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

a=b=c => 3²⁰²⁰ = 3.a²⁰¹⁹ <=> 3²⁰¹⁹ = a²⁰¹⁹ => a=b=c=3

A= 1²⁰¹⁷ + 0²⁰¹⁸ + (-1)²⁰¹⁹ = 0

Đề đúng: \(M=\left(a^2+b^2-c^2\right)^2-4a^2b^2\)

a) Ta có:

\(M=\left(a^2+b^2-c^2\right)^2-4a^2b^2\)

\(M=\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)\)

\(M=\left[\left(a^2-2ab+b^2\right)-c^2\right]\left[\left(a^2+2ab+b^2\right)-c^2\right]\)

\(M=\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]\)

\(M=\left(a-b-c\right)\left(a-b+c\right)\left(a+b-c\right)\left(a+b+c\right)\)

b) Nếu a,b,c là độ dài 3 cạnh của tam giác thì:

\(\hept{\begin{cases}a+b>c\\c+a>b\\b+c>a\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b-c>0\\a-b+c>0\\a-b-c< 0\end{cases}}\) , mà a + b + c > 0

=> \(M< 0\)

\(x^4+2016x^2+2017x+2016\)

\(=x^4+2016x^2+2016x+x+2016\)

\(=\left(x^4+x\right)+\left(2016x^2+2016x+2016\right)\)

\(=x\left(x^3+1\right)+2016\left(x^2+x+1\right)\)

\(=x\left(x+1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2+x+2016\right)\)

=> \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4< =>\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4< =>\)2 + \(2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4< =>\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1< =>\frac{a+b+c}{abc}=1< =>\)a+b+c=abc

điều kiện ban đầu <=> (x-1)²+(y-2)²+(z-3)² \(\le1\)

áp dụng bdt sau (ax+ by+ cz)²\(\le\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)(bunhiacopxky với 3 số)

[ x-1 + 2(y-2) + 2(z-3)]² \(\le\left(1^2+2^2+2^2\right)\left[\left(x-1\right)^2+\left(y-2\right)^2+\left(z-2\right)^2\right]\le9.1=9\)

=>\(-3\le\) x-1 +2(y-2) +2(z-3) \(\le3\) <=> 8\(\le x+2y+2z\le14\)

a) 4x² - 20x + 25 - 36y²

= (2x - 5)² - 36y²

= (2x - 5 - 6y)(2x - 5 + 6y)

b) x³ + x² - 2x - 8

= (x³ - 8) + (x² - 2x)

= (x - 2)(x² + 2x + 4) + x(x - 2)

= (x - 2)(x² + 2x + 4 + x)

= (x - 2)(x² + 3x + 4)

d) x⁴ + 6x³ + 9x² - 16

= x²(x² + 6x + 9) - 16

= x²(x + 3)² - 16

= (x² + 3x)² - 16

= (x² + 3x - 4)(x² + 3x + 4)

= (x² + 4x - x - 4)(x² + 3x + 4)

= [x(x + 4) - (x + 4)](x² + 3x + 4)

= (x - 1)(x + 4)(x² + 3x + 4)

\(4x^2-\left(x+3\right).\left(x-3\right)+x\)\(=4x^2-\left(x^2-3^2\right)+x\)

                                                             \(=4x^2-\left(x^2-9\right)+x\)

                                                             \(=4x^2-x^2+9+x\)  

                                                             \(=3x^2+x+9\)  

#hok tốt#