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Ta có :
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+3=1\)
\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2=0\)
\(\Rightarrow abc\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2\right)=abc.0\)
\(\Rightarrow a^2b+b^2c+a^2c+b^2a+c^2a+c^2b+2abc=0\)
\(\Rightarrow\left(a^2b+ab^2\right)+\left(b^2c+abc\right)+\left(a^2c+abc\right)+\left(c^2a+c^2b\right)=0\)
\(\Rightarrow ab\left(a+b\right)+bc\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Rightarrow\left(ab+bc+ac+c^2\right)\left(a+b\right)=0\)
\(\Rightarrow\left[\left(ab+bc\right)+\left(ac+c^2\right)\right]\left(a+b\right)=0\)
\(\Rightarrow\left[b\left(a+c\right)+c\left(a+c\right)\right]\left(a+b\right)=0\)
\(\Rightarrow\left(a+c\right)\left(b+c\right)\left(a+b\right)=0\)
TH1 : \(a+c=0\)
\(\Rightarrow a=-c\)
\(\Rightarrow c^{2006}=a^{2006}\)
\(\Rightarrow P=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)\left(c^{2006}-a^{2006}\right)\)
\(=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)0\)
\(=0\)
CMTT đều có \(P=0\)
Vậy ...
\(1=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+\left(b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{a}{bc}\right)\)
\(\Leftrightarrow\left(b+c\right)\left(\dfrac{bc+ac+ab+a^2}{abc}\right)=0\)
\(\dfrac{\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(a+c\right)}{abc}=0\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)
Xét 3 TH
=> P=0 ( đề bài BT ở giữa có 1 số mũ sai nha )
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+ac+bc\right)=abc\)
\(\Leftrightarrow a\left(ab+ac+bc\right)+\left(b+c\right)\left(ab+ac+bc\right)-abc=0\)
\(\Leftrightarrow a\left(ab+ac+bc-bc\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow a^2\left(b+c\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow\left(a^2+ab+ac+bc\right)\left(b+c\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)
- Nếu \(a=-c\Rightarrow a^{2006}=c^{2006}\Rightarrow c^{2006}-a^{2006}=0\Rightarrow P=0\)
- Nếu \(a=-b\Rightarrow a^{2004}=b^{2004}\Rightarrow a^{2004}-b^{2004}=0\Rightarrow P=0\)
- Nếu \(b=-c\Rightarrow b^{2005}=-c^{2005}\Rightarrow b^{2005}+c^{2005}=0\Rightarrow P=0\)
Vậy \(P=0\)
tìm trc khi hỏi Câu hỏi của mai - Toán lớp 9 | Học trực tuyến
Bạn ghi nhầm đề bài thì phải, \(a,b,c\in Z\)* mới đúng
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
\(\Leftrightarrow\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}=0\Leftrightarrow2\left(\dfrac{a+b+c}{abc}\right)=0\Leftrightarrow a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^3=0\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3=-3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
Mà \(-3\left(a+b\right)\left(a+c\right)\left(b+c\right)⋮3\Rightarrow a^3+b^3+c^3⋮3\)
bài 28
\(P=\frac{\left[a^2-\left(b+c\right)^2\right]\left(a+b-c\right)}{\left(a+b+c\right)\left[\left(a-c\right)^2-b^2\right]}\)
=>\(P=\frac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-c-b\right)\left(a-c+b\right)}\)
=>\(P=1\)
Bài 30 phải là xy+y+x=3.
Ta có: xy+y+x=3 => (x+1)(y+1)=4(1)
yz+y+z=8 => (y+1)(z+1)=9(2)
zx+x+z=15 => (x+1)(z+1)=16(3)
Nhân (1), (2) và (3) theo vế, ta có:
[(x+1)(y+1)(z+1)]2=576
=> (x+1)(y+1)(z+1)=24(I) hoặc (x+1)(y+1)(z+1)=-24(II)
Lần lượt thay (1),(2),(3) vào (I),(II), tính x,y,z.
Kết quả: P=43/6 hoặc P=-79/6
Từ giả thiết suy ra: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a+b+c}\right)=0\)
\(\Rightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)
\(\Rightarrow\) (a + b)[c(a + b + c) + ab] = 0
\(\Rightarrow\) (a + b)(ac + ab + bc + c2) = 0
\(\Rightarrow\) (a + b)(b + c)(a + c) = 0
P = (a2004 - b2004)(b2005 + c2005)(c2006 - a2006)
= (a + b)(b + c)(a + c) = 0