f(x1+x2)= a(x1+x2)+b=ax1+ax2+b. Tương tự ta có f(x1)+f(x2)=ax1+b+ax2+b. => để 2 vế bằng nhau thì b=0

đúng ko bạn

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)

\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+...+\frac{1}{20}.\frac{20\left(20+1\right)}{2}\)

\(=\frac{2}{2}+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{20+1}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{20}{2}\)

\(=\frac{2+3+4+...+20}{2}=\frac{\frac{20\left(20+1\right)}{2}-1}{2}=\frac{209}{2}\)

Đinh Đức Hùng trả lời sai rồi

Đặt \(\sqrt{x+2016}=y\ge0\)\(\Rightarrow y^2=x+2016\)\(\Rightarrow x=y^2-2016\)

\(\Rightarrow M=y^2-2016+y\)\(=y^2+2.\frac{1}{2}.y+\frac{1}{4}-\frac{8065}{4}=\left(y+\frac{1}{2}\right)^2-\frac{8065}{4}\ge\)\(\left(\frac{1}{2}\right)^2-\frac{8065}{4}=-2016\)\(\forall y\ge0\)

Dấu "=" xảy ra khi \(\sqrt{x+2016}=y=0\Leftrightarrow\)\(x+2016=0\Leftrightarrow x=-2016\)

Vậy ...

• -6x³ + x² + 5x - 2 = 0

=> -6x³ - 6x² + 7x² + 7x - 2x - 2 = 0

=> -6x²(x+1) + 7x(x+1) - 2(x+1) = 0

=> (x+1)(-6x²+7x-2) = 0

=> (x+1)(x²-\(\frac{7}{6}x+\frac{1}{3}\)) = 0

\(\Rightarrow\left(x+1\right)\left(x-\frac{1}{2}\right)\left(x-\frac{2}{3}\right)=0\)

=> x = -1 hoặc x = 1/2 hoặc x = 2/3

• 3x³ + 19x² + 4x - 12 = 0

=> 3x³ + 3x² + 16x² + 16x - 12x - 12 = 0

=> (x+1)(3x²+16x-12)=0

=> (x+1)\(\left(x^2+\frac{16}{3}x-4\right)=0\)

=> (x+1) \(\left(x-\frac{2}{3}\right)\left(x+6\right)=0\)

=> x = -1 hoặcx = 2/3 hoặc x = -6

• 2x³ - 11x² + 10x + 8 = 0

=> 2x³ - 4x² - 7x² + 14x - 4x + 8 = 0

=> 2x²(x - 2) - 7x(x - 2) - 4(x - 2) = 0

=> (x - 2)(2x² - 7x - 4)=0    

=> (x - 2)(\(x^2-\frac{7}{2}x-2\)) = 0

=> \(\left(x-2\right)\left(x-4\right)\left(x+\frac{1}{2}\right)=0\)

=> x = 2 hoặc x = 4 hoặc x = -1/2