A=(\(\dfrac{1}{a+\sqrt{a}}\)+\(\dfrac{1}{\sqrt{a}+1}\)):\(\dfrac{\sqrt{a}-1}{a-2\sqrt{a}+1}\)(với a>0,a khác 1)
a)Rút gọn A
b)Tìm giá trị của a để A=-2
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TN
20 tháng 12 2022
a)A=\(\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{-2}\)
=\(\dfrac{-2\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{-2}\)
=\(\dfrac{2\sqrt{x}+1}{\sqrt{x}-1}\)
b)Ta có A = \(\dfrac{2\sqrt{x}+1}{\sqrt{x}-1}\)=2+\(\dfrac{2}{\sqrt{x}-1}\)
Để A nguyên thì \(\sqrt{x}-1\)∈Ư(2)
⇒x∈{4;0;9}
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
a) \(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
\(A=\left[\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
\(A=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right]\)
\(A=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left[\dfrac{\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(A=\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(A=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\left(\sqrt{a}-1\right)\)
\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
\(A=\dfrac{a-1}{\sqrt{a}}\)
b) Ta có:
\(a=4+2\sqrt{3}=\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2=\left(\sqrt{3}+1\right)^2\)
Thay vào A ta có:
\(A=\dfrac{\left(\sqrt{3}+1\right)^2-1}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\dfrac{4+2\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{3+2\sqrt{3}}{\sqrt{3}+1}\)
c) \(A< 0\) khi:
\(\dfrac{a-1}{\sqrt{a}}< 0\)
Mà: \(\sqrt{a}\ge0\forall x\) (xác định)
\(\Leftrightarrow a-1< 0\)
\(\Leftrightarrow a< 1\)
Kết hợp với đk:
\(0< a< 1\)
13 tháng 9 2023
a: \(Q=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\cdot\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{a-1}\)
\(=\dfrac{\left(a-1\right)^2}{4a}\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)
=\(\dfrac{\left(a-1\right)^2\cdot\left(-4\sqrt{a}\right)}{\left(a-1\right)\cdot4a}=\dfrac{-\left(a-1\right)}{\sqrt{a}}\)
b: Q<0
=>-(a-1)<0
=>a-1>0
=>a>1
c: Q=2
=>\(a-1=-2\sqrt{a}\)
=>\(a+2\sqrt{a}-1=0\)
=>\(\left[{}\begin{matrix}\sqrt{a}=-1+\sqrt{2}\left(nhận\right)\\\sqrt{a}=-1-\sqrt{2}\left(loại\right)\end{matrix}\right.\Leftrightarrow a=3-2\sqrt{2}\)
26 tháng 12 2021
a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
27 tháng 12 2021
\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
17 tháng 12 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)
\(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}-2}\right)\cdot\dfrac{a-4}{\sqrt{4a}}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{2a}\)
\(=\sqrt{a}+2\)
b: A-2<0
=>\(\sqrt{a}+2-2< 0\)
=>\(\sqrt{a}< 0\)
=>\(a\in\varnothing\)
c: Bạn ghi đầy đủ đề đi bạn
6 tháng 1 2023
a: \(P=\dfrac{a-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\sqrt{a}-1}{1}=\sqrt{a}-1\)
b: Để P<0 thì căn a-1<0
=>căn a<1
=>0<a<1
M
29 tháng 12 2022
\(P=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}\left(a>0;a\ne1\right)\)
\(a,P=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(=\left[\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\dfrac{\sqrt{a}+1}{a-1}\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(=1:\dfrac{\sqrt{a}+1}{a-1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+1}\)
\(=\sqrt{a}-1\)
\(b,P< 0\Rightarrow\sqrt{a}-1< 0\Leftrightarrow\sqrt{a}< 1\Leftrightarrow a< 1\)
Kết hợp điều kiện \(a>0;a\ne1\)
\(\Rightarrow0< a< 1\)
15 tháng 6 2023
a:
Sửa đề: a+2căn a+8
\(=\dfrac{5a+10\sqrt{a}-3\sqrt{a}-6+3a-6\sqrt{a}-a-2\sqrt{a}-8}{\left(a-4\right)}\)
\(=\dfrac{7a-\sqrt{a}-14}{\left(a-4\right)}\)
b: A>0
=>(7a-căn a-14)/(a-4)>0
=>a>4 hoặc 0<a<(1+căn 393)/14
\(a,A=\left(\dfrac{1}{a+\sqrt{a}}+\dfrac{1}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}-1}{a-2\sqrt{a}+1}\left(dk:a>0,a\ne1\right)\\ =\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{1}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)^2}\\ =\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}:\dfrac{1}{\sqrt{a}-1}\\ =\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-1}{1}\\ =\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(b,A=-2\Leftrightarrow\dfrac{\sqrt{a}-1}{\sqrt{a}}=-2\Leftrightarrow\dfrac{\sqrt{a}-1+2\sqrt{a}}{\sqrt{a}}=0\Leftrightarrow3\sqrt{a}=1\Leftrightarrow\sqrt{a}=\dfrac{1}{3}\Leftrightarrow a=\dfrac{1}{9}\left(tmdk\right)\)
Vậy \(a=\dfrac{1}{9}\) thì \(A=-2\)