Cho \(x+\frac{1}{x}=a\). Tính biểu thức sau theo a: \(x^5+\frac{1}{x^5}\)
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ĐKXĐ : \(x\ne\pm3\)
a) \(A=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)
\(A=\left(\frac{-2x\left(3+x\right)}{\left(3-x\right)\left(3+x\right)}-\frac{\left(x+1\right)\left(3-x\right)}{\left(x+3\right)\left(3-x\right)}+\frac{x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3}{x+3}-\frac{x-1}{x+3}\right)\)
\(A=\left(\frac{-2x^2-6x+x^2-2x-3+x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3-x+1}{x+3}\right)\)
\(A=\left(\frac{-8x-2}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{4}{x+3}\right)\)
\(A=\frac{-2\left(4x+1\right)\left(x+3\right)}{\left(3-x\right)\left(3+x\right)4}\)
\(A=\frac{-\left(4x+1\right)}{2\left(3-x\right)}\)
\(A=\frac{4x+1}{2\left(x-3\right)}\)
b) \(\left|x-5\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}}\)
Mà ĐKXĐ x khác 3 => ta xét x = 7
\(A=\frac{4\cdot7+1}{2\cdot\left(7-3\right)}=\frac{29}{8}\)
c) Để A nguyên thì 4x + 1 ⋮ 2x - 3
<=> 4x - 6 + 7 ⋮ 2x - 3
<=> 2 ( 2x - 3 ) + 7 ⋮ 2x - 3
Mà 2 ( 2x - 3 ) ⋮ ( 2x - 3 ) => 7 ⋮ 2x - 3
=> 2x - 3 thuộc Ư(7) = { 1; -1; 7; -7 }
=> x thuộc { 2; 1; 5; -2 }
Vậy .....
a) ĐKXĐ: \(x\ne\pm3\)
\(A=\frac{2x\left(x+3\right)-\left(x+1\right)\left(x-3\right)-\left(x^2+1\right)}{x^2-9} : \frac{x+3-\left(x-1\right)}{x+3}\)
\(A=\frac{2x^2-6x-x^2+2x+3-x^2-1}{x^2-9} : \frac{4}{x+3}\)
\(A=\frac{-4x+2}{x^2+9} : \frac{4}{x+3}\)
\(A=\frac{2\left(1-2x\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{4}=\frac{1-2x}{2x-6}\)
b)
Có 2 trường hợp:
T.Hợp 1:
\(x-5=2\Leftrightarrow x=7\)(thỏa mã ĐKXĐ)
thay vào A ta được: A=\(-\frac{13}{8}\)
T.Hợp 2:
\(x-5=-2\Leftrightarrow x=3\)(Không thỏa mãn ĐKXĐ)
Vậy không tồn tại giá trị của A tại x=3
Vậy với x=7 thì A=-13/8
c)
\(\frac{1-2x}{2x-6}=\frac{1-\left(2x-6\right)-6}{2x-6}=-1-\frac{5}{2x-6}\)
Do -1 nguyên, để A nguyên thì \(-\frac{5}{2x-6}\inℤ\)
Để \(-\frac{5}{2x-6}\inℤ\)thì \(2x-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Do 2x-6 chẵn, để x nguyên thì 2x-6 là 1 số chẵn .
