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

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

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

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

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

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

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

<=> \(abc\left(abc-4\right)+4\)

<=> \(a^2b^2c^2-4abc+4\)

<=> \(\left(abc\right)^2-4abc+4\)

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

P=(x-2012)^2 +(x+2013)^2

đặt x-2012=t ta được:

P=t^2+(t+4025)^2

  =t^2+t^2+8050t+4025^2

  =2t^2+8050t+4025^2

  =2(t^2+4024t)+4025^2

  =2(t+4025/2)^2+4025^2-4025^2/2

Dấu '=' xảy ra khi t+4025/2=0 =>t=-4025/2

=>x-2012=-4025/2

=>x=-1/2

Vậy GTNN của P là P=4025^2-4025^2/2 với x=-1/2

Chúc bn hok tốt

Đúng thì k nha

Áp dụng BĐT cô si cho 2 số dương ta được

 (x-2012)^2+(x+2013)^2>=2(x-2012)(x+2013)

=> P >= 2(x^2+x-4050156)

= 2(x^2+1/4)-8100312,5 >= -8100312,5

Min P=8100312,5

Dấu "=" xảy ra <=> x= -1/2