Giải thích các bước giải:
Bài `1` : Tìm `x`
`a)`
`(x-2,5)^4=(x-2,5)^2`
`=>(x-2,5)^4-(x-2,5)^2=0`
`=>(x-2,5)^2 .(x-2,5)^2 -(x-2,5)^2 .1=0`
`=>(x-2,5)^2 .[(x-2,5)^2-1]=0`
`=>`\(\left[ \begin{array}{l}(x-2,5)^2=0\\(x-2,5)^2-1=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x-2,5=0\\(x-2,5)^2=0+1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=0+2,5\\(x-2,5)^2=1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=2,5\\(x-2,5)^2=1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=2,5\\ \left[ \begin{array}{l}x-2,5=1\\x-2,5=-1\end{array} \right.\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=2,5\\ \left[ \begin{array}{l}x=1+2,5\\x=-1+2,5\end{array} \right.\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=2,5\\ \left[ \begin{array}{l}x=3,5\\x=1,5\end{array} \right.\end{array} \right.\)
Vậy `x\in{2,5;3,5;1,5}`
`b)`
`x.sqrt{x}-2=0`
`=>sqrt{x}.sqrt{x}.sqrt{x}-2=0`
`=>sqrt{x.x.x}=0+2`
`=>sqrt{x^3}=2`
`=>x^3=2^2`
`=>x^3=4`
`=>x=`$\sqrt[3]{4}$
Vậy `x=`$\sqrt[3]{4}$
Bài `2` : So sánh
`a)`
`sqrt{6}` và `3`
Ta có:
`sqrt{6}<sqrt{9}(sqrt{9}=sqrt{3^2}=3)`
`=>sqrt{6}<3`
Vậy `sqrt{6}<3`
`b)`
`sqrt{7}+sqrt{15}` và `7`
Ta có:
`7=3+4` mà `3=sqrt{9};4=sqrt{16}`
`=>7=sqrt{9}+sqrt{16}`
Lại có:
`sqrt{7}<sqrt{9}(text{vì}\quad 7<9)`
`sqrt{15}<sqrt{16}(text{vì}\quad 15<16)`
`=>sqrt{7}+sqrt{15}<sqrt{9}+sqrt{16}`
`=>sqrt{7}+sqrt{15}<3+4`
`=>sqrt{7}+sqrt{15}<7`
Vậy `sqrt{7}+sqrt{15}<7`
