Bài 5: Phương Trình Chứa Dấu Giá Trị Tuyệt Đối
Hướng dẫn giải Bài 37 (Trang 51 SGK Toán Đại số 8, Tập 2)
<p>Giải các phương trình;</p>
<p>a) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>|</mo><mi>x</mi><mo>-</mo><mn>7</mn><mo>|</mo><mo>=</mo><mn>2</mn><mi>x</mi><mo>+</mo><mn>3</mn></math><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span>b) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>|</mo><mi>x</mi><mo>+</mo><mn>4</mn><mo>|</mo><mo>=</mo><mn>2</mn><mi>x</mi><mo>-</mo><mn>5</mn></math></p>
<p>c) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>|</mo><mi>x</mi><mo>+</mo><mn>3</mn><mo>|</mo><mo>=</mo><mn>3</mn><mi>x</mi><mo>-</mo><mn>1</mn></math><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span>d) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>|</mo><mi>x</mi><mo>-</mo><mn>4</mn><mo>|</mo><mo>+</mo><mn>3</mn><mi>x</mi><mo>=</mo><mn>5</mn></math></p>
<p><em><strong>Hướng dẫn:</strong></em></p>
<p>a) - Với <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>≥</mo><mn>7</mn></math> ta có: <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>-</mo><mn>7</mn><mo>=</mo><mn>2</mn><mi>x</mi><mo>+</mo><mn>3</mn><mo> </mo><mo>⇔</mo><mo> </mo><mi>x</mi><mo>=</mo><mo>-</mo><mn>10</mn><mo> </mo></math>(loại</p>
<p><span class="mce-nbsp-wrap" contenteditable="false"> </span> - Với x<7 ta có: <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mi>x</mi><mo>+</mo><mn>7</mn><mo>=</mo><mn>2</mn><mi>x</mi><mo>+</mo><mn>3</mn><mo>⇔</mo><mi>x</mi><mo>=</mo><mfrac><mn>4</mn><mn>3</mn></mfrac></math></p>
<p><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span>Tập nghiệm <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mo>=</mo><mo>{</mo><mfrac><mn>4</mn><mn>3</mn></mfrac><mo>}</mo></math></p>
<p>b) - Với <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>≥</mo><mo>-</mo><mn>4</mn></math> ta có: <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>+</mo><mn>4</mn><mo>=</mo><mn>2</mn><mi>x</mi><mo>-</mo><mn>5</mn><mo>⇔</mo><mi>x</mi><mo>=</mo><mn>9</mn></math> (nhận)</p>
<p><span class="mce-nbsp-wrap" contenteditable="false"> </span> - Với x<-3 ta có: <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mi>x</mi><mo>-</mo><mn>4</mn><mo>=</mo><mn>2</mn><mi>x</mi><mo>-</mo><mn>5</mn><mo>⇔</mo><mn>3</mn><mi>x</mi><mo>=</mo><mn>1</mn><mo>⇔</mo><mi>x</mi><mo>=</mo><mfrac><mn>1</mn><mn>3</mn></mfrac></math> (loại)</p>
<p><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span>Tập nghiệm phương trình có nghiệm <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mo>=</mo><mo>{</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mo>}</mo></math></p>
<p>c) - Với <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>≥</mo><mo>-</mo><mn>3</mn></math> ta có: <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>+</mo><mn>3</mn><mo>=</mo><mn>3</mn><mi>x</mi><mo>-</mo><mn>1</mn><mo>⇔</mo><mi>x</mi><mo>=</mo><mn>2</mn></math> (nhận)</p>
<p><span class="mce-nbsp-wrap" contenteditable="false"> </span> - Với x<-3 ta có: <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mi>x</mi><mo>-</mo><mn>3</mn><mo>=</mo><mn>3</mn><mi>x</mi><mo>-</mo><mn>1</mn><mo>⇔</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mo>-</mo><mn>1</mn></mrow><mn>2</mn></mfrac></math> (loại)</p>
<p><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span>Tập nghiệm phương trình đã cho S={2}</p>
<p>d) - Với <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>≥</mo><mn>4</mn></math> ta có: <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>-</mo><mn>4</mn><mo>+</mo><mn>3</mn><mi>x</mi><mo>=</mo><mn>5</mn><mo>⇔</mo><mn>4</mn><mi>x</mi><mo>=</mo><mn>9</mn><mo>⇔</mo><mi>x</mi><mo>=</mo><mfrac><mn>9</mn><mn>4</mn></mfrac></math> (loại)</p>
<p><span class="mce-nbsp-wrap" contenteditable="false"> </span> - Với x<4 ta có: <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mi>x</mi><mo>+</mo><mn>4</mn><mo>+</mo><mn>3</mn><mi>x</mi><mo>=</mo><mn>5</mn><mo>⇔</mo><mn>2</mn><mi>x</mi><mo>=</mo><mn>1</mn><mo>⇔</mo><mfrac><mn>1</mn><mn>2</mn></mfrac></math> (nhận)</p>
<p><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span><span class="mce-nbsp-wrap" contenteditable="false"> </span>Tập nghiệm cần tìm: <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mo>=</mo><mo>{</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>}</mo></math></p>
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