\(x^4+4\)

\(=x^{4^{ }}+4x^{2^{ }}+4-4x^2\)

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

\(=\left(x^2+2\right)^2-\left(2x\right)^2\)

\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

a)

\(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)

\(=\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+2x^2+6x^2+12x+5x+10}\)

\(=\dfrac{\left(x+1\right)\left(x^2-4\right)}{x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+6x+5\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left[x\left(x+5\right)+\left(x+5\right)\right]}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{x-2}{x+5}\)

b)

\(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)

\(=\dfrac{x^4+3x^3+x^2+3x^3+9x^2+3x-x^2-3x-1}{x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1}\)

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

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

\(=\dfrac{x^2+3x+1}{x^2+3x-1}\)

3)

Sửa đề \(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)

Đặt b + c - a = x , a+c-b = y , a+b-c= z

\(\Rightarrow\left[{}\begin{matrix}2a=y+z\\2b=x+z\\2c=x+y\end{matrix}\right.\)

Có :

\(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)

\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\)

\(\Rightarrow\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)

\(\Rightarrow\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\)

Áp dụng bất đẳng thức \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\forall a,b>0\)

\(\Rightarrow\) \(\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)

\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\ge6\)

\(\Rightarrow2\left(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\right)\ge6\)

\(\Rightarrow\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\) \(\left(đpcm\right)\)

=48

tick nha 

1)

Có a+ b+ c = 0

\(\Rightarrow\left[{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left(a+b\right)^2=c^2\\\left(a+c\right)^2=b^2\\\left(b+c\right)^2=a^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a^2+2ab+b^2=c^2\\a^2+2ac+c^2=b^2\\b^2+2bc+c^2=a^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a^2+b^2-c^2=-2ab\\a^2+c^2-b^2=-2ac\\b^2+c^2-a^2=-2bc\end{matrix}\right.\)

Có \(B=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ac}{a^2+c^2-b^2}\)

\(\Rightarrow B=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ac}{-2ac}\)

\(\Rightarrow B=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}\)

\(\Rightarrow B=-\dfrac{3}{2}\)

Có : \(a^2+b^2\le2\) ^{\(\left(1\right)\)}

Áp dụng bất đẳng thức AM - GM ta được :

\(a^2+b^2\ge2ab\)

\(\Rightarrow2ab\le a^2+b^{2^{ }}\le2\) ^{\(\left(2\right)\)}

Cộng ^{\(\left(1\right)\)} và \(\)^{\(\left(2\right)\)} :

\(a^2+2ab+b^2\le4\)

\(\Rightarrow\left(a+b\right)^2\le4\)

\(\Rightarrow-2\le a+b\le2\)

Có \(89^6\) \(⋮89\)

\(\Rightarrow89^6-1⋮89-1\)

\(\Rightarrow89^6-1⋮88\)

\(\Rightarrow89^6-1⋮11\)

Có : \(89^6-1\)

\(=\left(89^3+1\right)\left(89^3-1\right)\)

\(\Rightarrow89^6-1⋮89^3+1\)

\(\Rightarrow89^6-1⋮\left(89+1\right)\left(89^2-89+1\right)\)

\(\Rightarrow89^6-1⋮89+1\)

\(\Rightarrow89^6-1⋮90\)

\(\Rightarrow89^6-1⋮9\)

Đặt \(89^6-1=F\)

\(\Rightarrow F=\overline{4969xy290961}-1\)

\(\Rightarrow F=\overline{4969xy290961}\) \(⋮9;11\)

Có tổng các chữ số hàng lẻ từ trái sang phải của F = 36 + y

Có tổng các chữ số hàng chẵn của F = 18 + x

Có F \(⋮\) 9

\(\Rightarrow54+x+y⋮9\)

\(\Rightarrow x+y\in\left\{0;9;18\right\}\)

Vì A chia hết cho 11 nên

(36+y) - ( 18+x) chia hết cho 11

\(\Rightarrow x-y\in\left\{-4;7\right\}\)

Xét x + y = 0

\(\Rightarrow x=y=0\left(loại\right)\)

Xét x + y= 18

\(\Rightarrow x=y=9\left(loại\right)\)

Xét x + y = 9 : có x+ y và x -y cùng chẵn hoặc cùng lẻ

\(\Rightarrow x-y=7\)

\(\Rightarrow x=8,y=1\)

Vậy \(89^6=496981290961\)

Đặt A(x) = x-2 = 0

\(\Rightarrow x=2\)

\(\Rightarrow\) nghiệm của A(x) là 2

Thay x = 2 vào f(x) ta được

\(\Rightarrow f\left(2\right)=\left(4-6+1\right)^{31}-\left(4-8+5\right)^{30}+2\)

\(\Rightarrow f\left(2\right)=\left(-1\right)^{31}-1^{30}+2\)

\(\Rightarrow f\left(2\right)=-2+2\)

\(\Rightarrow f\left(2\right)=0\)

\(\Rightarrow2\) là nghiệm của \(f\left(x\right)\)

Mà theo định lí Bê - đu ta có :

Đa thức f(x) chia hết cho x - a khi và chỉ khi f(a) = 0 ( tức là khi và chỉ khi a là nghiệm của đa thức)

\(\Rightarrow f\left(x\right)=\left(x^2-3x+1\right)^{31}-\left(x^2-4x+5\right)^{30}+2⋮x-2\)