Toán Lớp 9: nếu x lớn hơn 0 thì biểu thức x- √(x-1) ² cs giá trị bằng
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Toán Lớp 9: nếu x lớn hơn 0 thì biểu thức x- √(x-1) ² cs giá trị bằng
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x – \sqrt {{{\left( {x – 1} \right)}^2}} = 1\,khi:x \ge 1\\
x – \sqrt {{{\left( {x – 1} \right)}^2}} = 2x – 1\,khi:0 < x < 1
\end{array} \right.$
x – \sqrt {{{\left( {x – 1} \right)}^2}} \\
= x – \left| {x – 1} \right|\\
= \left[ \begin{array}{l}
x – \left( {x – 1} \right) = x – x + 1 = 1\left( {khi:x \ge 1} \right)\\
x – \left( {1 – x} \right) = x – 1 + x = 2x – 1\left( {khi:0 < x < 1} \right)
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x – \sqrt {{{\left( {x – 1} \right)}^2}} = 1\,khi:x \ge 1\\
x – \sqrt {{{\left( {x – 1} \right)}^2}} = 2x – 1\,khi:0 < x < 1
\end{array} \right.
\end{array}$