\(A=\left(\dfrac{\left(\sqrt{x}+\sqrt{3}\right)\cdot\left(x-\sqrt{3x}+3\right)}{x-\sqrt{3x}+3}-2\sqrt{x}\right)\cdot\left(\sqrt{3}-\sqrt{x}\right)\)
\(=\left(\sqrt{3}-\sqrt{x}\right)^2=x-2\sqrt{3x}+3\)
1.\(\sqrt{-4x^2+25}=x\)
2.\(\sqrt{3x^2-4x+3}=1-2x\)
3. \(\sqrt{4\left(1-x\right)^2}-\sqrt{3}=0\)
4.\(\dfrac{3\sqrt{x+5}}{\sqrt{ }x-1}< 0\)
5. \(\dfrac{3\sqrt{x-5}}{\sqrt{x+1}}\ge0\)
cho A=\(\left(\dfrac{\sqrt{3}}{x^2+x\sqrt{3}+3}+\dfrac{3}{x^3-\sqrt{27}}\right)\)\(\left(\dfrac{x}{\sqrt{3}}+\dfrac{\sqrt{3}}{x}+1\right)\)
a) rút gọn A
Mong mng giúp ạ
câu1 rút gọn
a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
b)\(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\left(x\ne\sqrt{2},x\ne-\sqrt{2}\right)\)
c)\(\sqrt{9\text{x}^2}-2\text{x}\left(x< 0\right)\)
d)\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
e)\(\dfrac{x^2-5}{x+\sqrt{5}}\left(x\ne-\sqrt{5}\right)\)
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 +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
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
\(K=\left[\dfrac{x+3\sqrt{x}+2}{x+\sqrt{x}-2}-\dfrac{x+\sqrt{x}}{x-1}\right]:\left[\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right]\)
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 :
\(\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
b)\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
c)\(\left(\sqrt{5}+1\right)\left(\sqrt{7}+1\right)\left(\sqrt{35}+1\right)\left(34-4\sqrt{7}-6\sqrt{5}\right)\)
d) \(\left(\sqrt{7}+1\right)\left(2\sqrt{2}-1\right)\left(2\sqrt{14}-1\right)\left(55+12\sqrt{2}-7\sqrt{7}\right)\)
e)\(\left(3\sqrt{2}+1\right)\left(2\sqrt{3}+1\right)\left(6\sqrt{6}+1\right)\left(215-34\sqrt{3}-33\sqrt{2}\right)\)
2.4 Rút gọn biểu thức
\(a,\dfrac{3-\sqrt{x}}{x-9}\) ( vs x ≥ 0, x≠ 9)
b, \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}\)( vs x ≥ 0 ; x ≠ 9)
c, \(6-2x-\sqrt{9-6x+x^2}\left(x< 3\right)\)
