\(A=\dfrac{\sqrt{y}-1}{y\left(\sqrt{y}-1\right)\left(y+\sqrt{y}+1\right)}:\dfrac{\sqrt{y}+1-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}+1\right)}\)
\(=\dfrac{1}{y\left(y+\sqrt{y}+1\right)}\cdot\dfrac{\sqrt{y}\left(\sqrt{y}+1\right)}{1}=\dfrac{\sqrt{y}+1}{\sqrt{y}\left(y+\sqrt{y}+1\right)}\)
b: Thay \(y=3+2\sqrt{2}\) vào A, ta được:
\(A=\dfrac{\sqrt{2}+1+1}{\left(\sqrt{2}+1\right)\left(3+2\sqrt{2}+\sqrt{2}+1+1\right)}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)\left(5+3\sqrt{2}\right)}=\dfrac{-6+5\sqrt{2}}{7}\)
`a)` Với `y > 0,y \ne 1` có:
`A=[\sqrt{y}-1]/[y^2-y]:(1/\sqrt{y}-1/[\sqrt{y}+1])`
`A=[\sqrt{y}-1]/[y(\sqrt{y}-1)(\sqrt{y}+1)]:[\sqrt{y}+1-\sqrt{y}]/[\sqrt{y}(\sqrt{y}+1)]`
`A=1/[y(\sqrt{y}+1)].[\sqrt{y}(\sqrt{y}+1)]/1`
`A=1/\sqrt{y}`
`b)` Có:`y=3+2\sqrt{2}=2+2\sqrt{2}+1=(\sqrt{2}+1)^2`
`=>\sqrt{y}=|\sqrt{2}+1|=\sqrt{2}+1`
Thay `\sqrt{y}=\sqrt{2}+1` vào `A` thu gọn có:
`A=1/[\sqrt{2}+1]=[\sqrt{2}-1]/[2-1]=\sqrt{2}-1`
a) \(A=\dfrac{\sqrt{y}-1}{y^2-y}:\left(\dfrac{1}{\sqrt{y}}-\dfrac{1}{\sqrt{y}+1}\right)\left(ĐK:y>0;y\ne1\right)\\ A=\dfrac{\sqrt{y}-1}{y\left(y-1\right)}:\left(\dfrac{\sqrt{y}+1-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}+1\right)}\right)\\ A=\dfrac{\sqrt{y}-1}{y\left(\sqrt{y}-1\right)\left(\sqrt{y}+1\right)}:\dfrac{1}{\sqrt{y}\left(\sqrt{y}+1\right)}\\ A=\dfrac{1}{y\left(\sqrt{y}+1\right)}\cdot\left[\sqrt{y}\left(\sqrt{y+1}\right)\right]\\ A=\dfrac{1}{\sqrt{y}}\)
b) Với \(y=3+2\sqrt{2}\left(tmđk\right)\) ta có:
\(A=\dfrac{1}{\sqrt{3+2\sqrt{2}}}\\ A=\dfrac{1}{\sqrt{\left(\sqrt{2}+1\right)^2}}\\ A=\dfrac{1}{\left|\sqrt{2}+1\right|}\\ A=\dfrac{1}{\sqrt{2}+1}\\ A=\dfrac{\sqrt{2}-1}{2-1}\\ A=\sqrt{2}-1\)
Vậy khi \(y=3+2\sqrt{2}\) thì \(A=\sqrt{2}-1\)
