Rút gọn biểu thức sau
A=\(\frac{1-\sqrt{x-1}}{\sqrt{x-2\sqrt{x-1}}}\)
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6 tháng 2 2022
Câu 1: \(\sqrt{8}\) − \(\sqrt{18}\) + \(2\sqrt{32}\) = \(\sqrt{4\text{×}2}\) − \(\sqrt{\text{9×2}}\) + 2\(\sqrt{\text{16×2}}\)
=2\(\sqrt{2}\) − 3\(\sqrt{2}\) + 2×4\(\sqrt{2}\)
=(2− 3+ 8)\(\sqrt{2}\)
=7\(\sqrt{2}\)
Câu 2: Mik ko chắc làm đúng hay ko 
nên ko làm
10 tháng 1 2021
a) Ta có: \(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b) Ta có: \(B=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}-2x+2\sqrt{x}-3}{x\sqrt{x}+1}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{2x\sqrt{x}-2x+2\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\left(\dfrac{2x\sqrt{x}-3x+3\sqrt{x}-1+3x+2\sqrt{x}-1-2x\sqrt{x}+2x-2\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}+1}{x-\sqrt{x}+1}\)
9 tháng 9 2021
\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}+1}\)
9 tháng 9 2021
\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0;x\ne25\right)\\ A=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\\ A=\dfrac{5+\sqrt{x}}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)
H9
HT.Phong (9A5)
CTVHS
5 tháng 8 2023
\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\) (ĐK: \(x>1\))
\(A=\left(\dfrac{2}{\sqrt{x-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\)
\(A=\dfrac{4}{x-1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{2}-\sqrt{x^2-1}\)
\(A=2\left(x+1\right)-\sqrt{\left(x+1\right)\left(x-1\right)}\)
\(A=\sqrt{x+1}\left(2\sqrt{x+1}-\sqrt{x-1}\right)\)
HQ
Hà Quang Minh
Giáo viên
5 tháng 8 2023
\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x+1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\left(\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{2x+2\sqrt{x^2-1}-2\sqrt{x^2-1}}{2}\\ \Rightarrow A=x\)
16 tháng 8 2023
a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
29 tháng 3 2020
Ta có : A = \(\left(\frac{x+2}{x.\sqrt{x}-1}+\frac{\sqrt{x}+2}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
= \(\frac{x+2+x+\sqrt{x}-2-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
= \(\frac{x-1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}=1\)
Vậy A = 1
28 tháng 7 2017
\(P=\frac{x+2}{\sqrt{x}^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(P=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
2,
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}+\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{7\sqrt{7}}{7}\)
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}+\frac{\left(\sqrt{2}+1\right)}{2-1}-\sqrt{7}\)
\(A=\sqrt{7}-\sqrt{2}+\sqrt{2}+1-\sqrt{7}=1\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(A=\frac{1-\sqrt{x-1}}{\sqrt{x-2\sqrt{x-1}}}\)
\(A=\frac{1-\sqrt{x-1}}{\sqrt{x-1-2\sqrt{x-1}+1}}\)
\(A=\frac{1-\sqrt{x-1}}{\sqrt{\left(\sqrt{x-1}-1\right)^2}}\)
\(A=\frac{1-\sqrt{x-1}}{\left|\sqrt{x-1}-1\right|}\)
\(TH1:1\le x\le2\)
\(A=\frac{1-\sqrt{x-1}}{1-\sqrt{x-1}}=1\)
\(TH2:2< x\)
\(A=\frac{\sqrt{x-1}-1}{\sqrt{x-1}-1}=1\)