Ta có:\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)

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

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

\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)=\(\frac{a+b+c+d+a+b+c+d+a+b+c+d+a+b+c+d}{a+b+c+d}=4\)(T/C)

Xét a+b+c+d=0

\(\Rightarrow a+c=-\left(b+d\right),a+b=-\left(c+d\right),b+c=-\left(a+d\right)\)

\(\Rightarrow M=-1+-1+-1+-1=-4\)

Xét \(a+b+c+d\ne0\Rightarrow a=b=c=d\)

\(\Rightarrow M=1+1+1+1=4\)

Vậy M=-4 hoặc M=4

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 ...

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\)\(\frac{d}{a+b+c}\)

\(\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}\)

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

Mà: \(a+b+c+d\ne0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c\)

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

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

\(\Rightarrow A=1+1+1+1=4\)

số đo slaf

4 

nhe sbn

bài dài 

lắm mình

vhir tiện ghi

thế này thôi

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)

\(\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}\)

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

Mà :\(a+b+c+d=0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c\)

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

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

\(\Rightarrow A=1+1+1+1=4\)

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

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

Do a + b + c + d khác 0 nên: b+c+d = a+c+d = a+b+d = a+b+c  => a = b = c = d

\(\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)\(\left(a=b=c=d\right)\)

\(\Rightarrow A=1+1+1+1=4\)

b phan d

Theo tính chất của dãy tỉ só bằng nhau ta có:

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

=> a=b=c=d

=> M=1+1+1+1=4

Đặt \(b+c+d=x;c+d+a=y;a+b+d=z;a+b+c=t\)

Có \(a=\frac{y+z+t-2x}{3}\)

Tương tự :\(b=\frac{x+z+t-2y}{3}\)

\(c=\frac{x+y+t-2z}{3}\)

\(d=\frac{y+x+z-2t}{3}\)

Đặt \(M=\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}\)

Thay vào biểu thức ta có :

\(M=\frac{\frac{y+z+t-2x}{3}}{x}+\frac{\frac{x+z+t-2y}{3}}{y}+\frac{\frac{x+y+t-2z}{3}}{z}+\frac{\frac{y+x+z-2t}{3}}{t}\)

\(=\frac{1}{3}\left(\frac{y+z+t-2x}{x}+\frac{x+z+t-2y}{y}+\frac{x+y+t-2z}{z}+\frac{x+z+y-2t}{t}\right)\)

\(=\frac{1}{3}\left[\left(\frac{y}{x}+\frac{x}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{t}{x}+\frac{x}{t}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)+\left(\frac{t}{y}+\frac{y}{t}\right)+\left(\frac{t}{z}+\frac{z}{t}\right)-8\right]\)

Sử dụng BĐT Cô-si suy ra \(Min_M=\frac{1}{3}.\left(12-8\right)=\frac{4}{3}\)

Dấu bằng xảy ra khi x = y = z = t hay \(b+c+d=a+b+c=c+d+a=b+d+a\) ( tự giải ra a=b=c=d)

Đặt \(N=\frac{b+c+d}{a}+\frac{c+a+d}{b}+\frac{d+a+b}{c}+\frac{a+b+c}{d}\)

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

Sử dụng Cô-si ra \(N\ge12\)

Dấu bằng xảy ra khi a=b=c=d ( tự giải ).

Do đó \(S=M+N\ge\frac{4}{3}+12=13\frac{1}{3}\)

Dấu bằng xảy ra khi \(a=b=c=d\)

\(\)

Áp dụng bđt cô - si cho 2 số không âm, ta được:

\(S=\text{ Σ}_{a,b,c,d}\left(\frac{a}{b+c+d}+\frac{b+c+d}{9a}\right)+\text{ Σ}_{a,b,c,d}\frac{8}{9}.\frac{b+c+d}{9a}\)

\(\ge8\sqrt[8]{\frac{a}{b+c+d}.\frac{b}{c+d+a}.\frac{c}{a+b+d}.\frac{d}{a+b+c}}\)\(\sqrt{\frac{b+c+d}{9a}.\frac{c+d+a}{9b}.\frac{a+b+d}{9c}.\frac{a+b+c}{9d}}\)

\(+\frac{8}{9}\left(\frac{b}{a}+\frac{c}{a}+\frac{d}{a}+\frac{c}{b}+\frac{d}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{d}{c}+\frac{a}{d}+\frac{b}{d}+\frac{c}{d}\right)\)

\(\ge\frac{8}{3}+\frac{8}{9}.12=\frac{40}{3}\)

Đẳng thức xảy ra khi a = b = c = d

Trl:

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