Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+ac+bc\right)=abc\)
\(\Leftrightarrow a\left(ab+ac+bc\right)+\left(b+c\right)\left(ab+ac+bc\right)-abc=0\)
\(\Leftrightarrow a\left(ab+ac+bc-bc\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow a^2\left(b+c\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow\left(a^2+ab+ac+bc\right)\left(b+c\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)
- 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 \(P=0\)
a) phương trình \(x^3-3x^2+1\) có 3 nghiệm thực phân biệt là a,b,c(đề bài). Áp dụng Định lí Vi-ét cho đa thức bậc 3 ta có:\(\left\{{}\begin{matrix}a+b+c=3\\ab+bc+ac=0\\a.b.c=-1\end{matrix}\right.\)
ta có
a+b+c=3
<=>\(\left(a+b+c\right)^2=9\)
<=>\(a^2+b^2+c^2+2ab+2bc+2ac=9\)
<=>\(a^2+b^2+c^2=9\)
<=>\(\left(a^2+b^2+c^2\right)^2=81\)
<=>\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=81\)(1)
ta có ab+bc+ac=0
<=>\(\left(ab+bc+ac\right)^2=0\)
<=>\(a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=0\)
<=>\(a^2b^2+b^2c^2+a^2c^2-2.1.3=0\)
<=>\(a^2b^2+b^2c^2+a^2c^2=6\)(2)
Thay (2) vào (1) ta có \(a^4+b^4+c^4+2.6=81\)
<=>\(a^4+b^4+c^4=69\)
b) \(\dfrac{a+1}{\left(b+c\right)\left(1-a\right)+1}=\dfrac{a+1}{\left(3-a\right)\left(1-a\right)+1}=\dfrac{a+1}{3+a^2-4a+1}=\dfrac{a+1}{a^2-4a+4}=\dfrac{a+1}{\left(a-2\right)^2}\)
cmtt =>\(B=\dfrac{a+1}{\left(a-2\right)^2}+\dfrac{b+1}{\left(b-2\right)^2}+\dfrac{c+1}{\left(c-2\right)^2}\)=\(\dfrac{1}{a-2}+\dfrac{1}{b-2}+\dfrac{1}{c-2}+3\left[\dfrac{1}{\left(a-2\right)^2}+\dfrac{1}{\left(b-2\right)^2}+\dfrac{1}{\left(c-2\right)^2}\right]\)=\(\dfrac{3\left[\left(a-2\right)\left(b-2\right)\right]^2+3\left[\left(b-2\right)\left(c-a\right)\right]^2+3\left[\left(c-2\right)\left(a-2\right)\right]^2}{\left[\left(a-2\right)\left(b-2\right)\left(c-2\right)\right]^2}\)
đặt t=(a-2)(b-2);u=(b-2)(c-2);v=(c-2)(a-2) =>t+u+v=0
B thành \(\dfrac{3\left(t^2+u^2+v^2\right)}{t.u.v}\) bạn biến đổi để xuất hiện t+u+v
=>B=\(\dfrac{3\left(t+u+v\right)^2-6\left(t.u+u.v+t.v\right)}{t.u.v}=\dfrac{-6.\left(a-2\right)\left(b-2\right)\left(c-2\right)\left(a-2+b-2+c-2\right)}{t.u.v}=\dfrac{18}{\left(a-2\right)\left(b-2\right)\left(c-2\right)}\)
(a-2)(b-2)(c-2)= abc-2(ab+bc+ac)+4(a+b+c)-8=12-9=3
Vậy B=3
\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)
