Cho:
\(A = {{\sqrt {4{x^2} - 4x + 1} } \over {4x - 2}}.\)
Chứng minh: \(\left| A \right| = 0,5\) với \(x \ne 0,5.\)
Gợi ý làm bài
Ta có:
\(A = {{\sqrt {4{x^2} - 4x + 1} } \over {4x - 2}} = {{\sqrt {{{\left( {2x - 1} \right)}^2}} } \over {4x - 2}} = {{\left| {2x - 1} \right|} \over {2\left( {2x - 1} \right)}}\)
- Nếu : \(\eqalign{
& 2x - 1 \ge 0 \Leftrightarrow 2x \ge 1 \cr
& \Leftrightarrow x \ge {1 \over 2} \Leftrightarrow x \ge 0,5 \cr} \)
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Suy ra: \(\left| {2x - 1} \right| = 2x - 1\)
Ta có: \(A = {{\left| {2x - 1} \right|} \over {2\left( {2x - 1} \right)}} = {{2x - 1} \over {2\left( {2x - 1} \right)}} = {1 \over 2} = 0,5\)
- Nếu: \(\eqalign{
& 2x - 1 < 0 \Leftrightarrow 2x < 1 \cr
& \Leftrightarrow x < {1 \over 2} \Leftrightarrow x < 0,5 \cr} \)
Suy ra: \(\left| {2x - 1} \right| = - (2x - 1)\)
Ta có:
\(\eqalign{
& A = {{\left| {2x - 1} \right|} \over {2\left( {2x - 1} \right)}} = {{ - \left( {2x - 1} \right)} \over {2\left( {2x - 1} \right)}} = {1 \over 2} = - 0,5 \cr
& \Rightarrow \left| A \right| = \left| { - 0,5} \right| = 0,5 \cr} \)