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\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)=8\)
\(\Leftrightarrow\dfrac{a+b}{a}\times\dfrac{b+c}{b}\times\dfrac{a+c}{c}=8\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\)
~*~*~*~*~
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}\)
\(=\dfrac{3}{4}+\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\) (1)
\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{b}{b+c}-\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{c}{c+a}-\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{a+b}\left(1-\dfrac{b}{b+c}\right)+\dfrac{b}{b+c}\left(1-\dfrac{c}{c+a}\right)+\dfrac{c}{a+c}\left(1-\dfrac{a}{a+b}\right)\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{a+b}\times\dfrac{c}{b+c}+\dfrac{b}{b+c}\times\dfrac{a}{a+c}+\dfrac{c}{a+c}\times\dfrac{b}{a+b}\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}=\dfrac{3}{4}\)
\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)=\dfrac{3}{4}\times8abc\)
\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)+2abc=8abc\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\) luôn đúng
=> (1) đúng
Bạn cũng có thể giải bằng cách đặt \(x=\dfrac{a}{a+b};y=\dfrac{b}{b+c};z=\dfrac{c}{a+c}\).
Ta có: \(A=a\left(a^2-bc\right)+b\left(b^2-ac\right)+c\left(c^2-ab\right)=0\)
\(\Rightarrow A=a^3+b^3+c^3-3abc=0\) \(\Rightarrow A=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow A=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)
Xét \(M=a^2+b^2+c^2-ab-ac-bc=0\)
\(\Rightarrow2M=2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Rightarrow2M=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Rightarrow a-b=0;b-c=0;c-a=0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
Ta có:
\(\dfrac{b-c}{1\left(a-b\right)\left(a-c\right)}+\dfrac{c-a}{\left(b-c\right)\left(b-a\right)}+\dfrac{a-b}{\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{c-b}{1\left(a-b\right)\left(c-a\right)}+\dfrac{a-c}{\left(b-c\right)\left(a-b\right)}+\dfrac{b-a}{\left(c-a\right)\left(b-c\right)}\)
Quy đồng rút gọn ta được
\(=\dfrac{2\left(ab+bc+ca-a^2-b^2-c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{2\left[\left(a-b\right)\left(b-c\right)+\left(b-c\right)\left(c-a\right)+\left(c-a\right)\left(a-b\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=2\left(\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{c-a}\right)\)
PS: Hôm qua đi chơi nên nay mới giải nhé.
Mấy bài này đăng nhiều rồi bạn ;v
Bài 1: Nhân cả 2 vế cho a+b+c rồi rút gọn được đpcm
Bài 2: Thêm 1 rồi bớt 1 :v (x+y+xy+1-1)
Từ \(\dfrac{a-\left(c-b\right)}{b-c}+\dfrac{b-\left(a-c\right)}{c-a}+\dfrac{c-\left(b-a\right)}{a-b}=3\)
\(=>\dfrac{a}{b-c}+1+\dfrac{b}{c-a}+1+\dfrac{c}{a-b}+1=3\)
\(=>\dfrac{a}{b-c}-\dfrac{b}{a-c}-\dfrac{c}{b-a}=0\)
\(=>\dfrac{a}{b-c}=\dfrac{b}{a-c}+\dfrac{c}{b-a}=\dfrac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)
Nhân cả 2 vế với \(\dfrac{1}{b-c}\) ta được
\(\dfrac{a}{\left(b-c\right)^2}=\dfrac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(1\right)\)
Tương tự ta có:
\(\dfrac{b}{\left(c-a\right)^2}=\dfrac{c^2-bc+bc-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(2\right)\)
\(\dfrac{c}{\left(a-b\right)^2}=\dfrac{a^2-ca+cb-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(3\right)\)
Cộng theo vế (1);(2);(3) ta có ĐPCM
CHÚC BẠN HỌC TỐT.........
Đặt x = a - b, y = b - c, z = c - a
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=0\\ay+bz+cx=ab-ac+bc-ab+ac-bc=0\end{matrix}\right.\)
+ \(ay+bz+cx=0\)
\(\Rightarrow\dfrac{1}{y}\left(\dfrac{a}{y}+\dfrac{b}{z}+\dfrac{c}{x}\right)=0\)
\(\Rightarrow\dfrac{a}{y^2}+\dfrac{bx}{xyz}+\dfrac{cz}{xyz}=0\)
\(\Rightarrow\dfrac{a}{y^2}=\dfrac{-bx-cz}{xyz}\)
+ Tương tự : \(\dfrac{b}{z^2}=\dfrac{-cy-ax}{xyz}\)
\(\dfrac{c}{x^2}=\dfrac{-az-by}{xyz}\)
Do đó : \(\dfrac{a}{y^2}+\dfrac{b}{z^2}+\dfrac{c}{x^2}=\dfrac{-a\left(x+z\right)-b\left(x+y\right)-c\left(y+z\right)}{xyz}\)
\(=\dfrac{ay+bz+cx}{xyz}\) ( do x + y + z = 0)
\(=0\) ( do ay + bz + cx = 0 )