Question

[Đinh Thị Thảo Oanh]

1.a) Tìm cặp số nguyên (x; y) thỏa mãn: |y+2015|+32=\(\frac{2016}{\left(2x-6\right)^2+63}\).

b) Cho các số thực dương a, b, c thỏa mãn \(b^2\)=ac. Chứng minh rằng: \(\frac{a}{c}\)=\(\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\)

Answer 1

[Diệu Huyền]

\(1a,\) Ta có: \(\left(2x-6\right)^2\ge0\forall x\Rightarrow\left(2x-6\right)^2+36\ge36\forall x\)

\(\Rightarrow\frac{2016}{\left(2x-6\right)^2+63}\le\frac{2016}{63}=32\)

\(\Rightarrow\left|y+2015\right|+32\le32\)

\(\Rightarrow\left|y+2015\right|\le0\)

\(\Rightarrow\left|y+2015\right|=0\)

\(\Rightarrow y=-2015\)

\(\Rightarrow2x-6=0\Rightarrow x=3\)

Vậy \(x=3;y=-2015\)

Answer 2

[Vũ Minh Tuấn]

b)

Ta có: \(b^2=ac.\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c}.\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{2017b}{2017c}.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{a}{b}=\frac{b}{c}=\frac{2017b}{2017c}=\frac{a+2017b}{b+2017c}.\)

\(\Rightarrow\frac{a}{b}=\frac{a+2017b}{b+2017c}\)

\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{a+2017b}{b+2017c}\right)^2\)

\(\Rightarrow\left(\frac{a}{b}\right)^2=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}.\)

\(\Rightarrow\frac{a}{b}.\frac{a}{b}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\)

\(\Rightarrow\frac{a}{b}.\frac{b}{c}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}.\)

\(\Rightarrow\frac{a}{c}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\left(đpcm\right).\)

Chúc bạn học tốt!