Yêu cầu để bài là gì vậy bạn?
ĐKXĐ: 1/(-1+x)>=0
=>x-1>0
=>x>1
Rút gọn:
1) \(P=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
2) \(P=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
Giúp mk nhé :3
a : \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)với a ≥ 0 x ≠ 4
b : \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
c : \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
d : \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\)
Bài 1: Rút gọn:
a) \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\) với x>1
b) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right).\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{x-1}\right)\)với x>1
c) \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\) với x>1
d) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)với x ≠ 4, x ≠ 16,x >0
Mng giúp mk nha
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}
\)
\(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\)
\(\left(1-\dfrac{4}{\sqrt{x}-1}+\dfrac{1}{x-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
TÌM ĐKXĐ VÀ RÚT GỌN
RG: B = [\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) + \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\) - \(\dfrac{\sqrt{x}}{1-x}\) ] : [ \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) + \(\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\) ] ; ĐKXĐ: x ≥ 0
Ch.minh
a) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)-\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) = \(\dfrac{3\sqrt{x}-1}{x-1}\)
b) (\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)-\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) ) : \(\dfrac{1}{x-4}\) = -4\(\sqrt{x}\)
Rút gọn các biểu thức sau:
1.\(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
2.\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
Rút gọn các biểu thức sau:
1.\(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
2.\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
Cho biểu thức: B = \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{1-x}\right)\) : \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\) . Tính gtri của B biết x = \(\dfrac{2-\sqrt{3}}{2}\)
1.A=\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}}{2}\)
2.B=\(\left(\dfrac{2\sqrt{x+x+1}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\)
3.C=\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
Làm chi tiết hộ mình với ak mình đang cần gấp!!!
