rút gon biểu thức
B = \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
với a > 0 , a\(\ne1,a\ne4\)
giải nhanh giúp e ạ mai e thi r huhu
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Những câu hỏi liên quan
25 tháng 2 2022
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(\Rightarrow A=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}}\)
22 tháng 12 2020
Bài 1:
a) Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Để Q dương thì \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)
mà \(3\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ
nên \(\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\)
hay a>4
Kết hợp ĐKXĐ,ta được: a>4
Vậy: Để Q dương thì a>4
9 tháng 2 2022
b. \(=\left(\dfrac{\sqrt{a}-a+a\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right):\left(\dfrac{2\sqrt{a}}{1+\sqrt{a}}\right)\)
\(=\left(\dfrac{2\sqrt{a}}{1-\sqrt{a}}\right):\left(\dfrac{2\sqrt{a}}{1+\sqrt{a}}\right)\)
\(=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\)
\(=1-a\)
HP
9 tháng 2 2022
\(a.\sqrt{8}-2\sqrt{50}+\sqrt{18}=2\sqrt{2}-10\sqrt{2}+3\sqrt{2}=\sqrt{2}\left(2-10+3\right)=-5\sqrt{2}\)
\(b.\left(\dfrac{\sqrt{a}-a}{1-\sqrt{a}}+\sqrt{a}\right):\dfrac{2\sqrt{a}}{1+\sqrt{a}}\left(đk:a\ge0;a\ne1\right)\)
\(=\left(\sqrt{a}+\sqrt{a}\right).\dfrac{1+\sqrt{a}}{2\sqrt{a}}\)
\(=2\sqrt{a}.\dfrac{1+\sqrt{a}}{2\sqrt{a}}\)
\(=1+\sqrt{a}\)
(Chỗ điều kiện bài b mik thấy a = 0 cũng có thể là nghiệm nên mik sửa lại nhé)
6 tháng 8 2022
Câu 1:
a: \(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: Để Q>0 thì \(\sqrt{a}-2>0\)
=>a>4
18 tháng 8 2021
a: Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
18 tháng 8 2021
b: Ta có: \(\left(\sqrt{x}+1\right)\cdot A=x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\cdot\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=x\)
\(\Leftrightarrow x-2\sqrt{x}+1=0\)
\(\Leftrightarrow x=1\left(loại\right)\)
HP
17 tháng 12 2020
\(P=\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{a-\sqrt{a}}\right):\dfrac{1}{\sqrt{a}-1}\)
\(=\left[\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\sqrt{a}}+\dfrac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}\right].\left(\sqrt{a}-1\right)\)
\(=\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\left(\sqrt{a}-1\right)=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
5 tháng 5 2021
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
5 tháng 5 2021
Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 12 2023
P = (\(\dfrac{1}{\sqrt{x}-1}\) - \(\dfrac{1}{\sqrt{x}}\)) : (\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)) với 0 < \(x\) ≠ 1; 4
P = \(\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}\): (\(\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right).\left(\sqrt{x-2}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\))
P = \(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\): \(\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)
P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) : \(\dfrac{3}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)
P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) \(\times\) \(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}{3}\)
P = \(\dfrac{\sqrt{x}-2}{3.\sqrt{x}}\)
P = \(\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{3x}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 12 2023
b, P = \(\dfrac{1}{4}\)
⇒ \(\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{3x}\) = \(\dfrac{1}{4}\)
⇒4\(x\) - 8\(\sqrt{x}\) = 3\(x\)
⇒ 4\(x\) - 8\(\sqrt{x}\) - 3\(x\) = 0
\(x\) - 8\(\sqrt{x}\) = 0
\(\sqrt{x}\).(\(\sqrt{x}\) - 8) = 0
\(\left[{}\begin{matrix}x=0\\\sqrt{x}=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=64\end{matrix}\right.\)
\(x=0\) (loại)
\(x\) = 64
B=\(\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\):\(\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
B=\(\left(\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{a-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
=\(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+2}\)
=\(\dfrac{\sqrt{a}-2}{\sqrt{a}}\)