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[ 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 />

⇔ (x + 3)(x – 2) = 0....<br />

* y + 3 = 0 ⇔ x 2 + x – 2 + 3 = 0 ⇔ x 2 + x + 1 = 0 (vô nghiệm)<br />

b) (x – 4)( x – 5)( x – 6)( x – 7) = 1680 ⇔ (x 2 – 11x + 28)( x 2 – 11x + 30) = 1680<br />

Đặt x 2 – 11x + 29 = y , ta có:<br />

(x 2 – 11x + 28)( x 2 – 11x + 30) = 1680 ⇔ (y + 1)(y – 1) = 1680 ⇔ y 2 = 1681 ⇔ y = ± 41<br />

y = 41 ⇔ x 2 – 11x + 29 = 41 ⇔ x 2 – 11x – 12 = 0 (x 2 – x) + (12x – 12) = 0<br />

⇔ (x – 1)(x + 12) = 0.....<br />

* y = - 41 ⇔ x 2 – 11x + 29 = - 41 ⇔ x 2 – 11x + 70 = 0 ⇔ (x 2 – 2x. 11 2 +121<br />

c) (x 2 – 6x + 9) 2 – 15(x 2 – 6x + 10) = 1 (3)<br />

Đặt x 2 – 6x + 9 = (x – 3) 2 = y ≥ 0, ta có<br />

(3) ⇔ y 2 – 15(y + 1) – 1 = 0 ⇔ y 2 – 15y – 16 = 0 ⇔ (y + 1)(y – 15) = 0<br />

Với y + 1 = 0 ⇔ y = -1 (loại)<br />

Với y – 15 = 0 ⇔ y = 15 ⇒ (x – 3) 2 = 16 ⇔ x – 3 = ± 4<br />

+ x – 3 = 4 ⇔ x = 7<br />

+ x – 3 = - 4 ⇔ x = - 1<br />

d) (x 2 + 1) 2 + 3x(x 2 + 1) + 2x 2 = 0 (4)<br />

Đặt x 2 + 1 = y thì<br />

4 )+159<br />

(4) ⇔ y 2 + 3xy + 2x 2 = 0 ⇔ (y 2 + xy) + (2xy + 2x 2 ) = 0 ⇔ (y + x)(y + 2x) = 0<br />

+) x + y = 0 ⇔ x 2 + x + 1 = 0 : Vô nghiệm<br />

+) y + 2x = 0 ⇔ x 2 + 2x + 1 = 0 ⇔ (x + 1) 2 = 0 ⇔ x = - 1<br />

Bài 3:<br />

a) (2x + 1)(x + 1) 2 (2x + 3) = 18 ⇔ (2x + 1)(2x + 2) 2 (2x + 3) = 72. (1)<br />

Đặt 2x + 2 = y, ta có<br />

(1) ⇔ (y – 1)y 2 (y + 1) = 72 ⇔ y 2 (y 2 – 1) = 72<br />

⇔ y 4 – y 2 – 72 = 0<br />

Đặt y 2 = z ≥ 0 Thì y 4 – y 2 – 72 = 0 ⇔ z 2 – z – 72 = 0 ⇔ (z + 8)( z – 9) = 0<br />

* z + 8 = 0 ⇔ z = - 8 (loại)<br />

4 = 0<br />

DIỄN ĐÀN <strong>TOÁN</strong> - LÍ - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN<br />

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