Đáp án + giải thích các bước giải:
1)
`2\sqrt{(3x-1)/x}=(4x-1)/(3x-1)`
`->2\sqrt{(3x-1)/x}=(3x-1+x)/(3x-1)`
`->2\sqrt{(3x-1)/x}=1+x/(3x-1)`
Điều kiện: `1/3<x<=1/2`
Đặt `\sqrt{(3x-1)/x}=a>0`
`->2a=1+1/a^2`
`->2a^3-a^2-1=0`
`->2a^3-2a^2+a^2-a+a-1=0`
`->2a^2(a-1)+a(a-1)+(a-1)=0`
`->(2a^2+a+1)(a-1)=0`
`->a-1=0`
`->a=1(TM)`
`->\sqrt{(3x-1)/x}=1`
`->(3x-1)/x=1`
`->3x-1=x`
`->2x=1`
`->x=1/2(TM)`
2)
`5\sqrt{x}+5/(2\sqrt{x})=2x+1/(2x)+4`
Điều kiện: `x>0`
Đặt `\sqrt{x}+1/(2\sqrt{x})=a>=\sqrt{2}` (AM-GM)
`->(\sqrt{x}+1/(2\sqrt{x}))^2=a^2`
`->x+1/(4x)+1=a^2`
`->x+1/(4x)=a^2-1`
`->5a=2(a^2-1)+4`
`->5a=2a^2-2+4`
`->2a^2-5a+2=0`
`->2a^2-4a-a+2=0`
`->2a(a-2)-(a-2)=0`
`->(2a-1)(a-2)=0` `(a>=\sqrt{2})`
`->a-2=0`
`->a=2`
`->\sqrt{x}+1/(2\sqrt{x})=2`
`->2x+1=4\sqrt{x}`
`->2x-4\sqrt{x}+1=0`
Đặt `t=\sqrt{x}>0`
`->2t^2-4t+1=0`
`Δ=8>0`
`->`\(\left[ \begin{array}{l}t=\dfrac{2-\sqrt{2}}{2}\\t=\dfrac{2+\sqrt{2}}{2}\end{array} \right.(TM)\)
`->`\(\left[ \begin{array}{l}x=\dfrac{3-2\sqrt{2}}{2}\\x=\dfrac{3+2\sqrt{2}}{2}\end{array} \right.(TM)\)
