Bài 7: Phương trình quy về phương trình bậc hai
Hướng dẫn giải Bài 39 (Trang 57 SGK Toán Đại số 9, Tập 2)
<p>Giải phương trình bằng cách đưa về phương trình tích.</p>
<p>a) (3x<sup>2</sup> − 7x − 10)<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mrow><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo> </mo><mo>+</mo><mo> </mo><mo>(</mo><mn>1</mn><mo> </mo><mo>−</mo><mo> </mo><msqrt><mn>5</mn></msqrt><mo>)</mo><mi>x</mi><mo> </mo><mo>+</mo><mo> </mo><msqrt><mn>5</mn></msqrt><mo> </mo><mo>−</mo><mn>3</mn></mrow></mfenced></math> = 0;</p>
<p>b) x<sup>3</sup> + 3x<sup>2</sup> − 2x − 6 = 0 ; c) (x<sup>2</sup> − 1)(0,6x + 1) = 0,6x<sup>2</sup> +x ;</p>
<p>d) (x<sup>2</sup> + 2x − 5)<sup>2</sup> = (x<sup>2</sup> − x + 5)<sup>2</sup> .</p>
<p><strong>Giải</strong></p>
<p>a) (3x<sup>2</sup> − 7x − 10)<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mrow><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo> </mo><mo>+</mo><mo> </mo><mo>(</mo><mn>1</mn><mo> </mo><mo>−</mo><mo> </mo><msqrt><mn>5</mn></msqrt><mo>)</mo><mi>x</mi><mo> </mo><mo>+</mo><mo> </mo><msqrt><mn>5</mn></msqrt><mo> </mo><mo>−</mo><mo> </mo><mn>3</mn></mrow></mfenced></math> = 0</p>
<p>⇔ 3x<sup>2</sup> − 7x − 10 = 0 (1) hoặc 2x<sup>2</sup> + (1 − <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mn>5</mn></msqrt></math>)x + <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mn>5</mn></msqrt></math> − 3 = 0 (2)</p>
<p>Giải (1): Phương trình có a <strong>−</strong> b + c = 3 + 7 − 10 = 0</p>
<p>nên x<sub>1</sub> = −1, x<sub>2</sub> = −<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mo>−</mo><mn>10</mn></mrow><mn>3</mn></mfrac><mo> </mo><mo>=</mo><mo> </mo><mfrac><mn>10</mn><mn>3</mn></mfrac></math> </p>
<p>Giải (2): Phương trình có a + b + c = 2 + (1 −<math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mn>5</mn></msqrt></math> ) + <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mn>5</mn></msqrt></math> − 3 = 0</p>
<p>nên x<sub>3 </sub>= 1, x<sub>4 </sub>= <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><msqrt><mn>5</mn></msqrt><mo> </mo><mo>−</mo><mo> </mo><mn>3</mn></mrow><mn>2</mn></mfrac></math> </p>
<p>b) x<sup>3</sup> + 3x<sup>2 </sup>− 2x − 6 = 0 ⇔ x<sup>2</sup>(x + 3) − 2(x + 3) = 0 ⇔ (x + 3)(x<sup>2</sup> − 2) = 0</p>
<p> ⇔ x + 3 = 0 (1) hoặc x<sup>2</sup> − 2 = 0 (2)</p>
<p>Giải (1) ta được: x<sub>1 </sub>= −3</p>
<p>Giải (2) ta được: x<sub>2</sub> = −<math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mn>2</mn></msqrt></math> , x<sub>3</sub> = <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mn>2</mn></msqrt></math> </p>
<p>c) (x<sup>2</sup> − 1)(0,6x + 1) = 0,6x<sup>2 </sup>+ x ⇔ (0,6x + 1)(x<sup>2</sup> − x − 1) = 0</p>
<p>⇔ 0,6x + 1 = 0 (1) hoặc x<sup>2</sup> − x − 1 = 0 (2)</p>
<p>(1) ⇔ 0,6x + 1 = 0 ⇔ x<sub>1</sub> = −<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mrow><mn>0</mn><mo>,</mo><mn>6</mn></mrow></mfrac><mo> </mo><mo>=</mo><mo> </mo><mo>−</mo><mo> </mo><mfrac><mn>5</mn><mn>3</mn></mfrac></math> </p>
<p>(2): ∆ = (−1)<sup>2</sup> − 4.1. (−1) = 1 + 4 = 5, <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mo>∆</mo></msqrt></math> = <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mn>5</mn></msqrt></math> </p>
<p> x<sub>2 </sub>= <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>1</mn><mo> </mo><mo>−</mo><mo> </mo><msqrt><mn>5</mn></msqrt></mrow><mn>2</mn></mfrac></math>, x<sub>3</sub> = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>1</mn><mo> </mo><mo>+</mo><mo> </mo><msqrt><mn>5</mn></msqrt></mrow><mn>2</mn></mfrac></math></p>
<p>Vậy phương trình có ba nghiệm</p>
<p> x<sub>1 </sub>= −<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>5</mn><mn>3</mn></mfrac></math> , x<sub>2</sub> = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>1</mn><mo> </mo><mo>+</mo><mo> </mo><msqrt><mn>5</mn></msqrt></mrow><mn>2</mn></mfrac></math> , x<sub>3</sub> = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>1</mn><mo> </mo><mo>−</mo><mo> </mo><msqrt><mn>5</mn></msqrt></mrow><mn>2</mn></mfrac></math> </p>
<p>d) (x<sup>2</sup> + 2x − 5)<sup>2</sup> = (x<sup>2</sup> − x + 5)<sup>2</sup> ⇔ (x<sup>2</sup> + 2x − 5)<sup>2</sup> − (x<sup>2</sup> − x +5)<sup>2</sup> = 0</p>
<p> ⇔ (x<sup>2</sup> + 2x − 5 + x<sup>2</sup> − x + 5)(x<sup>2</sup> + 2x<sup>2</sup> − 5 − x<sup>2</sup> + x − 5) = 0</p>
<p> ⇔ (2x<sup>2</sup> +x )(3x − 10) = 0</p>
<p> ⇔ x(2x + 1)(3x − 10) = 0 </p>
<p> ⇔ x<sub>1 </sub>= 0, x<sub>2 </sub>= −<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>2</mn></mfrac></math>, x<sub>3</sub> = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>10</mn><mn>3</mn></mfrac></math></p>
<p>Vậy phương trình có 3 nghiệm:</p>
<p> x<sub>1</sub> = 0, x<sub>2</sub> = −<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>2</mn></mfrac></math> , x<sub>3</sub> = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>10</mn><mn>3</mn></mfrac></math></p>
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