\(A-1=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-1=\dfrac{\sqrt{x}-2-\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+1}\)
Do \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1>0;\forall x\in D\)
\(\Rightarrow\dfrac{-3}{\sqrt{x}+1}< 0\)
\(\Rightarrow A-1< 0\Rightarrow A< 1\)
Bài 1: Cho A=\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)với x≥0; y≥0; x≠y
a) Rút gọn A
b) Chứng minh A≥0
Bài 2:Cho A= \(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
với x>0; x≠1
a) Rút gọn A
b)Tìm x để A=6
1. \(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\)
Rút gọn biểu thức A
a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
Câu 1 tìm đkxđ của các căn thức bậc hai sau
a)\(\sqrt{1-x}\)
b)\(\sqrt{\dfrac{2}{x}}\)
c)\(\sqrt{\dfrac{4}{x+1}}\)
d)\(\sqrt{x^2+2}\)
Câu 2 rút gọn
a)\(\sqrt{\left(-\sqrt{2-1}\right)^2}\)
b)\(\sqrt{\left(4+\sqrt{2}\right)^2}\)
rút gọn A
A = \(\left(\dfrac{2}{x-\sqrt{x}}+\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{2\sqrt{x}-x}\)
Cho A=\(\sqrt{\dfrac{x-2\sqrt{x-1}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}}\left(1-\dfrac{1}{x-1}\right)\)
Tìm x để A có nghiã
\(K=\left[\dfrac{x+3\sqrt{x}+2}{x+\sqrt{x}-2}-\dfrac{x+\sqrt{x}}{x-1}\right]:\left[\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right]\)
a,Rút gọn K
b,Tính K khi x=\(24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
c,Tìm x để \(\dfrac{1}{K}-\dfrac{\sqrt{x}+1}{8}\)≥1
\(\dfrac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\) \(\dfrac{1}{\left(\sqrt{X}-1\right)\left(3-\sqrt{X}\right)}\)
GIUP MIK VS NHA,CẢM ƠN
Rút gọn
A=\(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
với x>0 và x\(\ne1\)
