Cho các số a, b, c dương nguyên đôi một khác nhau và thỏa mãn:
\(\dfrac{2a+b}{a+b}+\dfrac{2b+c}{b+c}+\dfrac{2c+d}{c+d}+\dfrac{2d+a}{d+a}=6\)
Chứng minh A=abcd là số chính phương
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2 tháng 8 2023
Điều kiện đã cho có thể được viết lại thành \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)
hay \(1-\dfrac{a}{a+b}-\dfrac{b}{b+c}+1-\dfrac{c}{c+d}-\dfrac{d}{d+a}=0\)
\(\Leftrightarrow\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)
\(\Leftrightarrow\dfrac{b^2+bc-ab-b^2}{\left(a+b\right)\left(b+c\right)}+\dfrac{d^2+da-cd-d^2}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow\left(c-a\right)\left[\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right]=0\)
\(\Leftrightarrow\dfrac{b}{\left(a+b\right)\left(b+c\right)}=\dfrac{d}{\left(c+d\right)\left(d+a\right)}\) (do \(c\ne a\))
\(\Leftrightarrow b\left(cd+ca+d^2+da\right)=d\left(ab+ac+b^2+bc\right)\)
\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)
\(\Leftrightarrow abc+bd^2-acd-b^2d=0\)
\(\Leftrightarrow ac\left(b-d\right)-bd\left(b-d\right)=0\)
\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow ac=bd\) (do \(b\ne d\))
Do đó \(A=abcd=ac.ac=\left(ac\right)^2\), mà \(a,c\inℕ^∗\) nên A là SCP (đpcm)
HN
7 tháng 3 2021
Ta có:
\(\dfrac{2a+b}{a+b}+\dfrac{2c+d}{c+d}+\dfrac{2b+c}{b+c}+\dfrac{2d+a}{d+a}=6\)
⇔ \(\left(\dfrac{2a+b}{a+b}-1\right)+\left(\dfrac{2c+d}{c+d}-1\right)+\left(\dfrac{2b+c}{b+c}-1\right)+\left(\dfrac{2d+a}{d+a}-1\right)=2\)
⇔ \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)
⇔ \(\left(1-\dfrac{a}{a+b}\right)-\dfrac{b}{b+c}+\left(1-\dfrac{c}{c+d}\right)-\dfrac{d}{d+a}=0\)
⇔ \(\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)
⇔ \(\dfrac{b\left(b+c\right)-b\left(a+b\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(d+a\right)-d\left(c+d\right)}{\left(c+d\right)\left(d+a\right)}=0\)
⇔ \(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
⇔ \(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}-\dfrac{d\left(c-a\right)}{\left(c+d\right)\left(d+a\right)}=0\)
⇔ \(\left(c-a\right)\left(\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right)=0\)
⇒ \(\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}=0\) \(\left(a\ne c\right)\)
⇒ \(b\left(c+d\right)\left(d+a\right)-d\left(a+b\right)\left(b+c\right)=0\)
⇔ \(\left(bc+bd\right)\left(d+a\right)-\left(ad+bd\right)\left(b+c\right)=0\)
⇔ \(bcd+abc+bd^2+abd-abd-acd-b^2d-bcd=0\)
⇔ \(abc+bd^2-acd-b^2d=0\)
⇔ \(ac\left(b-d\right)-bd\left(b-d\right)=0\)
⇔ \(\left(b-d\right)\left(ac-bd\right)=0\)
⇒ \(ac-bd=0\) \(\left(b\ne d\right)\)
⇔ \(ac=bd\)
Khi đó:
\(A=abcd=\left(ac\right)^2\)
⇒ \(ĐPCM\)
25 tháng 7 2020
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\Leftrightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)
\(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow b\left(c-a\right)\left(a+b\right)\left(b+c\right)-d\left(c-a\right)\left(c+d\right)\left(d+a\right)=0\)
\(\Leftrightarrow b\left(a+b\right)\left(b+c\right)-d\left(c+d\right)\left(d+a\right)=0\)
\(\Leftrightarrow bad+bd^2+bca+bcd-dab-dac-db^2-cbd=0\)
\(\Leftrightarrow bca-dca+bd^2-db^2=0\)
\(\Leftrightarrow\left(b-d\right)\left(ca-bd\right)=0\)
\(\Rightarrow ca=bd\Rightarrow abcd=bd^2\)
25 tháng 7 2020
Tách ra bạn có: \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\Leftrightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)
Quy đồng: \(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow b\left(c-a\right)\left(a+b\right)\left(b+c\right)-d\left(c-a\right)\left(c+d\right)\left(d+a\right)=0\)
Do a<>c:
\(\Leftrightarrow b\left(a+b\right)\left(b+c\right)-d\left(c+d\right)\left(d+a\right)=0\)
Phá ngoặc:
\(\Leftrightarrow bad+bd^2+bca+bcd-dab-dac-db^2-cbd=0\)
\(\Leftrightarrow bca-dca+bd^2-db^2=0\)
Phân tích đa thức thành nhân tử:
\(\Leftrightarrow\left(b-d\right)\left(ca-bd\right)=0\)
Do b<>d:
\(\Rightarrow ca=bd\Rightarrow abcd=bd^2\)