a. \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)

\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{x-1}{-\left(x+4\right)}=\dfrac{4\sqrt{x}}{x+4}\)

b. \(\:B=\dfrac{4\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}+4}=\dfrac{4+4\sqrt{2}}{7+2\sqrt{2}}=\dfrac{\left(4+4\sqrt{2}\right).\left(7-2\sqrt{2}\right)}{\left(7+2\sqrt{2}\right).\left(7-2\sqrt{2}\right)}=\dfrac{12+20\sqrt{2}}{41}\)

\(B=\dfrac{2}{xy}:\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2-\dfrac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\dfrac{2xy}{\left(x-y\right)^2}-\dfrac{x^2+y^2}{\left(x-y\right)^2}=\dfrac{-\left(x-y\right)^2}{\left(x-y\right)^2}=-1\)

Các bạn giải giùm mình.Mình đang cần gấp!

They said to him:"It is raining, so we can't go out for a walk."

-> They told him....................................................................

a. H=\(\dfrac{1}{\sqrt{x-1}-\sqrt{x}}+\dfrac{1}{\sqrt{x-1}+1}\dfrac{\sqrt{x^3}-x}{\sqrt{x}-1}\left(đkxđ:x\ge1\right)\)

H=\(-2\sqrt{x-1}+x\)

b. Với x=\(\dfrac{53}{9-2\sqrt{7}}:\)

H=\(-2\sqrt{\dfrac{53}{9-2\sqrt{7}}-1}+\dfrac{53}{9-2\sqrt{7}}\)

H\(=7\)

c. \(-2\sqrt{x-1}+x=16\)

\(\sqrt{x-1}=\dfrac{x-16}{2}\)

\(4x-4=x^2-32x+256\)

\(x^2-36x+260=0\)

x=26

d. Để H>1 thì x>3

\(\sqrt{\dfrac{3}{7}}=\sqrt{\dfrac{3.7}{7^2}}=\dfrac{\sqrt{21}}{7}\)

1. \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)

vt=\(\left(1+\sqrt{a}\right).\left(1-\sqrt{a}\right)=1-a=vp\left(đpcm\right)\)

2. a. Q=\(\dfrac{a}{\sqrt{a^2-b^2}}-\left(1+\dfrac{a}{\sqrt{a^2-b^2}}\right):\dfrac{b}{a-\sqrt{a^2-b^2}}\left(a>b>0\right)\)

=\(\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{a^2-\left(a^2-b^2\right)}{b.\sqrt{a^2-b^2}}\)

=\(\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{b^2}{b\sqrt{a^2-b^2}}\)

=\(\dfrac{a-b}{\sqrt{a^2-b^2}}=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}\)

b. Với a=3b thay vào biểu thức đã cho, ta được:

\(Q=\dfrac{\sqrt{3b-b}}{\sqrt{3b+b}}=\dfrac{\sqrt{2b}}{\sqrt{4b}}=\dfrac{\sqrt{2}}{2}\)

\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\dfrac{-3}{2}\)

\(vt=\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)

=\(\left(\dfrac{\sqrt{6}}{2}-\dfrac{6\sqrt{6}}{3}\right).\dfrac{1}{\sqrt{6}}\)

=\(\dfrac{-9\sqrt{6}}{6}.\dfrac{1}{\sqrt{6}}=-\dfrac{3}{2}\)

\(vt=\dfrac{-3}{2}=vp\left(đpcm\right)\)

a. \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

=\(8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)

=\(\left(8-2-6\right)\sqrt{5\sqrt{3}}\)

=0

b. \(22\sqrt{40\sqrt{22}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}\)

=\(88\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-12\sqrt{5\sqrt{3}}\)

=\(80\sqrt{5\sqrt{3}}\)

c. \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

=\(4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)

=\(4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)

d. \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10.\left(2+\sqrt{3}\right)}}}\)

=\(\sqrt{4+\sqrt{5\sqrt{3}-5.\left(5-\sqrt{3}\right)}}\)

=\(\sqrt{4+\sqrt{25}}\)

=\(\sqrt{4+5}=\sqrt{9}=3\)

x³ - x² - x = 1/3
<=> x³ = x² + x + 1/3
<=> 3x³ = 3(x² + x + 1/3)
<=> 3x³ = 3x² + 3x + 1
<=> 3x³ + x³ = x³ + 3x² + 3x + 1
<=> 4x³ = (x + 1)³
<=> ³√(4x³) = ³√(x + 1)³
<=> ³√4.x = x + 1
<=> ³√4.x - x = 1
<=> x(³√4 - 1) = 1
<=> x = 1/(³√4 - 1)