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Áp dụng tính chất của dãy tỉ số = 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
=> \(D=\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}\)
D = 1 + 1 + 1 + 1 = 4
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
=>a=b=c=d=>\(a+b=\frac{1}{2}\left(a+b+c+d\right)\)
\(\Rightarrow\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-b+2b-c+2c-d+2d-a}{a+b}\)
\(=\frac{2\left(a+b+c+d\right)-\left(a+b+c+d\right)}{\frac{1}{2}\left(a+b+c+d\right)}=\frac{a+b+c+d}{\frac{1}{2}\left(a+b+c+d\right)}=\frac{1}{\frac{1}{2}}=2\)
vậy A=2
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}\)
\(\frac{2a-a}{a+a}.4=\frac{a}{2a}.4=\frac{4a}{2a}=2\)
vậy A=2
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Leftrightarrow\)\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Leftrightarrow\)\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
+) Xét \(a+b+c+d=0\)
Suy ra :
\(a+b=-\left(c+d\right)\)
\(b+c=-\left(d+a\right)\)
\(c+a=-\left(b+d\right)\)
\(d+a=-\left(b+c\right)\)
Do đó : \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{c+b}\)
\(M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(d+a\right)}{d+a}+\frac{-\left(a+b\right)}{a+b}+\frac{-\left(b+c\right)}{b+c}\)
\(M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(M=-4\)
+) Xét \(a+b+c+d\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}=4\)
Do đó :
\(\frac{a+b+c+d}{a}=4\)\(\Leftrightarrow\)\(a+b+c+d=4a\) \(\left(1\right)\)
\(\frac{a+b+c+d}{b}=4\)\(\Leftrightarrow\)\(a+b+c+d=4b\) \(\left(2\right)\)
\(\frac{a+b+c+d}{c}=4\)\(\Leftrightarrow\)\(a+b+c+d=4c\) \(\left(3\right)\)
\(\frac{a+b+c+d}{d}=4\)\(\Leftrightarrow\)\(a+b+c+d=4d\) \(\left(4\right)\)
Từ (1), (2), (3) và (4) suy ra \(4a=4b=4c=4d\) \(\left(=a+b+c+d\right)\)
\(\Leftrightarrow\)\(a=b=c=d\)
\(\Rightarrow\)\(M=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)
\(\Rightarrow\)\(M=1+1+1+1=4\)
Vậy \(M=-4\) hoặc \(M=4\)
Chúc bạn học tốt ~
Ta có :
\(2a+2b+2c=by+cz+ax+cz+ax+by\)
\(\Leftrightarrow\)\(2\left(a+b+c\right)=2\left(ax+by+cz\right)\)
\(\Leftrightarrow\)\(a+b+c=ax+by+cz\)
+) \(a+b+c=ax+\left(by+cz\right)=ax+2a=a\left(x+2\right)\)
\(\Rightarrow\)\(\frac{1}{x+2}=\frac{a}{a+b+c}\) \(\left(1\right)\)
+) \(a+b+c=by+\left(ax+cz\right)=by+2b=b\left(y+2\right)\)
\(\Rightarrow\)\(\frac{1}{y+2}=\frac{b}{a+b+c}\) \(\left(2\right)\)
+) \(a+b+c=cz+\left(ax+by\right)=cz+2c=c\left(z+2\right)\)
\(\Rightarrow\)\(\frac{1}{z+2}=\frac{c}{a+b+c}\) \(\left(3\right)\)
Từ (1), (2) và (3) suy ra \(M=\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\)
\(M=\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\)
\(M=\frac{a+b+c}{a+b+c}=1\)
Vậy \(M=1\)
Chúc bạn học tốt ~
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\) (vì a + b + c + d khác 0) nên a = b = c = d
\(\Rightarrow\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}\)
\(=\frac{1}{2}.4=2\)
Áp dụng tính chất của dãy tỉ số bằng nhau vao A ,ta dc :
A = (2011a - 2010b + 2011b - 2010c + 2011c - 2010d + 2011d - 2010a) / (c + d + a + d + a + b + b + c)
A = (a + b + c + d) / (2a + 2b + 2c + 2d)
Ta có
a/2b = b/2c = c/2d = d/2a = (a + b + c + d) / (2a + 2b + 2c + 2d)
Vay : A = a/2b = b/2c = c/2d = d/2a
Phải sửa đề thành\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)
Ta có :\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\Rightarrow a=b=c=d\)
\(\Rightarrow P=\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}=\frac{a}{2a}.4=2\)
mình nói hướng làm cho bạn thôi nhé
nếu bạn đặt \(\frac{a}{b}\)= \(\frac{b}{c}\)=\(\frac{c}{d}\)=\(\frac{d}{a}\)=k vào thay vào rùi sẽ ra