrix}\right.\)
Kết hợp với ĐKXĐ\(\Rightarrow\) x ∈ ∅
ĐKXĐ: x≥9, x∈R
Ta có:
A= \(\left[\dfrac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)
= \(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\dfrac{1}{1+\sqrt{x}}\):\(\dfrac{1}{\sqrt{x}-2}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(1.\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)=8\) \(2.a,b.A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\) ( x # 0 ; x # -1 ; x # 1 )
\(A=\dfrac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{x^2-1}.\dfrac{x+2003}{x}\)
\(A=\dfrac{x^2-1}{x^2-1}.\dfrac{x+2003}{x}=\dfrac{x+2003}{x}\)
c. \(A=1+\dfrac{2003}{x}\)
Để A ∈ Z ⇒ x ∈ { 1 ; -1 ; 2003 ; - 2003 )
KL...............
\(b,P< 0\Leftrightarrow\dfrac{x-1}{\sqrt{x}}< 0\)
Mà: \(\sqrt{x}\ge0\)
\(\Rightarrow x-1< 0\\ \Leftrightarrow x< 1\)
Ta có: \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow P=\dfrac{x-1}{\sqrt{x}}=\dfrac{\left(\sqrt{3}-1\right)^2-1}{\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\dfrac{\left(\sqrt{3}-1+1\right)\left(\sqrt{3}-1-1\right)}{\sqrt{3}-1}\\ =\dfrac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-1}\)
a: ĐKXĐ: x>0; x<>1
\(P=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{x-1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\)
b: Để P<0 thì x-1<0
=>0<x<1
c: Khi x=4-2 căn 3 thì \(P=\dfrac{4-2\sqrt{3}-1}{\sqrt{3}-1}=\dfrac{3-2\sqrt{3}}{\sqrt{3}-1}\)
`a)ĐKXĐ:{(x > 0),(x \ne 4):}`
`b)` Với `x > 0,x \ne 4` có:
`A=[\sqrt{x}(\sqrt{x}+2)+\sqrt{x}(\sqrt{x}-2)]/[x-4].[x-4]/[\sqrt{4x}]`
`A=[x-2\sqrt{x}+x-2\sqrt{x}]/[2\sqrt{x}]`
`A=[2\sqrt{x}(\sqrt{x}-2)]/[2\sqrt{x}]=\sqrt{x}-2`
`c)` Với `x > 0,x \ne 4` có:
`A < 3 <=>\sqrt{x}-2 < 3<=>\sqrt{x} < 5<=>x < 25`
Kết hợp đk
`=>0 < x < 25 ,x \ne 4`