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(f) \( \quad x-5 ; 2 x-4 ; 6 x \) are the first three terms of a geometric sequence. De value(s) of \( x \).

Ask by Vaughn Vega. in South Africa
Jan 23,2025

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The values of \( x \) that make the terms \( x-5, 2x-4, 6x \) form a geometric sequence are \( x = -1 \) and \( x = 8 \).

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The Deep Dive

In a geometric sequence, the ratio of consecutive terms is constant. So, we can set up the equation by comparing the ratios of the first two terms and the last two terms: \[ \frac{2x - 4}{x - 5} = \frac{6x}{2x - 4} \] Cross-multiplying gives: \[ (2x - 4)(2x - 4) = 6x(x - 5) \] Expanding both sides leads to a quadratic equation. Solving this quadratic will yield the values of \( x \). Considering some troubleshooting tips, when checking for errors, ensure you've correctly expanded and simplified each axis in the equation. Watch out for negative signs and factor carefully if the quadratic seems unyielding; it's easy to misplace a term in the process!

Related Questions

PHIẾU: PHÉP CỘNG, TRỪ PHÂN SỐ HỌ VÀ TÊN:.............................LỚP 8A4 Bài 1. Thực hiện phép tính: a) \( \frac{x-5}{5}+\frac{1-x}{5} \) b) \( \frac{x-y}{8}+\frac{2 y}{8} \) c) \( \frac{x^{2}+1}{x-2}-\frac{1-2 x}{x-2} \) d) \( \frac{4 x+1}{3}-\frac{x-2}{3} \) e) \( \frac{4 x-1}{3 x^{2} y}-\frac{7 x-1}{3 x^{2} y} \) f) \( \frac{3 x+2 y}{x-y}-\frac{2 x+3 y}{x-y} \) d) \( \frac{5 x y^{2}-x^{2} y}{3 x y}-\frac{4 x y^{2}+x^{2} y}{3 x y} \) e) \( \frac{x+1}{a-b}+\frac{x-1}{a-b}-\frac{x+3}{a-b} \) f) \( \frac{5 x y-4 y}{2 x^{2} y^{3}}+\frac{3 x y+4 y}{2 x^{2} y^{3}} \) h) \( \frac{x^{2}+4}{x-2}+\frac{4 x}{2-x} \) i) \( \frac{2 x^{2}-x y}{x-y}+\frac{x y+y^{2}}{y-x}-\frac{2 y^{2}-x^{2}}{x-y} \) Bài 2: Thực hiện phép tính: a) \( \frac{2 x+4}{10}+\frac{2-x}{15} \) b) \( \frac{x^{2}}{x^{2}+3 x}+\frac{3}{x+3}+\frac{3}{x} \) c) \( \frac{2}{x+y}-\frac{1}{y-x}+\frac{-3 x}{x^{2}-y^{2}} \) d) \( \frac{4}{x+2}+\frac{2}{x-2}+\frac{5 x-6}{4-x^{2}} \); e) \( \frac{1-3 x}{2 x}+\frac{3 x-2}{2 x-1}+\frac{3 x-2}{2 x-4 x^{2}} \); f) \( \frac{x^{2}+2}{x^{3}-1}+\frac{2}{x^{2}+x+1}+\frac{1}{1-x} \) Bài 3. Làm tính trừ các phân thức sau: a) \( \frac{4 x+1}{3}-\frac{x-2}{3} \) b) \( \frac{4 x-1}{3 x^{2} y}-\frac{7 x-1}{3 x^{2} y} \) c) \( \frac{3 x+2 y}{x-y}-\frac{2 x+3 y}{x-y} \) Bài 4. Làm các phép tính a) \( \frac{x y-1}{2 x-y}-\frac{1-2 x^{2}}{y-2 x} \) b) \( \frac{3 x y^{2}+x^{2} y}{x^{2} y-x y^{2}}-\frac{3 x^{2} y+x y^{2}}{x y(x-y)} \) c) \( \frac{x+9}{x^{2}-9}-\frac{3}{x^{2}+3 x} \) Bài 5. Thực hiện phép tính a) \( \frac{5 x^{2}}{6 x-6 y}-\frac{2 x^{2}}{3 y-3 x} \) b) \( \frac{y}{x y-5 x^{2}}-\frac{25 x-15 y}{25 x^{2}-y^{2}} \) c) \( \frac{1}{2 x-3}-\frac{2}{2 x+3}-\frac{6}{4 x^{2}-9} \) Bài 6. Rút gọn rồi tính giá trị của biểu thức a) \( \frac{x+1}{x-1}-\frac{4 x}{x^{2}-1} \) với \( x=-2 \) b) \( \frac{1}{x y-x^{2}}-\frac{1}{y^{2}-x y} \) với \( x=-5 ; y=\frac{-1}{5} \) Bài 7. Chứng minh biểu thức sau không phụ thuộc vào x : \[ A=\frac{11 x}{2 x-3}-\frac{x-18}{3-2 x} \quad B=\frac{1}{x+1}-\frac{2 x}{x-1}+\frac{x+3}{x^{2}-1} \] Bài 8. Chứng minh biểu thức sau không phụ thuộc vào \( \mathrm{x}, \mathrm{y} \) : \[ D=\frac{2 x+1}{x-2 y}-\frac{5 y+2}{x-2 y}-\frac{y+1}{2 y-x} \quad K=\frac{x+y}{x}-\frac{x}{x-y}+\frac{y^{2}}{x^{2}-x y} \]
Algebra Vietnam Jan 24, 2025
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