Đặt: \(A=2x^2+5x+4\)

\(2A=4x^2+10x+8\)

\(2A=4x^2+10x+\frac{25}{4}+\frac{7}{4}\)

\(2A=\left(2x+\frac{5}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)

\(A\ge\frac{7}{8}\rightarrow M\le\frac{5}{\frac{7}{8}}=\frac{40}{7}\)

Pt có nghiệm x = 2 tức là 

\(\left(m^2-1\right).2^2+2\left(m-1\right)-3m^2+m=0\)

\(\Leftrightarrow4m^2-4+2m-2-3m^2+m=0\)

\(\Leftrightarrow m^2+3m-6=0\)

\(\Delta=33>0\Rightarrow x=\frac{-3\pm\sqrt{33}}{2}\)

Thử lại (tự thử)

Nhầm , cái kết quả là m = ... chứ ko phải x nhá

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

      \(=-\left(4x^2+5x-4\right)\)

       \(=-\left(4x^2+4x+1-5\right)\)

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

         \(=-\left(2x+1\right)^2+4\)

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

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

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

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

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