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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
Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\) (đề bài)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
\(\Rightarrow\begin{cases}\frac{a}{b}=1\\\frac{b}{c}=1\\\frac{c}{d}=1\\\frac{d}{a}=1\end{cases}\Rightarrow\begin{cases}a=b\\b=c\\c=d\\d=a\end{cases}\)
\(\Rightarrow a=b=c=d\)
Thay \(b=a\) ; \(c=a\) ; \(d=a\) vào biểu thức \(M=\frac{2a-b}{c+d}=\frac{2b-c}{d+a}=\frac{2c-d}{a+b}=\frac{2d-a}{b+c}\) ta có :
\(M=\frac{2a-a}{a+a}=\frac{2a-a}{a+a}=\frac{2a-a}{a+a}=\frac{2a-a}{a+a}\)
\(M=\frac{1a}{2a}=\frac{1a}{2a}=\frac{1a}{2a}=\frac{1a}{2a}=\frac{1}{2}\)
Vậy \(M=\frac{1}{2}\)
Tham khảo nhé
https://olm.vn/hoi-dap/tim-kiem?id=1164587&subject=1&q=+++++++++++2a+b+c+da+=a+2b+c+db+=a+b+2c+dc+=a+b+c+2dd+T%C3%ADnh+M=a+bc+d++b+cd+a++c+da+b++d+ab+c+++++++++++
Ta có: \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d}{a+b+c+d}=4\)
=>2a+b+c+d=4a
=>2a=b+c+d
Tương tự ta có:2b=a+c+d 2c=a+b+d 2d=a+b+c
=>2a+2b=b+c+d+a+c+d
=>a+b+2c+2d
=>a+b=2c+2d
\(\Rightarrow\frac{a+b}{c+d}=2\)
Tương tự ta có:\(b+\frac{c}{d}+a=2\)
\(c+\frac{d}{a}+b=2\)
\(d+\frac{a}{b}+c=2\)
=>M=2+2+2+2=8
Bạn tham khảo tại đây:
Câu hỏi của Nguyễn Quỳnh Chi - Toán lớp 7 - Học toán với OnlineMath
Xét a+b+c+d = 0 ta có :
\(a+b=-c-d;b+c=-d-a;c+d=-a-b;d+a=-b-c\)
\(\Rightarrow M=\frac{-c-d}{c+d}+\frac{-d-a}{d+a}+\frac{-a-b}{a+b}+\frac{-b-c}{b+c}=-4\)
Xét a+b+c+d \(\ne0\) ta có :
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Leftrightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Leftrightarrow\frac{a+b+c+d}{a}=\frac{a+b+d+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)Thay vào M ta được : \(M=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=4\)
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Leftrightarrow\)\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Leftrightarrow\)\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
+) 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 ~
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\)
\(=\frac{a+b+2c+d+a+b+c+2d}{c+d}=\frac{2\left(a+b\right)}{c+d}+3=\)
Tương tự
\(=\frac{2\left(b+c\right)}{d+a}+3=\)
\(=\frac{2\left(c+d\right)}{a+b}+3=\)
\(=\frac{2\left(d+a\right)}{b+c}+3\)
\(\Rightarrow\frac{2\left(a+b\right)}{c+d}+3=\frac{2\left(b+c\right)}{d+a}+3=\frac{2\left(c+d\right)}{a+b}+3=\frac{2\left(d+a\right)}{b+c}+3\)
\(\Rightarrow\frac{2\left(a+b\right)}{c+d}=\frac{2\left(b+c\right)}{d+a}=\frac{2\left(c+d\right)}{a+b}=\frac{2\left(d+a\right)}{b+c}=\)
\(=\frac{2\left(a+b\right)+2\left(b+c\right)+2\left(c+d\right)+2\left(d+a\right)}{c+d+d+a+a+b+b+c}=\frac{4\left(a+b+c+d\right)}{2\left(a+b+c+d\right)}=2\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>2a+b+c+da =a+2b+c+db =a+b+2c+dc =a+b+c+2dd =2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2ca+b+c+d =4
=>2a+b+c+d=4a
=>2a=b+c+d
Tương tự ta có:2b=a+c+d
2c=a+b+d
2d=a+b+c
=>2a+2b=b+c+d+a+c+d=>a+b+2c+2d
=>a+b=2c+2d
=>a+b/c+d=2
Tương tự ta có:b+c/d+a=2
c+d/a+b=2
d+a/b+c=2
=>M=2+2+2+2=8
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