Rút gọn:
C= \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
K
Khách
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Những câu hỏi liên quan
17 tháng 11 2021
Sửa: \(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\)
\(C=\dfrac{\sqrt{x}+3\sqrt{x}+2-x+3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}{\sqrt{x}}\\ C=\dfrac{6\sqrt{x}}{\sqrt{x}}=6\)
UM
1
HT
24 tháng 6 2019
ĐKXĐ: \(x>0;x\ne1\)
\(C=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(C=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{x}\)
12 tháng 8 2023
1: \(=\left(1+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}-1}:\dfrac{x-9+x-4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{2x+\sqrt{x}-11}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(2x+\sqrt{x}-11\right)}\)
2: \(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\dfrac{\sqrt{x}+1-2}{x-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
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: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
14 tháng 8 2021
Ủa đáp số là\(\sqrt{2x-2}\) với \(\sqrt{2}\) mà bạn thử A2 đi
CT
26 tháng 5 2019
mik phân `1 đẳng thức
x+2 căn x-1= x-1+2 căn x-1+1= (căn x-1+1)^2
hằng đẳng thức số 1
AT
14 tháng 7 2021
\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{x-\sqrt{x}+6}{x+\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{x-\sqrt{x}-2}{x+\sqrt{x}-2}\right)\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{x-\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}+2+x-\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\dfrac{2x-\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2x-\sqrt{x}-3}=\dfrac{x+8}{2x-\sqrt{x}-3}\)
AH
Akai Haruma
Giáo viên
5 tháng 11 2023
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc đề dễ hiểu hơn bạn nhé.
23 tháng 8 2021
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
23 tháng 8 2021
2: Ta có: \(A=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
3: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
27 tháng 10 2021
\(ĐK:x>0;x\ne1\\ A=\dfrac{2+x-1-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\\ A=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}=\dfrac{\left(1-\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\)
27 tháng 10 2021
\(A=\left(\dfrac{2+x-1-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
C = \(\sqrt{x-1+2\sqrt{x-1}.1+1}\)+\(\sqrt{1-2\sqrt{x-1}.1+x-1}\)
C = \(\sqrt{\left(\sqrt{x-1}+1\right)^2}\)+ \(\sqrt{\left(1-\sqrt{x-1}\right)^2}\)
C = I \(\sqrt{x-1}+1\)I + I\(1-\sqrt{x-1}\)I
C = \(\sqrt{x-1}+1\) + \(1-\sqrt{x-1}\)
C = 2
\(C=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(C^2=\left(\sqrt{x+2\sqrt{x-1}}\right)^2+2\sqrt{\left(x+2\sqrt{x-1}\right)\left(x-2\sqrt{x-1}\right)}\) \(+\left(\sqrt{x-2\sqrt{x-1}}\right)^2\)
\(=x+2\sqrt{x-1}+2\sqrt{x^2-\left(2\sqrt{x-1}\right)^2}+x-2\sqrt{x-1}\)
\(=2x+2\sqrt{x^2-4x+1}\)
\(=2\left(x+\sqrt{x^2-4x+1}\right)\)
\(\Rightarrow C=\sqrt{2\left(x+\sqrt{x^2-4x+1}\right)}\)