tim 3 so a,b,c khac nhau va khac 0 thoa man : a/(b+c) = b/(a+c) = c/(a+b)
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Cho 3 so a,b,c khac 0 va doi mot khac nhau thoa man a^2.(b+c)=b^2.(a+c)=2015 Tinh c^2.(a+b)
cho 3 so a,b,c khac thuoc Q khac nhau tung doi mot va khac 0 thoa man a/b+c=b/a+c=c/a+b
Chung minh b+c/a+a+c/b+a+b/c khong phu thuoc vao cac gia tri cua a,b,c
Cho 3 so khac nhau va khac 0 thoa man \(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}\).Khi do gia tri cua \(P=\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\)
Theo bài ra:
\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b};a\ne b\ne c;a,b,c\ne0\)
\(P=\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c+a+c+a+b}=\dfrac{a+b+c}{2a+2b+2c}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(hay:\dfrac{a}{b+c}=\dfrac{1}{2}\Rightarrow a=\dfrac{b+c}{2}\)
Thay \(a=\dfrac{b+c}{2}\) vào \(P\), ta có:
\(P=\dfrac{b+c}{\dfrac{b+c}{2}}+\dfrac{b+c+c}{b}+\dfrac{b+c+b}{c}\\ P=\dfrac{2\left(b+c\right)}{b+c}+\dfrac{2c+b}{b}+\dfrac{2b+c}{c}\\ P=2+\dfrac{2c}{b}+\dfrac{b}{b}+\dfrac{2b}{c}+\dfrac{c}{c}\\ P=2+\dfrac{2c}{b}+1+\dfrac{2b}{c}+1\\ P=\left(2+1+1\right)+\dfrac{2c}{b}+\dfrac{2b}{c}\\ P=4+\dfrac{2c}{b}+\dfrac{2b}{c}\\ P=4+\dfrac{2c+2b}{b+c}\\ P=4+\dfrac{2\left(b+c\right)}{b+c}\\ P=4+2\\ P=6\)
Vậy: \(P=6\)
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cho a,b,c la 3 so thuc doi mot khac nhau va khac khong, thoa man: a+1/b=b+1/c=c+1/a. hay chung minh rang abc=1 hoac abc=-1
cho 3 so thuc abc khac 0 va mot doi so khac nhau thoa man
a2 . ( b+ c ) = b2 . ( a + c ) = 2018
Tinh gia tri bieu thuc H = c2 . ( a+b)
Tim cac so a, b, c khac 0 thoa man: a + b - 2 / c = b + c +1 / a = c + a +1 / b = a + b +c / 2
Cho ba so a , b, c thuoc Q khac nhau tung doi mot va khac 0 thoa man \(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}\). Chung minh \(\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\) khong phu thuoc vao cac so a , b, c
Tim a,b,c thoa man khac 0 thoa man:
a+b-2/c= b+c+1/a= c+a+1/b= a+b+c/ 2
cho 3 so a,b,c khac 0 va thoa man a+b-c/c=a+c-b/b=b+c-a/a
tinh gia tri bieu thuc P=(a+b)(b+c)(c+a)=abc
Ta có : \(\frac{a+b-c}{c}=\frac{a+c-b}{b}=\frac{b+c-a}{a}\)
\(\Rightarrow\frac{a+b}{c}-\frac{c}{c}=\frac{a+c}{b}-\frac{b}{b}=\frac{b+c}{a}-\frac{a}{a}\)
\(\frac{a+b}{c}-1=\frac{c+b}{a}-1=\frac{a+c}{b}-1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)
Vậy \(P=\left(a+b\right)\left(b+c\right)\left(c+a\right)=2c.2a.2b=8abc\)
mà \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=abc\Rightarrow8abc=abc\Rightarrow abc=0\Rightarrow P=0\)
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