x^2 + x + 1 >0

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

\(x^2+x+1=\left[x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]+1-\left(\frac{1}{2}\right)^2\)\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Mà \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

co 37 hoc sinh

Gọi r(x) = ax + b là dư trong phép chia f(x) cho (x-1)(x-2)

Theo đề bài ta có :

f(x) = (x-1).A(x) + 2 [ A(x) là thương trong phép chia f(x) cho (x-1) ](1)

f(x) = (x+2).B(x) + 4 [ B(x) ___________________________ (x+2) ](2)

f(x) = (x-1)(x-2).C(x) + ax + b [ C(x) ___________________ (x-1)(x+2) ](3)

Với x = 1 ta có \(\hept{\begin{cases}\left(1\right)=2\\\left(3\right)=a+b\end{cases}}\Rightarrow a+b=2\)(*)

Với x = -2 ta có \(\hept{\begin{cases}\left(2\right)=4\\\left(3\right)=-2a+b\end{cases}}\Rightarrow-2a+b=4\)(**)

Từ (*) và (**) \(\Rightarrow\hept{\begin{cases}a+b=2\\-2a+b=4\end{cases}}\Leftrightarrow\hept{\begin{cases}3a=-2\\a+b=2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-\frac{2}{3}\\b=\frac{8}{3}\end{cases}}\)

Vậy dư là -2/3x + 8/3

4x⁴-4x²+1 
= (2x)² - 2 (2x .1 ) +1²
= (2x+1)²

E = ( 2x - 1 )( 4x² + 2x + 1 ) - 2x( 2x - 7 )( 2x + 7 )

= ( 2x )³ - 1³ - 2x( 4x² - 49 )

= 8x³ - 1 - 8x³ + 98x = 98x - 1

a, \(x-xy+y-y^2=x\left(1-y\right)+y\left(1-y\right)=\left(x+y\right)\left(1-y\right)\)

b, \(x^2-4x+4=\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)

c, \(x^2-2x-3=\left(x-1\right)^2-4=\left(x-3\right)\left(x+1\right)\)

c, Đặt \(x^2-3x-1=t\)

\(t^2-12t+27=t^2-3t-9t+27=\left(t-9\right)\left(t-3\right)\)

Theo cách đặt \(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)=\left(x-5\right)\left(x+2\right)\left(x-4\right)\left(x+1\right)\)