a: \(Q=\dfrac{\sqrt{x}-2+7}{x-4}:\dfrac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{1}=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
b:
Trường hợp 1: \(16x^2-625=0\)
=>(4x-25)(4x+25)=0
=>x=25/4(nhận) hoặc x=-25/4(loại)
Thay x=25/4 vào Q, ta được:
\(Q=\left(\dfrac{5}{2}+5\right):\left(\dfrac{5}{2}+2\right)=\dfrac{15}{2}:\dfrac{9}{2}=\dfrac{5}{3}\)
TRường hợp 2: \(x=5+\sqrt{2}-4-\sqrt{2}=1\)
Thay x=1 vào Q, ta được:
\(Q=\dfrac{1+5}{1+2}=\dfrac{6}{3}=2\)
a.\(Q=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{7}{x-4}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
\(Q=\left(\dfrac{\sqrt{x}-2+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
\(Q=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\sqrt{x}-2\)
\(Q=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
b.
i.\(16x^2-625=0\)
\(\Leftrightarrow x^2=\dfrac{625}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{25}{4}\left(tm\right)\\x=-\dfrac{25}{4}\left(ktm\right)\end{matrix}\right.\)
Thế \(x=\dfrac{25}{4}\) vào `Q` ta được:
\(Q=\dfrac{\sqrt{\dfrac{25}{4}}+5}{\sqrt{\dfrac{25}{4}}+2}\)
\(Q=\dfrac{15}{2}:\dfrac{9}{2}\)
\(Q=\dfrac{5}{3}\)
ii.\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(x=\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{\left(\sqrt{2}+4\right)^2}\)
\(x=\left|\sqrt{2}+5\right|-\left|\sqrt{2}+4\right|\)
\(x=\sqrt{2}+5-\sqrt{2}-4\)
\(x=1\)
Thế `x=1` vào `Q` ta được:
\(Q=\dfrac{\sqrt{1}+5}{\sqrt{1}+2}=\dfrac{6}{3}=2\)
`a)` Với `x >= 0,x \ne 4` có:
`Q=(1/[\sqrt{x}+2]+7/[x-4]):([\sqrt{x}-1]/[\sqrt{x}-2]-1)`
`Q=[\sqrt{x}-2+7]/[\sqrt{x}+2)(\sqrt{x}-2)]:[\sqrt{x}-1-\sqrt{x}+2]/[\sqrt{x}-2]`
`Q=[\sqrt{x}+5]/[(\sqrt{x}+2)(\sqrt{x}-2)] .[\sqrt{x}-2]/1`
`Q=[\sqrt{x}+5]/[\sqrt{x}+2]`
__________________________________________________________
`b)`
`i // 16x^2-625=0`
`<=>(4x-25)(4x+25)=0`
`<=>x=[+-25]/4`
Mà `x >= 0`
`=>x=25/4`
`=>\sqrt{x}=5/2`
`=>Q=[5/2+5]/[5/2+2]=5/3`
`ii // x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}`
`<=>x=\sqrt{25+2.5.\sqrt{2}+2}-\sqrt{16+2.4.\sqrt{2}+2}`
`<=>x=\sqrt{(5+\sqrt{2})^2}-\sqrt{(4+\sqrt{2})^2}`
`<=>x=|5+\sqrt{2}|-|4+\sqrt{2}|=5+\sqrt{2}-4-\sqrt{2}=1` (t/m)
`=>\sqrt{x}=1`
`=>Q=[1+5]/[1+2]=2`
