a: \(x^2-8x+5\)

\(=x^2-8x+16-11\)

\(=\left(x-4\right)^2-11\ge-11\forall x\)

Dấu '=' xảy ra khi x-4=0

=>x=4

b: \(a^3+b^3+c^3=3bac\)

=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

=>\(\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)=0\)

=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

=>\(a^2+b^2+c^2-ab-ac-bc=0\)

=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

=>a=b=c

\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)

\(=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3}{3^2}=\frac13\)

học sinh giỏi toán đâu hết rồi

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{b+c-5}{a}=\frac{a+c+2}{b}=\frac{a+b+3}{c}=\frac{b+c-5+a+c+2+a+b+3}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

=>b+c-5=2a; a+c+2=2b; a+b+3=2c

=>b+c=2a+5; a+c=2b-2; a+b=2c-3

\(\frac{1}{a+b+c}=\frac{b+c-5}{a}\)

=>\(\frac{1}{a+b+c}=2\)

=>\(a+b+c=\frac12\)

=>a+2a+5=0,5

=>3a=0,5-5=-4,5

=>a=-1,5

a+b+c=0,5

=>2b-2+b=0,5

=>3b=2,5

=>\(b=\frac{2.5}{3}=\frac56\)

a+b+c=0,5

=>c+2c-3=0,5

=>3c=3,5

=>\(c=\frac{3.5}{3}=\frac76\)

M=(a-3b)(b-c)(3c-a)

\(=\left(-1,5-3\cdot\frac56\right)\left(\frac56-\frac76\right)\left(3\cdot\frac76+1,5\right)\)

\(=\left(-1,5-\frac52\right)\cdot\frac{-2}{6}\left(\frac72+1,5\right)=-4\cdot5\cdot\frac{-2}{6}=20\cdot\frac26=\frac{20}{3}\)

a) Có:

 \(a+b+c=0\\\Leftrightarrow\left(a+b+c\right)^2=0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\\ \Leftrightarrow2ab+2bc+2ca=-1\\ \Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\\ \Leftrightarrow\left(ab+bc+ca\right)^2=\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}-0=\dfrac{1}{4} \)

câu (b) cho đa thức P (x) = cái gì?

Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn (biểu tượng $\sum$ bên trái khung soạn thảo)

Ta có: \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)

\(\Rightarrow\frac{abc}{c\left(a+b\right)}=\frac{abc}{a\left(b+c\right)}=\frac{abc}{b\left(c+a\right)}\)

\(\Rightarrow c\left(a+b\right)=a\left(b+c\right)=b\left(c+a\right)\)

\(\Rightarrow ac+bc=ab+ac=bc+ab\)

Lại có: \(ac+bc=ab+ac\)\(\Rightarrow bc=ab\)\(\Rightarrow a=c\)   (1)

 \(ab+ac=bc+ab\)\(\Rightarrow ac=bc\)\(\Rightarrow a=b\)              (2)

Từ (1) và (2) \(\Rightarrow a=b=c\) 

Ta có: \(P=\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\frac{a.a^2+b.b^2+c.c^2}{a^3+b^3+c^3}=\frac{a^3+b^3+c^3}{a^3+b^3+c^3}=1\)

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

Mà  \(a+b+c\ne0\left(gt\right)\)

\(\Leftrightarrow a=b=c\)

Do đó:

\(A=\frac{a^2+2b^2+6c^2}{\left(a+b+c\right)^2}+2015=\frac{a^2+2a^2+6c^2}{\left(a+a+a\right)^2}+2015=\frac{9a^2}{9a^2}+2015=1+2015=2016\)

Ta có:  a³(b - c) + b³(c - a) + c³(a - b)

= a³(b - c) - b³(b - c) - b³(a - b) + c³(a - b)

= (b - c)(a³ - b³) - (b³ - c³)(a - b)

= (b - c)(a - b)(a² + ab + b²) - (a - b)(b - c)(b² + bc + c²)

= (a - b)(b - c)(a² + ab + b² - b² - bc - c²)

= (a - b)(b - c)(a² + ab - bc - c²)

= (a - b)(b - c)[(a + c)(a - c) + b(a - c)]

= (a - b)(b - c)(a - c)(a + b + c) = 0   ( vì a + b + c = 0 )