Toán Lớp 9: a)căn x ² ·6x+9 =căn 4+2 √3 b)căn 9-12x+4x ²=4 c)căn 4x ²+4x+1 d)căn x ²-8x+16=x+2 (chữ căn nghĩa là kí hieuj căn kéo dài nha)
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a)\sqrt {{x^2} – 6x + 9} = \sqrt {4 + 2\sqrt 3 } \\
\Leftrightarrow \sqrt {{{\left( {x – 3} \right)}^2}} = \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
\Leftrightarrow \left| {x – 3} \right| = \left| {\sqrt 3 + 1} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
x – 3 = \sqrt 3 + 1\\
x – 3 = – \sqrt 3 – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \sqrt 3 + 4\\
x = 2 – \sqrt 3
\end{array} \right.\\
Vậy\,x = 4 + \sqrt 3 ;x = 2 – \sqrt 3 \\
b)\sqrt {9 – 12x + 4{x^2}} = 4\\
\Leftrightarrow 9 – 12x + 4{x^2} = 16\\
\Leftrightarrow {\left( {2x – 3} \right)^2} = {4^2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x – 3 = 4\\
2x – 3 = – 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 7\\
2x = – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x = \dfrac{{ – 1}}{2}
\end{array} \right.\\
Vậy\,x = \dfrac{7}{2};x = – \dfrac{1}{2}\\
c)\sqrt {4{x^2} + 4x + 1} = ???\\
d)DKxd:x + 2 \ge 0 \Leftrightarrow x \ge – 2\\
\sqrt {{x^2} – 8x + 16} = x + 2\\
\Leftrightarrow \sqrt {{{\left( {x – 4} \right)}^2}} = x + 2\\
\Leftrightarrow \left| {x – 4} \right| = x + 2\\
\Leftrightarrow \left[ \begin{array}{l}
x – 4 = x + 2\\
x – 4 = – x – 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
– 4 = 2\left( {ktm} \right)\\
2x = 2
\end{array} \right.\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1
\end{array}$