[ COMBO ] BỒI DƯỠNG TOÁN 8 NÂNG CAO VÀ PHÁT TRIỂN (VŨ HỮU BÌNH-NXBGD) & TUYỂN TẬP ĐỀ THI HSG TOÁN 8 (NGUYỄN VĂN TÚ-THCS THANH MỸ)
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Nơi bồi dưỡng kiến thức Toán - Lý - Hóa cho học sinh cấp 2+3 /<br />
Diễn Đàn Toán - Lý - Hóa 1000B Trần Hưng Đạo Tp.Quy Nhơn Tỉnh Bình Định<br />
(2) ⇔ (6x 3 – 18x 2 ) + (7x 2 – 21x) + (2x – 6) = 0<br />
⇔ 6x 2 (x – 3) + 7x(x – 3) + 2(x – 3) = 0 ⇔ (x – 3)(6x 2 + 7x + 2) = 0<br />
⇔ (x – 3)[(6x 2 + 3x) + (4x + 2)] = 0 ⇔ (x – 3)[3x(2x + 1) + 2(2x + 1)] = 0<br />
⇔ (x – 3)(2x + 1)(3x + 2) .....<br />
d) (x 2 + 5x) 2 – 2(x 2 + 5x) = 24 ⇔ [(x 2 + 5x) 2 – 2(x 2 + 5x) + 1] – 25 = 0<br />
⇔ (x 2 + 5x - 1) 2 – 25 = 0 ⇔ (x 2 + 5x - 1 + 5)( (x 2 + 5x - 1 – 5) = 0<br />
⇔ (x 2 + 5x + 4) (x 2 + 5x – 6) = 0 ⇔ [(x 2 + x) +(4x + 4)][(x 2 – x) + (6x – 6)] = 0<br />
⇔ (x + 1)(x + 4)(x – 1)(x + 6) = 0 ....<br />
e) (x 2 + x + 1) 2 = 3(x 4 + x 2 + 1) ⇔ (x 2 + x + 1) 2 - 3(x 4 + x 2 + 1) = 0<br />
⇔ (x 2 + x + 1) 2 – 3(x 2 + x + 1)( x 2 - x + 1) = 0<br />
⇔ ( x 2 + x + 1)[ x 2 + x + 1 – 3(x 2 - x + 1)] = 0 ⇔ ( x 2 + x + 1)( -2x 2 + 4x - 2) = 0<br />
⇔ (x 2 + x + 1)(x 2 – 2x + 1) = 0 ⇔ ( x 2 + x + 1)(x – 1) 2 = 0...<br />
f) x 5 = x 4 + x 3 + x 2 + x + 2 ⇔ (x 5 – 1) – (x 4 + x 3 + x 2 + x + 1) = 0<br />
⇔ (x – 1) (x 4 + x 3 + x 2 + x + 1) – (x 4 + x 3 + x 2 + x + 1) = 0<br />
⇔ (x – 2) (x 4 + x 3 + x 2 + x + 1) = 0<br />
+) x – 2 = 0 ⇔ x = 2<br />
+) x 4 + x 3 + x 2 + x + 1 = 0 ⇔ (x 4 + x 3 ) + (x + 1) + x 2 = 0 ⇔ (x + 1)(x 3 + 1) + x 2 = 0<br />
⇔ (x + 1) 2 (x 2 – x + 1) + x 2 = 0 ⇔ (x + 1) 2 [(x 2 – 2.x. 1 2 + 1 4 ) + 3 4 ] + x2 = 0<br />
⎡<br />
⎤<br />
2<br />
2<br />
⎛ 1 ⎞ 3<br />
⇔ (x + 1) 2 ⎢⎜<br />
x + ⎟ + ⎥<br />
⎢⎣ ⎝ 2 ⎠ 4 ⎥⎦ + ⎛ ⎞<br />
x2 = 0 Vô nghiệm vì (x + 1) 2 ⎢⎜<br />
⎟<br />
⎢⎣<br />
⎝ ⎠<br />
không xẩy ra dấu bằng<br />
Bài 2:<br />
⎡<br />
1 3<br />
x + +<br />
2 4<br />
a) (x 2 + x - 2)( x 2 + x – 3) = 12 ⇔ (x 2 + x – 2)[( x 2 + x – 2) – 1] – 12 = 0<br />
⇔ (x 2 + x – 2) 2 – (x 2 + x – 2) – 12 = 0<br />
Đặt x 2 + x – 2 = y Thì<br />
(x 2 + x – 2) 2 – (x 2 + x – 2) – 12 = 0 ⇔ y 2 – y – 12 = 0 ⇔ (y – 4)(y + 3) = 0<br />
* y – 4 = 0 ⇔ x 2 + x – 2 – 4 = 0 ⇔ x 2 + x – 6 = 0 ⇔ (x 2 + 3x) – (2x + 6) = 0<br />
⎤<br />
⎥<br />
⎥<br />
⎦ ≥ 0 nhưng<br />
DIỄN ĐÀN <strong>TOÁN</strong> - LÍ - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN<br />
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