\(x\sqrt{x}+\sqrt{x}-x-1=\sqrt{x}\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(\sqrt{x}-1\right)\)
\(x\sqrt{x}+\sqrt{x}-x-1=\sqrt{x}\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(\sqrt{x}-1\right)\)
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 biểu thức \(A=\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x^2}-1+x}-\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)
Tính A khi \(x=\frac{\sqrt{3}+1}{2\sqrt{2}}\)
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
Gidipt 1) sqrt(x ^ 2 - x) = sqrt(3 - x)
2) sqrt(x ^ 2 - 4x + 3) = x - 2
3) sqrt(4 * (1 - x) ^ 2) - 6 = 0
4) sqrt(x ^ 2 - 4x + 4) = sqrt(4x ^ 2 - 12x + 9)
5) sqrt(x ^ 2 - 4) + sqrt(x ^ 2 + 4x + 4) = 0
6) 1sqrt(x + 2sqrt(x - 1)) + sqrt(x - 2sqrt(x - 1)) = 2
a. \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-2}}=5\)
b. \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=x-1\)
c. \(\sqrt{3x+8+6\sqrt{3x-1}}+\sqrt{3x+8-6\sqrt{3x-1}}=3x+4\)
d. \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-2\sqrt{2x-5}}=2\sqrt{2}\)
Tìm x, biết:
a, \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\)
b, \(\sqrt{1-12x+36x^2}=5\)
c, \(\sqrt{x+2\sqrt{x-1}}=2\)
cho biểu thức M = \(\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) . Khi x > 0 ; x≠1.
Rút gon biểu thức M
rút gọn biểu thức
a) A= \(2\sqrt{\frac{1}{2}}+\sqrt{18}\)
b) B= \(\frac{5+3\sqrt{5}}{\sqrt{5}}+\frac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5+3}\right)\)
c) C= \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\left(x>0,x\ne1\right)\)
d) D = \(\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x-2}}{x-1}\right)\left(x+\sqrt{x}\right)\left(x>0,x\ne1\right)\)
e) E = \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
1. A= \(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Chứng minh: A<1
D= \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\) với x>0 và x≠ 1