Bài 6: Bất phương trình mũ và bất phương trình lôgarit
Hướng dẫn giải Bài 1 (Trang 89 SGK Toán Giải Tích 12)
<p><strong>Bài 1 (Trang 89 SGK Toán Giải Tích 12):</strong></p>
<p>Giải các bất phương trình mũ:</p>
<p>a) <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mn>2</mn><mrow><mo>-</mo><msup><mi mathvariant="normal">x</mi><mn>2</mn></msup><mo> </mo><mo>+</mo><mo> </mo><mn>3</mn><mi mathvariant="normal">x</mi></mrow></msup><mo> </mo><mo><</mo><mn>4</mn><mo> </mo></math></p>
<p>b) <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><mfrac><mn>7</mn><mn>9</mn></mfrac></mfenced><mrow><mn>2</mn><msup><mi mathvariant="normal">x</mi><mrow><mn>2</mn><mo> </mo></mrow></msup><mo> </mo><mo>-</mo><mo> </mo><mn>3</mn><mi mathvariant="normal">x</mi></mrow></msup><mo> </mo><mo> </mo><mo> </mo><mo>≥</mo><mo> </mo><mfrac><mn>9</mn><mn>7</mn></mfrac></math></p>
<p>c) <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mn>3</mn><mrow><mi mathvariant="normal">x</mi><mo> </mo><mo>+</mo><mo> </mo><mn>2</mn><mo> </mo></mrow></msup><mo> </mo><mo> </mo><mo>+</mo><mo> </mo><msup><mn>3</mn><mrow><mi mathvariant="normal">x</mi><mo> </mo><mo>-</mo><mn>1</mn></mrow></msup><mo> </mo><mo>≤</mo><mo> </mo><mn>28</mn></math></p>
<p>d) <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>4</mn><mi mathvariant="normal">x</mi><mo> </mo><mo> </mo><mo>-</mo><mo> </mo><mn>3</mn><mo>.</mo><msup><mn>2</mn><mi mathvariant="normal">x</mi></msup><mo> </mo><mo>+</mo><mo> </mo><mn>2</mn><mo> </mo><mo>></mo><mo> </mo><mn>0</mn></math>.</p>
<p><strong><em>Hướng dẫn giải:</em></strong></p>
<p>a) <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mn>2</mn><mrow><mo>-</mo><msup><mi mathvariant="normal">x</mi><mn>2</mn></msup><mo>+</mo><mn>3</mn><mi mathvariant="normal">x</mi></mrow></msup><mo> </mo><mo><</mo><mo> </mo><msup><mn>2</mn><mn>2</mn></msup><mo> </mo><mo>⇔</mo><mo> </mo><mo>-</mo><msup><mi mathvariant="normal">x</mi><mn>2</mn></msup><mo> </mo><mo>+</mo><mo> </mo><mn>3</mn><mi mathvariant="normal">x</mi><mo> </mo><mo><</mo><mo> </mo><mn>2</mn><mo> </mo><mo>⇔</mo><mo> </mo><msup><mi mathvariant="normal">x</mi><mn>2</mn></msup><mo> </mo><mo>-</mo><mo> </mo><mn>3</mn><mi mathvariant="normal">x</mi><mo> </mo><mo>+</mo><mo> </mo><mn>2</mn><mo> </mo><mo>></mo><mo> </mo><mn>0</mn><mo> </mo><mo>⇔</mo><mo> </mo><mi mathvariant="normal">x</mi><mo> </mo><mo><</mo><mo> </mo><mn>1</mn><mo> </mo><mi>hoặc</mi><mo> </mo><mi mathvariant="normal">x</mi><mo> </mo><mo>></mo><mo> </mo><mn>2</mn></math></p>
<p>Vậy <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">S</mi><mo> </mo><mo>=</mo><mo> </mo><mfenced><mrow><mo>-</mo><mo>∞</mo><mo>;</mo><mn>1</mn></mrow></mfenced><mo> </mo><mo>∪</mo><mo> </mo><mfenced><mrow><mn>2</mn><mo>;</mo><mo>+</mo><mo>∞</mo></mrow></mfenced></math>.</p>
<p>b)</p>
