Cho a/b=b/c=c/a với a+b+c khác 0. Hãy tính giá trị biểu thức M=a^2+b^2+c^2/(a+b+c)^2
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Lời giải:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
$\Rightarrow ab+bc+ac=0$
Đặt $ab=x, bc=y, ac=z$ thì $x+y+z=0$
Có:
$M=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}$
$=\frac{b^3c^3+a^3c^3+a^3b^3}{(abc)^2}$
$=\frac{x^3+y^3+z^3}{xyz}=\frac{(x+y)^3-3xy(x+y)+z^3}{xyz}$
$=\frac{(-z)^3-3xy(-z)+z^3}{xyz}$
$+\frac{-z^3+3xyz+z^3}{xyz}=\frac{3xyz}{xyz}=3$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{3}{a^2b}+\frac{3}{ab^2}+\frac{1}{b^3}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{b^3}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{-3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{-3}{ab}\cdot\frac{-1}{c}=\frac{3}{abc}\)
Ta có: \(M=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK : a;b;c khác 0
Thấy : \(a^2+b^2+c^2=\left(a+b+c\right)^2\Leftrightarrow ab+bc+ac=0\) (1)
Ta có : \(P=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\)
Từ (1) suy ra : \(\left(b+c\right)a=-bc\Leftrightarrow\dfrac{b+c}{a}=\dfrac{-bc}{a^2}\)
CMTT ; ta có : \(\dfrac{c+a}{b}=\dfrac{-ac}{b^2};\dfrac{a+b}{c}=\dfrac{-ab}{c^2}\)
Suy ra : \(P=-\left(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}\right)=-\dfrac{a^3b^3+b^3c^3+a^3c^3}{a^2b^2c^2}\) (2)
Đặt : ab = x ; bc = y ; ac = z ; ta có : x + y + z = 0 \(\Rightarrow x^3+y^3+z^3=3xyz\) (3)
Từ (2) và (3) suy ra : \(P=-\dfrac{3xyz}{xyz}=-3\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{a}=\dfrac{1}{b}\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(A=\frac{(bc)^3+(2ac)^3+(2ab)^3}{8a^2b^2c^2}=\frac{(bc)^3+(2ac+2ab)^3-3.2ac.2ab(2ac+2bc)}{8a^2b^2c^2}\)
\(=\frac{(bc)^3+(-bc)^3+12a^2b^2c^2}{8a^2b^2c^2}=\frac{12}{8}=1,5\)
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từ giả thiết ta có
a+b+c=0
<=> a=-(b+c0
a2=b2 +c2 +2bc
tương tự b2=a2+c2+2ac
c2=a2+b2+2ab
thay vào Q ta đc
\(Q=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\)
\(Q=\frac{1}{a^2+b^2-a^2-b^2-2ab}+\frac{1}{b^2+c^2-b^2-c^2-2bc}+\frac{1}{a^2+c^2-a^2-c^2-2ac}\)
\(Q=\frac{-1}{2ab}-\frac{1}{2bc}-\frac{1}{2ac}\)
\(Q=\frac{-b-a-c}{2abc}\)
\(Q=\frac{-\left(a+b+c\right)}{2abc}\)
\(Q=0\)
Vậy với a,b,c khác 0, a+b+c=0 thì Q=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}=\dfrac{ab}{\left(a+b\right)^2-2ab-c^2}+\dfrac{bc}{\left(b+c\right)^2-2bc-a^2}+\dfrac{ca}{\left(a+c\right)^2-2ac-b^2}=\dfrac{ab}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+\dfrac{bc}{\left(a+b+c\right)\left(b+c-a\right)-2bc}+\dfrac{ac}{\left(a+b+c\right)\left(a+c-b\right)-2ac}=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ca}{-2ca}=-\dfrac{1}{2}.3=-\dfrac{3}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)
Suy ra \(a=b=c\).
Khi đó: \(M=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\).