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Ta có: \(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{24}+x^{20}+x^{16}+...+1\right)\left(x^2+1\right)}\)
\(=\dfrac{1}{x^2+1}\)
x24+x20+x16+...+x4+1x26+x24+x22+...+x2+1x24+x20+x16+...+x4+1x26+x24+x22+...+x2+1
=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)
=x24+x20+x16+...+x4+1x2(x24+x20+...+1)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1x2(x24+x20+...+1)+(x24+x20+x16+...+x4+1)
=x24+x20+x16+...+x4+1(x24+x20+x16+...+1)(x2+1)
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\(\left(x^2+1\right)\left(x^4+1\right)\left(x^{16}+x^8+1\right)=\frac{x^{24}-x+1}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=x^6-1\)
đề như này à bạn ???
sao có 2 dấu = vậy
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Đề đọc khó hiểu quá. Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.
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\(\frac{x^{24}+x^{20}+...+x^4+1}{x^{26}+x^{24}+...+x^2+1}=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^{24}+x^{20}+...+x^4+1\right)+\left(x^{26}+x^{22}+...+x^2\right)}\)
\(=1-\frac{x^2\left(x^{24}+x^{20}+...+x^4+x^1\right)}{\left(1+x^2\right)\left(x^{24}+2^{20}+...+x^4+1\right)}=1-\frac{x^2}{1+x^2}\)
\(=\frac{1+x^2-x^2}{1+x^2}=\frac{1}{1+x^2}\)
Hoặc cách khác:
\(\frac{x^{24}+x^{20}+...+x^4+1}{x^{26}+x^{24}+...+x^2+1}=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^{24}+x^{20}+...+x^4+1\right)+x^2\left(x^4+x^{20}+...+x^4+1\right)}\)
\(=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^2+1\right)\left(x^{24}+x^{20}+...+x^4+1\right)}=\frac{1}{x^2+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>(x^2+x)(x^2+x-2)=24
=>(x^2+x)^2-2(x^2+x)-24=0
=>(x^2+x-6)(x^2+x+4)=0
=>(x+3)(x-2)=0
=>x=2 hoặc x=-3
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n: ĐKXĐ: x<>0
\(\left(x+\dfrac{1}{x}\right)^2-3\left(x+\dfrac{1}{x}\right)+2=0\)
=>\(\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)-\left(x+\dfrac{1}{x}\right)+2=0\)
=>\(\left(x+\dfrac{1}{x}-2\right)\left(x+\dfrac{1}{x}-1\right)=0\)
=>\(\dfrac{x^2+1-2x}{x}\cdot\dfrac{x^2+1-x}{x}=0\)
=>\(\left(x^2-2x+1\right)\left(x^2-x+1\right)=0\)
=>\(\left(x-1\right)^2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
p: \(x^4-4x^3+6x^2-4x+1=0\)
=>\(x^4-x^3-3x^3+3x^2+3x^2-3x-x+1=0\)
=>\(x^3\left(x-1\right)-3x^2\left(x-1\right)+3x\left(x-1\right)-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-3x^2+3x-1\right)=0\)
=>\(\left(x-1\right)^4=0\)
=>x-1=0
=>x=1
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\(a,x^3+x^2+x+1=0\\ \Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-1\right\}\)
\(b,x^3+x^2-x-1=0\\ \Rightarrow x^2\left(x+1\right)-\left(x+1\right)=0\\ \Rightarrow\left(x^2-1\right)\left(x+1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-1;1\right\}\)
\(c,\left(x+1\right)^2\left(x+2\right)+\left(x+1\right)^2\left(x-2\right)=-24\\ \Rightarrow\left(x+1\right)^2\left(x+2+x-2\right)=-24\\ \Rightarrow2x\left(x^2+2x+1\right)=-24\\ \Rightarrow x^3+2x^2+x+12=0\\ \Rightarrow\left(x^3+3x^2\right)-\left(x^2+3x\right)+\left(4x+12\right)=0\\ \Rightarrow x^2\left(x+3\right)-x\left(x+3\right)+4\left(x+3\right)=0\\ \Rightarrow\left(x^2-x+4\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\left(vô.lí\right)\\x=-3\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-3\right\}\)
Ta có : x(x - 1)(x + 1)(x + 2) = 24
<=> [(x - 1)(x + 2)][x(x + 1)] - 24 = 0
<=> (x2 + x - 2)(x2 + x) - 24 = 0
<=> (x2 + x - 1 - 1)(x2 + x - 1 + 1) - 24 = 0
<=> (x2 + x - 1)2 - 1 - 24 = 0
<=> (x2 + x - 1)2 - 25 = 0
<=> (x2 + x - 6)(x2 + x + 4) = 0
<=> (x2 - 2x + 3x - 6)(x2 + x + 4) = 0
<=> [x(x - 2) + 3(x - 2)](x2 + x + 4) = 0
<=> (x + 3)(x - 2)(x2 + x + 4) = 0
<=> x + 3 = 0 hoặc x - 2 = 0 hoặc x2 + x + 4 = 0
<=> x = -3 hoặc x = 2 hoặc x \(\in\varnothing\)
Vậy \(x\in\left\{-3;2\right\}\)là nghiệm phương trình
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt \(x^2+x=t\)
\(\Leftrightarrow t\left(t-2\right)-24=0\Leftrightarrow t^2-2t-24=0\)
\(\Leftrightarrow\left(t-6\right)\left(t+4\right)=0\Leftrightarrow t=6;t=-4\)
\(\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x-2\right)\left(x+3\right)\Leftrightarrow x=2;x=-3\)
\(\Leftrightarrow x^2+x+4=0\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\)
Vậy tập nghiệm của phương trình là S = { 2 ; -3 }