Advertisements (Quảng cáo)
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 : \(\eqalign{
& 2x – 1 \ge 0 \Leftrightarrow 2x \ge 1 \cr
& \Leftrightarrow x \ge {1 \over 2} \Leftrightarrow x \ge 0,5 \cr} \)
Suy ra: \(\left| {2x – 1} \right| = 2x – 1\)
Advertisements (Quảng cáo)
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:
\(\eqalign{
& A = {{\left| {2x – 1} \right|} \over {2\left( {2x – 1} \right)}} = {{ – \left( {2x – 1} \right)} \over {2\left( {2x – 1} \right)}} = {1 \over 2} = – 0,5 \cr
& \Rightarrow \left| A \right| = \left| { – 0,5} \right| = 0,5 \cr} \)
Mục lục môn Toán 9 (SBT)