Đáp án:
Giải thích các bước giải:
Bài 1 :
`a) sqrt { ( sqrt {7} - 2 )^2 } + sqrt { ( sqrt {7} - 3 )^2 }`
`= | sqrt {7} - 2 | + | sqrt {7} - 3 |`
`= sqrt {7} - 2 + ( 3 - sqrt {7} )`
`= sqrt {7} - 2 + 3 - sqrt {7}`
` = 1`
`b) sqrt { ( 4 - sqrt {17} )^2 } . ( 4 + sqrt {17} )`
`= | 4 - sqrt {17} | . ( 4 + sqrt {17} )`
`= ( sqrt {17} - 4 ) . ( 4 + sqrt {17} )`
`= ( sqrt {17} )^2 - 16`
`= 17 - 16`
`= 1`
`c) sqrt { ( sqrt {5} - 3 )^2 } + sqrt { ( 1 + sqrt {5} )^2 } - 2`
`= | sqrt {5} - 3 | + | 1 + sqrt {5} | - 2`
`= ( 3 - sqrt {5} ) + ( 1 + sqrt {5} ) - 2`
`= 3 - sqrt {5} + 1 + sqrt {5} - 2`
`= 2`
Bài 2 :
`a) sqrt { x^2 + 14x + 49 } - 3 = 2`
`<=> sqrt { ( x + 7 )^2 } = 5`
`<=> | x + 7 | = 5`
`<=>` \(\left[ \begin{array}{l}x+7=5\\x+7=-5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-2\\x=-12\end{array} \right.\)
Vậy `x ∈ { -2 ; -12 }`
`b) sqrt { 9x^2 - 6x + 1 } + 2 = 2x ( Đk: x ≥ 0 )`
`<=> sqrt { ( 3x - 1 )^2 } = 2x - 2`
`<=> | 3x - 1 | = 2x - 2`
`<=>` \(\left[ \begin{array}{l}3x-1=2x-2\\3x-1=-2x+2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x-2x=-2+1\\3x+2x=2+1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3\\5x=3\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3(tmđk)\\x=\dfrac{3}{5}(tmđk)\end{array} \right.\)
Vậy `x ∈ { 3/5 ; 3 }`
