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3 tháng 12 2021

TH1: \(a+b+c+d\ne0\)

\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)

\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)

\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+2d}{d}\)

\(\Rightarrow a=b=c=d\)

\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)

\(\Rightarrow P=1+1+1+1\)

\(\Rightarrow P=4\)

TH2: \(a+b+c+d=0\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)

\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)

\(\Rightarrow P=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(d+a\right)}{d+a}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}\)

\(\Rightarrow P=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)

\(\Rightarrow P=-4\)

4 tháng 12 2021

bn mình nền của bn là nôb team trưởng team là t gaming

24 tháng 10 2021

a: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\)

hay \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)

26 tháng 1 2022

:)

- Ta có: \(\dfrac{a}{b}< \dfrac{c}{d}\) (gt)

=>\(ad< bc\) 

=>\(ad+ab< bc+ab\)

=>\(a\left(b+d\right)< b\left(a+c\right)\)

=>\(\dfrac{a}{b}< \dfrac{a+c}{b+d}\) (1)

- Ta có: \(\dfrac{c}{d}>\dfrac{a}{b}\) (gt)

=>\(bc>ad\)

=>\(bc+cd>ad+cd\)

=>\(c\left(b+d\right)>d\left(a+c\right)\)

=>\(\dfrac{c}{d}>\dfrac{a+c}{b+d}\) (2)

- Từ (1) và (2) suy ra: \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)

26 tháng 1 2022

- Mình lỡ làm rồi bạn tanjiro kamado gì đó :)

6 tháng 11 2017

\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}\dfrac{1}{3}\)(vìa+b+c+d\(\ne\)0)

=>3a=b+c+d: 3b=a+c+d=>3a-3b=b-a

=>3(a-b)=-(a-b)=>4(a-b)=0=>a=b

Tương tự => a=b=c=d=> A=4

3 tháng 2 2018

Ta có: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)

Ta có: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{a+b}{a+b+2\left(c+d\right)}=\dfrac{1}{3}\)

\(\Rightarrow3\left(a+b\right)=\left(a+b\right)+2\left(c+d\right)\)

\(\Rightarrow2\left(a+b\right)=2\left(c+d\right)\)

\(\Rightarrow a+b=c+d\)

\(\Rightarrow\dfrac{a+b}{c+d}=1\)

Tương tự:\(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)

Vậy A=4.

AH
Akai Haruma
Giáo viên
22 tháng 1 2022

Bài 1: Ta có:

\(M=\frac{ad}{abcd+abd+ad+d}+\frac{bad}{bcd.ad+bc.ad+bad+ad}+\frac{c.abd}{cda.abd+cd.abd+cabd+abd}+\frac{d}{dab+da+d+1}\)

\(=\frac{ad}{1+abd+ad+d}+\frac{bad}{d+1+bad+ad}+\frac{1}{ad+d+1+abd}+\frac{d}{dab+da+d+1}\)

$=\frac{ad+abd+1+d}{ad+abd+1+d}=1$

AH
Akai Haruma
Giáo viên
22 tháng 1 2022

Bài 2:

Vì $a,b,c,d\in [0;1]$ nên

\(N\leq \frac{a}{abcd+1}+\frac{b}{abcd+1}+\frac{c}{abcd+1}+\frac{d}{abcd+1}=\frac{a+b+c+d}{abcd+1}\)

Ta cũng có:
$(a-1)(b-1)\geq 0\Rightarrow a+b\leq ab+1$

Tương tự:

$c+d\leq cd+1$

$(ab-1)(cd-1)\geq 0\Rightarrow ab+cd\leq abcd+1$

Cộng 3 BĐT trên lại và thu gọn thì $a+b+c+d\leq abcd+3$

$\Rightarrow N\leq \frac{abcd+3}{abcd+1}=\frac{3(abcd+1)-2abcd}{abcd+1}$

$=3-\frac{2abcd}{abcd+1}\leq 3$

Vậy $N_{\max}=3$

22 tháng 8 2021

a) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\) và \(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)

\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

 

22 tháng 8 2021

b) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-b}{b}=\dfrac{b\left(k-1\right)}{b}=k-1\\\dfrac{c-d}{d}=\dfrac{d\left(k-1\right)}{d}=k-1\end{matrix}\right.\)\(\Rightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\)

c) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a}{c}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)

d) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a}{c}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)

14 tháng 11 2021

\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}b+c+d=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=2a\\a+b+c+d=2b\\a+b+c+d=2c\\a+b+c+d=2d\end{matrix}\right.\\ \Rightarrow2a=2b=2c=2d\\ \Rightarrow a=b=c=d\\ \Rightarrow A=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)

30 tháng 11 2021

ab+c+d=ba+c+d=ca+b+d=da+b+c=a+b+c+d3(a+b+c+d)=13⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩b+c+d=3aa+c+d=3ba+b+d=3ca+b+c=3d⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩a+b+c+d=2aa+b+c+d=2ba+b+c+d=2ca+b+c+d=2d⇒2a=2b=2c=2d⇒a=b=c=d⇒A=a+aa+a+a+aa+a+a+aa+a+a+aa+a=1+1+1+1=4