Với x < 0, có:
\(2x^2\sqrt{\frac{9}{x^4}}\) = \(\frac{2x^2.9}{x^2}=18\)
Bài 8: Rút gọn biểu thức chứa căn bậc hai
Các câu hỏi tương tự
cho bieu thuc p=(x+1)(x+√x)/√x-x-√x, voi x>0
a/ rut gon bieu thuc
b/ tim gia tri cua x de gia tri cua bieu thuc p bang 2
Rut gon bieu thuc : \(\frac{2xy^2}{3ab}\sqrt{\frac{9a^3b^4}{8xy^3}}\)
M=\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)với x>0,\(x\ne1\)
Rut gon bieu thuc M
Tìm x để M=\(\dfrac{9}{2}\)
So sanh M va 4
A=(\(\frac{1}{\sqrt{x}-1}\)+\(\frac{1}{\sqrt{x}+1}\))2.\(\frac{x^2-1}{2}\)-\(\sqrt{1-x^2}\)
1)tim dk cua x de bieu thuc A co nghia
2)rut gon bieu thuc A
3) giai phuong trinh theo x khi A=-2
rut gon bieu thuc:\(\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\frac{a\sqrt{a^2-b}}{2}\)
Cho A = \(\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\)
a, Rut gon bieu thuc A
b, Tinh gia tri cua A khi x = \(\dfrac{1}{1+\sqrt{2}}\)
c, Tim Max A
\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
a, rut gon A
b, tinh A voi \(4-2\sqrt{3}\)
rut gon \(\frac{x-11}{\sqrt{x-2}-3}\)
CHO 2 BIET THUC A= x- can x binh/x-1 va
B= x-4/x+2 can x binh
a. rut gon a va b