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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 =.=
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) ta có:
(x-3)(x-5)(x-6)(x-10)=24x2
<=> \(\left[\left(x-3\right)\left(x-10\right)\right]\left[\left(x-5\right)-\left(x-6\right)\right]=24x^2\)
<=> \(\left(x^2-13x+30\right)\left(x^2-11x+30\right)=24x^2\)
<=> \(\left(x^2-12x+20-x\right)\left(x^2-12x+30+x\right)=24x^2\)
<=> \(\left(x^2-12x+30\right)^2-x^2=24x^2\)
<=> \(\left(x^2-12x+30\right)^2-x^2-24x^2=0\)
<=> \(\left(x^2-12x+30\right)^2-25x^2=0\)
<=> \(\left(x^2-17x+30\right)\left(x^2-7x+30\right)=0\)
mà x2-7x+30=(x-\(\dfrac{7}{2}\))2+\(\dfrac{71}{4}\)> 0
=> x2-17x+30=0
<=> (x-15)(x-2)=0
=>\(\left[{}\begin{matrix}x-15=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=2\end{matrix}\right.\)
Vậy S=\(\left\{15;2\right\}\)
b) ta có:
(x+1)(x+2)(x+4)(x+5)=40
<=> \(\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]=40\)
<=> (x2+6x+5)(x2+6x+8)=40
<=> (x2+6x+6,5-1,5)(x2+6x+6,5+1,5)=40
<=> (x2+6x+6,5)2 _ 2,25=40
<=> (x2+6x+6,5)2 _ 42,25=0
<=> (x2+6x+6,5-6,5)(x2+6x+6,5+6,5)=0
<=> (x2+6x)(x2+6x+13)=0
mà x2+6x+13=(x+3)2+4>0
=> x2+6x=0
<=> x(x+6)=0
=>\(\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy S=\(\left\{0;-6\right\}\)
b)
\(\Rightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)=40\)
\(\Rightarrow\left(x^2+5x+x+5\right)\left(x^2+4x+2x+8\right)=40\)
\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt: \(a=x^2+6x+5\)
\(\Rightarrow a.\left(a+3\right)=40\)
Mà:\(40=5.8\)
\(\Rightarrow a=5\)
Học tốt !!! :)
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
Ta có x(x - 10) + x(2- x) = -40
=>x2- 10x + 2x - x2 = -40
=> -8x = -40
=> x = 5
Vậy x = 5
\(x\left(x-10\right)+x\left(2-x\right)=40\)
\(x^2-10x+2x-x^2=40\)
\(-8x=40\)
\(x=-5\)