<p><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><mfrac><mn>7</mn><mn>9</mn></mfrac></mfenced><mrow><mn>2</mn><msup><mi mathvariant="normal">x</mi><mn>2</mn></msup><mo>-</mo><mn>3</mn><mi mathvariant="normal">x</mi></mrow></msup><mo> </mo><mo>≥</mo><mo> </mo><mfrac><mn>9</mn><mn>7</mn></mfrac><mo> </mo><mo>⇔</mo><mo> </mo><mn>2</mn><msup><mi mathvariant="normal">x</mi><mn>2</mn></msup><mo> </mo><mo>-</mo><mo> </mo><mn>3</mn><mi mathvariant="normal">x</mi><mo> </mo><mo>≤</mo><mo> </mo><mo>-</mo><mn>1</mn><mo> </mo><mo>⇔</mo><mo> </mo><mn>2</mn><msup><mi mathvariant="normal">x</mi><mn>2</mn></msup><mo> </mo><mo>-</mo><mo> </mo><mn>3</mn><mi mathvariant="normal">x</mi><mo> </mo><mo>+</mo><mo> </mo><mn>1</mn><mo> </mo><mo>≤</mo><mo> </mo><mn>0</mn><mo> </mo><mo>⇔</mo><mo> </mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>≤</mo><mi mathvariant="normal">x</mi><mo>≤</mo><mn>1</mn></math></p>
<p>Vậy <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">S</mi><mo> </mo><mo>=</mo><mo> </mo><mfenced open="[" close="]"><mrow><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>;</mo><mn>1</mn></mrow></mfenced></math></p>
<p>c)</p>
<p><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mn>3</mn><mrow><mi mathvariant="normal">x</mi><mo>+</mo><mn>2</mn></mrow></msup><mo> </mo><mo>+</mo><mo> </mo><msup><mn>3</mn><mrow><mi mathvariant="normal">x</mi><mo>-</mo><mn>1</mn></mrow></msup><mo> </mo><mo>≤</mo><mo> </mo><mn>28</mn><mo> </mo><mo>⇔</mo><mo> </mo><mn>9</mn><mo>.</mo><mn>3</mn><mi mathvariant="normal">x</mi><mo> </mo><mo>+</mo><mo> </mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mo>.</mo><mn>3</mn><mi mathvariant="normal">x</mi><mo> </mo><mo>≤</mo><mo> </mo><mn>28</mn><mo> </mo><mo>⇔</mo><mo> </mo><mn>3</mn><mi mathvariant="normal">x</mi><mo> </mo><mo>≤</mo><mo> </mo><mn>3</mn><mo> </mo><mo>⇔</mo><mo> </mo><mi mathvariant="normal">x</mi><mo> </mo><mo>≤</mo><mo> </mo><mn>1</mn></math></p>
<p>Vậy <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mo> </mo><mo>=</mo><mo> </mo><mo>(</mo><mo>-</mo><mo>∞</mo><mo>;</mo><mn>1</mn><mo>]</mo></math></p>
<p>d)</p>
<p>Đặt t = 2<sup>x</sup> ( t > 0), ta có bất phương trình t<sup>2</sup> - 3t + 2 > 0 <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇔</mo></math> 0 < t < 1 hoặc t > 2</p>
<p> <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇔</mo></math>2<sup>x</sup> < 1 hoặc 2<sup>x</sup> > 2 <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇔</mo></math>x < 0 hoặc x > 1.</p>
<p>Vậy <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">S</mi><mo> </mo><mo>=</mo><mo> </mo><mfenced><mrow><mo>-</mo><mo>∞</mo><mo>;</mo><mn>0</mn></mrow></mfenced><mo>∪</mo><mfenced><mrow><mn>1</mn><mo>;</mo><mo>+</mo><mo>∞</mo></mrow></mfenced></math></p>
Hướng dẫn Giải Bài 1 (Trang 89, SGK Toán Giải Tích 12)
GV: 
GV colearn
GV colearn
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Hướng dẫn Giải Bài 1 (Trang 89, SGK Toán Giải Tích 12)
GV: GV colearn
GV colearn