\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
Rút gọn bieur thức
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Những câu hỏi liên quan
5 tháng 5 2021
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
PH
2
H9
HT.Phong (9A5)
CTVHS
30 tháng 6 2023
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}\right)^2+2\cdot\sqrt{x-1}\cdot1+1^2}+\sqrt{\left(\sqrt{x-1}\right)^2-2\sqrt{x-1}\cdot1+1^2}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\sqrt{x-1}+1+\sqrt{x-1}-1\)
\(=2\sqrt{x-1}\)
30 tháng 6 2023
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
H9
HT.Phong (9A5)
CTVHS
4 tháng 9 2023
\(A=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\) (ĐK: \(x>0;x\ne1\))
\(A=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left[\left(\sqrt{x}\right)^2-1^2\right]}{\sqrt{x}-1}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(A=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(A=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(A=x-\sqrt{x}+1\)
29 tháng 5 2022
\(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\left(\sqrt{x}-1\right)\);\(ĐK:x\ge0;x\ne1\)
\(A=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(A=\left(\dfrac{x-\sqrt{x}+2\sqrt{x}-2-2\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(A=\left(\dfrac{-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
29 tháng 5 2022
\(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}\right).\left(\sqrt{x-1}\right)\left(đk:x\ne1\right)\\ A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\left(\sqrt{x}-1\right)\\ A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-2\right).\left(\sqrt{x}-1\right)\)
\(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\\ A=\left(\dfrac{\sqrt{x}+2-2\sqrt{x}-2}{\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\\ A=\dfrac{-\sqrt{x}}{\sqrt{x}+1}.\left(\sqrt{x}-1\right)\\ A=\dfrac{-x+\sqrt{x}}{\sqrt{x}+1}\)
AT
28 tháng 5 2021
ĐKXĐ: \(x>0;x\ne1,4\)
\(\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}\left(x-\sqrt{x}+2\right)-x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right).\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=-\dfrac{2}{\sqrt{x}+1}\)
G
9 tháng 8 2023
Đề nhầm à?
ĐK: \(x\ge0;x\ne1\)
\(B=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\sqrt{x}=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
9 tháng 8 2023
Sửa đề: \(B=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
H9
HT.Phong (9A5)
CTVHS
18 tháng 10 2023
1) \(A=\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)^2-2\)
\(A=\sqrt{x}\cdot\sqrt{x}+\sqrt{x}-\left(x-2\sqrt{x}+1\right)-2\)
\(A=x+\sqrt{x}-\left(x-2\sqrt{x}+1\right)-2\)
\(A=x+\sqrt{x}-x+2\sqrt{x}-1-2\)
\(A=3\sqrt{x}-3\)
Thay \(x=9\) vào A ta có:
\(A=3\cdot\sqrt{9}-3=3\cdot3-3=9-3=6\)
TA
1
20 tháng 2 2022
đk : x > 0 ; x khác 1
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)^2}=\dfrac{x-1}{x}\)
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(TH1:1\le x\le2\)
\(\sqrt{x-1}+1+1-\sqrt{x-1}\)
\(=2\)
\(TH2:x>2\)
\(\sqrt{x-1}+1+\sqrt{x-1}-1\)
\(2\sqrt{x-1}\)
ĐK : x >= 1
Đặt \(A=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(A^2=\left(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\right)^2\)
\(A^2=\left|x+2\sqrt{x-1}\right|+2\sqrt{\left(x+2\sqrt{x-1}\right)\left(x-2\sqrt{x-1}\right)}+\left|x-2\sqrt{x-1}\right|\)
\(A^2=x+2\sqrt{x-1}+2\sqrt{x^2-4x+4}+x-2\sqrt{x-1}\)
\(A^2=2x+2\sqrt{\left(x-2\right)^2}=2x+2\left|x-2\right|\)
Với 1 ≤ x < 2 => A2 = 2x - 2( x - 2 ) = 2x - 2x + 4 = 4 => A = 2
Với x ≥ 2 => A2 = 2x + 2x - 4 = 4x - 4 => A = 2√(x-1)