theo định lí Bơ du ta có

f(x)=f(1)=1\(^{80}\)+\(1^{40}+1^{20}+1^{10}+1^5+1\)=6

Vậy số dư trong phép chia trên là 6

a, x.(x-y) +y.(x+y)

=x²-xy+xy+y²

=x²+y²

b, (x²-5).(2x+3)-2x.(x-3)

=2x³+3x²-10x-15-2x²+6x

=2x³-x²-4x-15

c, 8-5x.(x+2) +4 .( x-2) . (x+1) +2.( x+2)+ 2.(x-2)+10

=8-5x²-10x+4.(x²+x-2x-2)+2x+4+2x-4+10

=18-6x-5x²+4x²+4x-8x-8

=10-10x-x²

ks bn nobita nha

Đặt biểu thức là A, ta có:

\(A=\frac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)

\(\Rightarrow A.x^5=\frac{x^{45}+x^{35}+x^{25}+x^{15}+x^5}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)

\(\Rightarrow A.x^5+A=\frac{x^{45}+x^{40}+x^{35}+x^{25}+x^{15}+x^5+x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)

\(\Rightarrow A.x^5+1=1\)

\(\Rightarrow A=\frac{1}{x^5+1}\)

a)

\(P=\dfrac{x^{10}-x^8+x^6-x^4+x^2-1}{x^4-1}\)

\(=\dfrac{x^8\left(x^2-1\right)+x^4\left(x^2-1\right)+\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)

\(=\dfrac{\left(x^2-1\right)\left(x^8+x^4+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)

\(=\dfrac{x^8+x^4+1}{x^2+1}\)

b)

\(Q=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)

\(=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{\left(x^{45}+x^{35}+...+x^5\right)+\left(x^{40}+x^{30}+...+1\right)}\)

\(=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^5\left(x^{40}+x^{30}+...+1\right)+\left(x^{40}+x^{30}+...+1\right)}\)

\(=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{\left(x^{40}+x^{30}+...+1\right)\left(x^5+1\right)}\)

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

cái câu b dòng cuối mẫu số đóng mở ngoặc chi cho mệt ei =.=

a)\(x^8+x^4+1\)

\(=\left(x^8+2x^4+1\right)-x^4\)

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

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

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

b)\(x^{10}+x^5+1\)

\(=\left(x^{10}+x^9+x^8\right)-\left(x^9+x^8+x^7\right)+\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)

\(=x^8\left(x^2+x+1\right)-x^7\left(x^2+x+1\right)+x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)

a) \(x^8+x^4+1\)

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

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

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

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

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

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

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

b) \(x^{10}+x^5+1\)

= ( x¹⁰+x⁹+x⁸) - (x⁹+x⁸+x⁷) + (x⁷+x⁶+x⁵) - (x⁶+x⁵+x⁴) + (x⁵+x⁴+x³) - (x³+x²+x) + (x²+x+1)

= (x²+x+1)(x⁸ - x⁷+x⁵-x⁴+x³-x+1)

\(k\left(x\right)=\dfrac{5x^2-22x+25}{x^2-4x+4}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{5x^2-20x+20-x+2-x+2+1}{x^2-4x+4}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{\left(5x^2-20x+20\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)

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

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

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

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

Đặt \(y=\dfrac{1}{x-2}\)

\(\Rightarrow k\left(x\right)=5-y-y+y^2=y^2-2y+1+4=\left(y-1\right)^2+4\ge4\)

Vậy GTNN của \(k\left(x\right)=4\) khi \(y=1\Rightarrow\dfrac{1}{x-2}=1\Leftrightarrow x=3\)

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

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

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

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

Đặt \(y=\dfrac{1}{x-1}\)

\(\Rightarrow h\left(x\right)=1+y+y^2\)

\(\Rightarrow h\left(x\right)=y^2+y+1\)

\(\Rightarrow h\left(x\right)=y^2+2.y.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(\Rightarrow h\left(x\right)=\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

=> GTNN của \(h\left(x\right)=\dfrac{3}{4}\) khi \(y+\dfrac{1}{2}=0\Leftrightarrow y=\dfrac{-1}{2}\)

\(\Leftrightarrow\dfrac{1}{x-1}=\dfrac{-1}{2}\)

\(\Leftrightarrow x=-1\)

 \(2x+k=x-1=0\)

\(\Rightarrow\hept{\begin{cases}x-1=0\\2x+k=0\end{cases}}\)

Xét x - 1 =0

=> x = 1

Thay vào ta có :

2 + k = 0

k = -2 

\(A=\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}+\dfrac{1}{\left(x+9\right)\left(x+11\right)}\)\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}+\dfrac{1}{x+9}-\dfrac{1}{x+11}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+11}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{x+11}{\left(x+1\right)\left(x+11\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+11\right)}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{x+11-x-1}{\left(x+1\right)\left(x+11\right)}\right)=\dfrac{1}{2}.\dfrac{10}{\left(x+1\right)\left(x+11\right)}=\dfrac{10}{2\left(x+1\right)\left(x+11\right)}\)