Question

[Nghia Pham]

a) Cho tỉ lệ thức a/b = c/d. Chứng minh rằng a.b/c.d =a^2-b^2/c^2-d^2.

b) Tìm x nguyên để biểu thức sau là số nguyên P= 2x-3/x+1

Giúp mình đi mai thi rồi

Answer 1

[Nguyễn Huy Tú]

a) Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)

Ta có:
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (1)

\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2.\left(k^2-1\right)}{d^2.\left(k^2-1\right)}=\frac{b^2}{d^2}\) (2)

Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)

b) Giải:
Để \(P\in Z\Rightarrow2x-3⋮x+1\)

Ta có:
\(2x-3⋮x+1\)

\(\Rightarrow\left(2x+2\right)-5⋮x+1\)

\(\Rightarrow5⋮x+1\)

\(\Rightarrow x+1\in\left\{1;-1;5;-5\right\}\)

+) \(x+1=1\Rightarrow x=0\)

+) \(x+1=-1\Rightarrow x=-2\)

+) \(x+1=5\Rightarrow x=4\)

+) \(x+1=-5\Rightarrow x=-6\)

Vậy \(x\in\left\{0;-2;4;-6\right\}\)

\(\Rightarrow5⋮x+1\)

Answer 2

[Phạm Nguyễn Tất Đạt]

1)Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)(tính chất dãy tỉ số bằng nhau)

\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)

2)\(P=\frac{2x-3}{x+1}=\frac{2x+2-5}{x+1}=\frac{2\left(x+1\right)-5}{x+1}=2-\frac{5}{x+1}\)

\(\Rightarrow P\in Z\Leftrightarrow2-\frac{5}{x+1}\in Z\Leftrightarrow\frac{5}{x+1}\in Z\Leftrightarrow5⋮x+1\Leftrightarrow x+1\inƯ\left(5\right)\)

\(\Rightarrow x+1\in\left\{-1;-5;1;5\right\}\)

\(\Rightarrow x\in\left\{-2;-6;0;4\right\}\)