\(M=4-5x-4x^2\)

      \(=-\left(4x^2+5x-4\right)\)

       \(=-\left(4x^2+4x+1-5\right)\)

       \(=-\left[\left(2x+1\right)^2-4\right]\)

         \(=-\left(2x+1\right)^2+4\)

Vì \(-\left(2x+1\right)^2\le0\)với mọi x

\(\Rightarrow-\left(2x+1\right)^2+4\le4\)với mọi x

Dấu "=" xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\) 

                          \(\Leftrightarrow2x+1=0\)

                         \(\Leftrightarrow x=\frac{-1}{2}\)

Vậy max M=4 khi \(x=-\frac{1}{2}\)

\(f\left(x\right)=\left(x-a\right)\left(x-b\right)\left(x-c\right)\)

\(=\left(x^2-ax-bx+ab\right)\left(x-c\right)\)

\(=x^3-\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)

\(=x^3+\left(ab+bc+ca\right)x+abc\)

\(-f\left(-x\right)=-\left[\left(-x-a\right)\left(-x-b\right)\left(-x-c\right)\right]\)

\(=-\left[\left(x^2+ax+bx+ab\right)\left(-x-c\right)\right]\)

\(=-\left[-x^3-\left(a+b+c\right)x^2-\left(ab+bc+ca\right)x-abc\right]\)

\(=-\left[-x^3-\left(ab+bc+ca\right)x-abc\right]\left(a+b+c=0\right)\)

\(=x^3+\left(ab+bc+ca\right)x+abc\)

Sửa đề:

\((2x^2+x-2015)^2+4(x^2-5x-2016)^2=4(2x^2+x-2015)(x^2-5x-2016)\)

\(\Rightarrow\left(2x^2+x-2015\right)^2-2.\left(2x^2+x-2015\right).2.\left(x^2-5x-2016\right)+[2.\left(x^2-5x-2016\right)]^2=0\)

\(\Rightarrow[2x^2+x-2015-2.\left(x^2-5x-2016\right)]^2=0\)

\(\Rightarrow11x+2017=0\)

\(\Rightarrow x=\frac{-2017}{11}\)

1,\(A=2x^2-6x+7\)

   \(=2\left(x^2-3x+\frac{9}{4}\right)+\frac{5}{2}\)

   \(=2\left(x-\frac{3}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)

Dấu "=" khi \(x=\frac{3}{2}\)

2,\(B=\frac{2x^2-6x+5}{x^2-2x+1}\left(ĐKXĐ:x\ne1\right)\)

\(\Leftrightarrow Bx^2-2Bx+B=2x^2-6x+5\)

\(\Leftrightarrow x^2\left(B-2\right)+2x\left(3-B\right)+B-5=0\)(1) 

*Với B = 2 thì \(\left(1\right)\Leftrightarrow x^2\left(2-2\right)+2x\left(3-2\right)+2-5=0\)

                                \(\Leftrightarrow2x-3=0\)

                                \(\Leftrightarrow x=\frac{3}{2}\left(TmĐKXĐ\right)\)

*Với \(B\ne2\)thì pt (1) là pt bậc 2 ẩn x tham số B

Pt (1) có nghiệm khi \(\Delta\ge0\)

                          \(\Leftrightarrow\left(3-B\right)^2-\left(B-2\right)\left(B-5\right)\ge0\)

                           \(\Leftrightarrow9-6B+B^2-B^2+7B-10\ge0\)

                           \(\Leftrightarrow B\ge1\)

Dấu "=" xảy ra khi \(\left(1\right)\Leftrightarrow-x^2+4x-4=0\)

                                        \(\Leftrightarrow-\left(x-2\right)^2=0\)

                                        \(\Leftrightarrow x=2\left(TmĐKXĐ\right)\)

Thấy 1 < 2 nên B_Min = 1<=> x = 2

Vậy ....

A=(9x²-6x+1)+(7x²+7)-1=(3x²+1)²+7(x²+7)-1

Vì: (3x²+1)²\(\ge\)0 và 7(x²+7)\(\ge\)0

Nên:A\(\ge\) -1

B=\(\frac{A-2}{\left(x-1\right)^2}\)\(\ge\)  -3

\(x^3-12x-16=0\Leftrightarrow x^2\left(x+2\right)-2x\left(x+2\right)-8\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x-8\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[x\left(x-4\right)+2\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)^2\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=4\end{cases}}\)

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

\(\Leftrightarrow2\left(ab+bc+ca\right)=2^2-2\)

\(\Leftrightarrow2\left(ab+bc+ca\right)=2\Leftrightarrow ab+bc+ca=1\)

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

\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(a+b\right)\right]\left[c\left(b+c\right)+a\left(b+c\right)\right]\)

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

Ta có : theo điều kiện cho trước:

 a + b + c =2

<=> \(\left(a+b+c\right)^2=4\)

<=> \(a^2+b^2+c^2+2ab+2ac+2bc=4\)

<=> \(2+2\left(ab+ac+bc\right)=4\)

<=> \(2\left(ab+ac+bc\right)=2\)

<=> \(ab+ac+bc=1\)

<=> \(\left(ab+ac+bc\right)^2=1\)

<=> \(a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+a^2bc+abc^2\right)=1\)

<=> \(a^2b^2+b^2c^2+a^2c^2=1-2\left(ab^2c+a^2bc+abc^2\right)\)

Theo đề bài ta có :

M = \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)

<=> \(\left(a^2b^2+a^2+b^2+1\right)\left(c^2+1\right)\)

<=> \(a^2b^2c^2+a^2b^2+a^2c^2+a^2+b^2c^2+b^2+c^2+1\)

<=> \(a^2b^2c^2+1-2ab^2c-2a^2bc-2abc^2+3\)

<=> \(a^2b^2c^2-2ab^2c-2a^2bc-2abc^2+4\)

<=> \(abc\left(abc-2b-2a-2c\right)+4\)

<=> \(abc\left\{abc-2\left(a+b+c\right)\right\}+4\)

<=> \(abc\left(abc-4\right)+4\)

<=> \(a^2b^2c^2-4abc+4\)

<=> \(\left(abc\right)^2-4abc+4\)

<=> \(\left(abc-2\right)^2\left(đpcm\right)\)