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ta có:\(\dfrac{a+b+c-d}{d}=\dfrac{b+c+d-a}{a}=\dfrac{a+d+a-b}{b}=\dfrac{d+a+b-c}{c}\)\(=>\dfrac{a+b+c-d}{d}+2=\dfrac{b+c+d-a}{a}+2=\dfrac{c+d+a-b}{b}+2=\dfrac{d+a+b-c}{c}+2\)\(=>\dfrac{a+b+c+d}{d}=\dfrac{b+c+d+a}{a}=\dfrac{c+d+a+b}{b}=\dfrac{d+a+b+c}{c}\)Nếu a+b+c+d=0=>a+b=-(c+d)
b+c=-(a+d)
c+d=-(a+b)
a+d=-(b+c)
thay vào bt M ta có:\(\dfrac{-\left(c+d\right)}{c+d}=\dfrac{-\left(d+a\right)}{d+a}=\dfrac{-\left(a+b\right)}{a+b}=\dfrac{-\left(b+c\right)}{b+c}\)=>-1-1-1-1=-4
Nếu a+b+c+d≠0
=>a=b=c=d thì lúc đó M=1+1+1+1=4
Vậy M=4 hoặc M=-4
trừ mỗi tỉ lệ cho 1 ta được:
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{2a+b+c+d}{a}-\frac{a}{a}=\frac{a+2b+c+d}{b}-\frac{b}{b}=\frac{a+b+2c+d}{c}-\frac{c}{c}=\frac{a+b+c+2d}{d}-\frac{d}{d}\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
+Nếu a+b+c+d\(\ne\)0 thì a=b=c=d lúc đó
M=1+1+1+1=4
+Nếu a+b+c+d=0 thì a+b=-(c+d);b+c=-(d+a);c+d=-(a+b);d+a=-(b+c) lúc đó:
M=(-1)+(-1)+(-1)+(-1)=-4
\(\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{a+b+2c+d+a+b+c+2d}{c+d}=\frac{2a+2b+3c+3d}{c+d}\)
\(=\frac{2\left(a+b\right)}{c+d}+\frac{3\left(c+d\right)}{c+d}=2.\frac{a+b}{c+d}+3\)
\(\frac{2a+b+c+d}{a}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d+a+b+c+2d}{a+d}=\frac{3a+3d+2c+2b}{a+d}\)
\(=\frac{3\left(a+d\right)}{a+d}+\frac{2\left(b+c\right)}{a+d}=3+2.\frac{b+c}{a+d}\)
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{2a+b+c+d+a+2b+c+d}{a+b}=\frac{3a+3b+2c+2d}{a+b}\)
\(=\frac{3\left(a+b\right)}{a+b}+\frac{2\left(c+d\right)}{a+b}=3+\frac{c+d}{a+b}.2\)
\(\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+2b+c+d+a+b+2c+d}{b+c}=\frac{3b+3c+2a+2d}{b+c}\)
\(=\frac{3\left(b+c\right)}{b+c}+\frac{2\left(a+d\right)}{b+c}=3+\frac{a+d}{b+c}.2\)
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)
\(\Rightarrow\frac{2a+b+c+d}{a}+\frac{a+2b+c+d}{b}+\frac{a+b+2c+d}{c}+\frac{a+b+c+2d}{d}=5.4=20\)
\(\Rightarrow3+\frac{a+b}{c+d}.2+3+\frac{b+c}{a+d}.2+3+\frac{c+d}{a+b}.2+3+\frac{d+a}{b+c}.2=20\)
\(\Rightarrow2.\left(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\right)=20-3-3-3-3\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{b+a}+\frac{d+a}{b+c}=8:2=4\)
vậy \(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=4\)
Cùng trừ mỗi tỉ số trên đi 1 đơn vị ta được:
\(\frac{2a+b+c+d}{a}-1=\frac{a+2a+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Suy ra 2 trường hợp:
TH1:
Nếu a+b+c+d ∉ 0 ⇒ a = b = c = d
➜ P= 1+1+1+1 = 4
TH2:
Nếu a+b+c+d =0
⇒ a+b = -(c+d)
b+c = -(d+a)
⇒ c+d = -(a+b)
d+a = -(b+c)
➜ P= (-1) + (-1) + (-1) + (-1) = -4
Mình thử nha :33
Ta có : \(\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}=\frac{1}{40}\)
\(\Leftrightarrow\left(a+b+c+d\right)\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}=\frac{1}{40}\cdot2000=50\) ( do \(a+b+c+d=2000\) )
\(\Rightarrow1+\frac{d}{a+b+c}+1+\frac{a}{b+c+d}+1+\frac{b}{c+d+a}+1+\frac{a}{b+c+d}=50\)
\(\Rightarrow S=50-4=46\)
Vậy : \(S=46\) với a,b,c,d thỏa mãn đề.
1, \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
Do đó \(\left\{{}\begin{matrix}3a=b+c+d\left(1\right)\\3b=a+c+d\left(2\right)\\3c=a+b+d\left(3\right)\\3d=a+b+c\left(4\right)\end{matrix}\right.\)
Từ (1) và (2) \(\Rightarrow3\left(a+b\right)=a+b+2c+2d\Leftrightarrow2\left(a+b\right)=2\left(c+d\right)\Leftrightarrow a+b=c+d\Leftrightarrow\dfrac{a+b}{c+d}=1\)
Tương tự cũng có: \(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)
\(\Rightarrow A=4\)
2, Có \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)\(\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
Do đó \(\dfrac{x^2}{4}=\dfrac{1}{4};\dfrac{y^2}{16}=\dfrac{1}{4};\dfrac{z^2}{36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(1;2;3\right),\left(-1;-2;-3\right)\)
Bài 2 :
a, Ta có : \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy ...
b, Ta có : \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{5+7}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\)
\(\Rightarrow x=2\)
\(\Rightarrow y=3\)
Vậy ...
Cộng thêm 1 vào mỗi đẳng thức, ta được:
\(\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
\(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)
Vì các tử số của mỗi tỉ số bằng nhau nên các mẫu số của mỗi tỉ số cũng bằng nhau
\(\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow M=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{a+d}{b+c}=1+1+1+1=4\)
