K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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)

 

 

9 tháng 3 2021

\(\orbr{\begin{cases}\\\end{cases}}\)

7 tháng 3 2021

7 tháng 3 2021

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

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\Leftrightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)

\(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)

\(\Leftrightarrow b\left(c-a\right)\left(a+b\right)\left(b+c\right)-d\left(c-a\right)\left(c+d\right)\left(d+a\right)=0\)

\(\Leftrightarrow b\left(a+b\right)\left(b+c\right)-d\left(c+d\right)\left(d+a\right)=0\)

\(\Leftrightarrow bad+bd^2+bca+bcd-dab-dac-db^2-cbd=0\)

\(\Leftrightarrow bca-dca+bd^2-db^2=0\)

\(\Leftrightarrow\left(b-d\right)\left(ca-bd\right)=0\)

\(\Rightarrow ca=bd\Rightarrow abcd=bd^2\)

30 tháng 3 2022

sao cái dấu tương đương thứ 4 bạn bỏ c-a v ạ

 

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:

\(\Rightarrow ca=bd\Rightarrow abcd=bd^2\)