Đáp án:
Câu 4:
$x=5; x=-3$
Giải thích các bước giải:
Câu 4:
\[\begin{array}{l}
a){\left( {x - 1} \right)^2} = 16\\
\Rightarrow {x^2} - 2x + 1 = 16\\
\Rightarrow {x^2} - 2x - 15 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 5\\
x = - 3
\end{array} \right.\\
c)x + 3 = 1\\
\Rightarrow x = - 2\\
b){\left( {x - 1} \right)^4} = 16 = {2^4}\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 2\\
x - 1 = - 2
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
d)\left[ \begin{array}{l}
x - 1 = 1\\
x - 1 = - 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.
\end{array}\]
Câu 5:
\[\begin{array}{l}
a)\left( {y - \sqrt 5 } \right).\left( {y + \sqrt 5 } \right)\\
b){x^2} - 2.\sqrt 7 .x + {\sqrt 7 ^2} = {\left( {x - \sqrt 7 } \right)^2}\\
c)4{a^2} + 4a\sqrt 3 + 3 = {\left( {2a} \right)^2} + 2.2a.\sqrt 3 + {\sqrt 3 ^2} = {\left( {2a + \sqrt 3 } \right)^2}
\end{array}\]
Câu 6:
\[\begin{array}{l}
a)\sqrt {{{\sqrt 3 }^2} + 2\sqrt 3 + 1} - \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + 2.2.\sqrt 3 + {1^2}} \\
= \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \sqrt {{{\left( {2\sqrt 3 + 1} \right)}^2}} \\
= \sqrt 3 + 1 - 2\sqrt 3 - 1 = - \sqrt 3 \\
b)\sqrt {{2^2} - 2.2.\sqrt 3 + {{\sqrt 3 }^2}} + \sqrt {{{\sqrt 3 }^2} - 2.\sqrt 3 + 1} \\
= \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
= 2 - \sqrt 3 + \sqrt 3 - 1 = 1\\
c)\sqrt {{4^2} + 2.4.\sqrt 5 + {{\sqrt 5 }^2}} + \sqrt {{4^2} - 2.4.\sqrt 5 + {{\sqrt 5 }^2}} \\
= \sqrt {{{\left( {4 + \sqrt 5 } \right)}^2}} + \sqrt {{{\left( {4 - \sqrt 5 } \right)}^2}} = 4 + \sqrt 5 + 4 - \sqrt 5 = 8\\
d)\sqrt {{{\sqrt 3 }^2} - 2.\sqrt 3 .\sqrt 5 + {{\sqrt 5 }^2}} + \sqrt {{{\sqrt 3 }^2} + 2.\sqrt 3 .\sqrt 5 + {{\sqrt 5 }^2}} \\
= \sqrt 5 - \sqrt 3 + \sqrt 3 + \sqrt 5 = 2\sqrt 5 \\
e)\sqrt {36 - 3} = \sqrt {33} \\
f)\sqrt {{x^2} + 2.x.\dfrac{1}{x} + {{\left( {\dfrac{1}{x}} \right)}^2}} + x = \sqrt {{{\left( {x + \dfrac{1}{x}} \right)}^2}} + \left( x \right) = - x - \dfrac{1}{x} + x = \dfrac{{ - 1}}{x}\left( {x < 0} \right)\\
g)A = \sqrt {\sqrt 5 - \sqrt {3 - \sqrt {{3^2} - 2.3.2.\sqrt 5 + {{\left( {2.\sqrt 5 } \right)}^2}} } } \\
= \sqrt {\sqrt 5 - \sqrt {3 - \left( {2\sqrt 5 - 3} \right)} } = \sqrt {\sqrt 5 - \sqrt {6 - 2\sqrt 5 } } = \sqrt {\sqrt 5 - \sqrt 5 + 1} = 1
\end{array}\]
