Rút gọn biểu thức \(P=\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{a-\sqrt{a}}\right):\dfrac{1}{\sqrt{a}-1}\left(0< a\in R,a\ne1\right)\)
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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)
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: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
19 tháng 1 2022
a, x > 0 ; x khác 1
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\dfrac{1}{\sqrt{x}-1}\)
\(=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{1}{\sqrt{x}-1}=\dfrac{x-2}{\sqrt{x}}\)
b, Ta có : \(P=\dfrac{x-2}{\sqrt{x}}=1\Rightarrow x-2=\sqrt{x}\)
\(\Leftrightarrow x-\sqrt{x}-2=0\Leftrightarrow\left(\sqrt{x}+1>0\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow x=4\)(tm)
19 tháng 1 2022
a: \(P=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{x-2}{\sqrt{x}}\)
b: Để P=1 thì \(x-\sqrt{x}-2=0\)
hay x=4
18 tháng 9 2021
a) \(P=\dfrac{x-1+4\left(\sqrt{x}+1\right)+1}{x-1}.\dfrac{x-1}{x+2\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}+4}{x+2\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
b) \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}}=1+\dfrac{2}{\sqrt{x}}\in Z\)
Do \(\sqrt{x}>0\)
\(\Rightarrow\sqrt{x}\inƯ\left(2\right)=\left\{1;2\right\}\)
\(\Rightarrow x\in\left\{1;4\right\}\)
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: \(\left(\sqrt{x}+1\right)\cdot A=x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\cdot\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=x\)
\(\Leftrightarrow x-2\sqrt{x}+1=0\)
\(\Leftrightarrow x=1\left(loại\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 7 2021
\(A=\left(\dfrac{\sqrt{x}+1+x+\sqrt{x}}{\sqrt{x}+1}\right)\left(\dfrac{\sqrt{x}-1-x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}\right)\left(\dfrac{-\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right)\left(\dfrac{-\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\right)\)
\(=-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)=1-x\)
\(A=-1\Leftrightarrow1-x=-1\Rightarrow x=2\)
11 tháng 7 2021
a) Ta có: \(A=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=1-x
b) Để A=-1 thì 1-x=-1
hay x=2
25 tháng 2 2022
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(\Rightarrow A=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}}\)
\(P=\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{a-\sqrt{a}}\right):\dfrac{1}{\sqrt{a}-1}\)
\(=\left[\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\sqrt{a}}+\dfrac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}\right].\left(\sqrt{a}-1\right)\)
\(=\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\left(\sqrt{a}-1\right)=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)