a) f(x)+g(x) = 2x⁴ -x³ -2x²+x+4

b) f(x)-g(x) =x³-4x²+x-5

x4 là x^4 hả bạn

Giải thích các bước giải:

a) f(x)+g(x)=x4x4 – 3x23x2 + x – 1 + x4x4 - x3x3 + x2x2 + 5

                  =2x42x4 - x3x3 -2x22x2 +x +4

b)f(x)-g(x)=x4x4 – 3x23x2 + x – 1 - x4x4 + x3x3 - x2x2 - 5

                = x3x3 - 4x24x2 +x -6

\(e,\left(x-2\right)^2-16=0\\ \Leftrightarrow\left(x-6\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\\ f,x^2-5x-14=0\\ \Leftrightarrow\left(x-7\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ g,8x\left(x-3\right)+x-3=0\\ \Leftrightarrow\left(8x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{8}\\x=3\end{matrix}\right.\)

e)\(\left(x-2-4\right)\left(x-2+4\right)=\left(x-6\right)\left(x+2\right)\)

f)\(x^2-5x-14=x^2-2.\dfrac{5}{2}x+\dfrac{25}{2}+\dfrac{3}{2}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{2}\)

e: Ta có: \(\left(x-2\right)^2-16=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

f: Ta có: \(x^2-5x-14=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\)

\(\Leftrightarrow4x^2+x-8x-2-4x^2-27x=1\)

=>-34x=3

hay x=-3/34

\(\left(x-2\right)\left(4x+1\right)-4x\left(x+7\right)=1\\ \Rightarrow4x^2-8x+x-2-4x^2-28x=1\\ \Rightarrow-35x=3\\ \Rightarrow x=\dfrac{-3}{35}\)

     (x - 2)(4x + 1) - 4x(x+7) = 1

⇔ 4x² + x - 8x - 2 - 4x² - 28x = 1

⇔ - 35x = 3

⇔ x = - 3/35

f(2)=g(0)

=> c=5

f(1)=g(1)

=> a+b+c=2 mà c=5 => a+b=-3 (1)

f(-1)=g(3)

=>9a+3b+c=2  mà c=5= > 9a+3b=-3=> 3a+b=-1(2)

(2)-(1) ta được:

2a=2=>a=1=> b=-4 

VẬy g(x)=x^2-4x+5

t i ck ủng hộ tui nha

Ta có: \(\left(x^{3n}+y^{3n}\right)\left(x^{3n}-y^{3n}\right)=-x^{6n}-y^{6n}\)

\(\Leftrightarrow x^{6n}-y^{6n}=-x^{6n}-y^{6n}\)

\(\Leftrightarrow n\in\varnothing\)

a) Ta có:

f(0) = -2.0³ + 3.0² - 0 + 5 = 0 + 0 - 0 + 5 = 5

g(-1) = 2.(-1)³ - 2.(-1)² + (-1) - 9 = -2 - 2 - 1 - 9 = -14

b) f(x) + g(x) = (-2x³ + 3x² - x + 5) + (2x³ - 2x² + x - 9)

                   = -2x³ + 3x² - x + 5 + 2x³ - 2x² + x - 9

                  = (-2x³ + 2x³) + (3x² - 2x²) - (x - x) + (5 - 9)

                 = x² - 4

f(x) - g(x) = (-2x³ + 3x² - x + 5) - (2x³ - 2x² + x - 9)

               = -2x³ + 3x² - x + 5 - 2x³ + 2x² - x + 9

              = -(2x³ + 2x³) + (3x² + 2x²) - (x + x) + (5 + 9)

             = -4x³ + 5x² - 2x + 14

\(D=4x^2-2x+3x\left(x-5\right)=4x^2-2x+3x^2-15x=7x^2-17x=7\left(-1\right)^2-17\left(-1\right)=24\)

\(E=x^{10}-2020x^9+2020x^8-2020x^7+...+2020x^2-2020x=x^9\left(x-2019\right)-x^8\left(x-2019\right)+x^7\left(x-2019\right)-...-x^2\left(x-2019\right)+x\left(x-2019\right)-x=x^9\left(2019-2019\right)-...+x\left(2019-2019\right)-2019=-2019\)

\(f\left(0\right)=2010\Rightarrow a.0^2+b.0+c=2010\Rightarrow c=2010\)

\(f\left(1\right)=2011\Rightarrow a.1^2+b.1+c=2011\Rightarrow a+b+c=2011\)

\(\Rightarrow a+b+2010=2011\Rightarrow a+b=1\) (1)

\(f\left(-1\right)=2012\Rightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=2012\)

\(\Rightarrow a-b+c=2012\Rightarrow a-b+2010=2012\)

\(\Rightarrow a-b=2\Rightarrow a=b+2\)

Thế vào (1) \(\Rightarrow b+2+b=1\Rightarrow2b=-1\Rightarrow b=-\dfrac{1}{2}\)

\(\Rightarrow a=b+2=-\dfrac{1}{2}+2=\dfrac{3}{2}\)

\(\Rightarrow f\left(x\right)=\dfrac{3}{2}x^2-\dfrac{1}{2}x+2010\)

\(\Rightarrow f\left(-2\right)=\dfrac{3}{2}.\left(-2\right)^2-\dfrac{1}{2}.\left(-2\right)+2010=2017\)