Vậy không có giá trị nguyên nào của x để A nguyên
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tương tự :
\(x+\frac{1}{x}=a\)
\(x^5+\frac{1}{x^5}+5x^3+10x+\frac{10}{x}+\frac{5}{x^3}=a^5\)
\(\Rightarrow x^5+\frac{1}{x^5}=a^5-5\left(x^3+\frac{1}{x^3}\right)-10\left(x+\frac{1}{x}\right)\)
Mà : \(x+\frac{1}{x}=a\Rightarrow x^3+\frac{1}{x^3}=a^3-3x-\frac{3}{x}=a^3-3a\)
\(\Rightarrow x^5+\frac{1}{x^5}=a^5-5\left(a^3-3a\right)-10a\)
\(\Rightarrow x^5+\frac{1}{x^5}=a^5-5a^3+15a-10a=a^5-5a^3+5a\)
nha
a) Ta có \(x+\frac{1}{x}=a\)
\(\Rightarrow x^4+4x^2+6+\frac{4}{x^2}+\frac{1}{x^4}=a^4\)
\(\Rightarrow x^4+\frac{1}{x^4}=a^4-6-4\left(x^2+\frac{1}{x^2}\right)\)
Mà \(x+\frac{1}{x}=a\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)
\(\Rightarrow x^4-\frac{1}{x^4}=a^4-6-4a^2+8=a^4-4a^2+2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-5x}{x^2-1}\right)\cdot\frac{x-3}{x}\left(x\ne\pm1;x\ne0\right)\)
\(\Leftrightarrow A=\left[\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-5x}{\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-5x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\frac{x\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x+1\right)x}=\frac{x-3}{x+1}\)
Vậy \(A=\frac{x-3}{x+1}\left(x\ne\pm1;x\ne0\right)\)
b) \(A=\frac{x-3}{x+1}\left(x\ne\pm1;x\ne0\right)\)
Để A nhận giá trị nguyên thì x-3 chia hết chi x+1
=> (x+1)-4 chia hết chi x+1
=> 4 chia hết cho x+1
x nguyên => x+1 nguyên => x+1 thuộc Ư (4)={-4;-2;-1;1;2;4}
Ta có bảng
x+1 | -4 | -2 | -1 | 1 | 2 | 4 |
x | -5 | -3 | -2 | 0 | 1 | 3 |
ĐCĐK | tm | tm | tm | ktm | ktm | tm |
Vậy x={-5;-3;-2;3} thì A đạt giá trị nguyên
c) I3x-1I=5
\(\Rightarrow\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=6\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{-4}{3}\end{cases}}}\)
Đên đây thay vào rồi tính nhé
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm1\\x\ne0\end{cases}}\)
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-5x}{x^2-1}\right)\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\frac{\left(x+1\right)^2-\left(x-1\right)^2+x^2-5x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\frac{x^2+2x+1-x^2+2x-1+x^2-5x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\frac{\left(x^2-x\right)\left(x-3\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow A=\frac{x-3}{x+1}\)
b) Để \(A\inℤ\)
\(\Leftrightarrow x-3⋮x+1\)
\(\Leftrightarrow x+1-4⋮x+1\)
\(\Leftrightarrow4⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;-3;1;3;-5\right\}\)
Mà \(x\ne0;x\ne1\)
\(\Leftrightarrow x\in\left\{-2;-3;3;-5\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{-2;-3;3;-5\right\}\)
c) Khi \(\left|3x-1\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=6\\3x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{4}{3}\end{cases}}\)
Vì khi x = 2 hoặc x = -4/3 thì x không thuộc tập hợp các giá trị làm cho A nguyên
Vậy khi |3x - 1| = 5 thì để cho A nguyên \(\Leftrightarrow x\in\varnothing\)
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Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+.....+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+....+\frac{1}{x+9}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{x+11-x+1}{\left(x+1\right)\left(x+11\right)}=\frac{12}{\left(x+1\right)\left(x+11\right)}\)
Ta có: \(\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3.1.\frac{1}{x}.\left(1+\frac{1}{x}\right)\)\(=a^3\)
\(< =>x^3+\frac{1}{x^3}+3.\left(1+\frac{1}{x}\right)=a^3\)
\(< =>x^3+\frac{1}{x^3}=a^3-3a\)
Lại có: \(\left(x+\frac{1}{x}\right)^5=x^5+\frac{1}{x^5}+5.\left(x^3+\frac{1}{x^3}\right)+10.\left(x+\frac{1}{x}\right)=a^5\)
\(< =>x^5+\frac{1}{x^5}+5.\left(a^3-3a\right)+10.a=a^5\)
\(< =>x^5+\frac{1}{x^5}+5a^3-15a+10a=a^5\)
\(< =>x^5+\frac{1}{x^5}=a^5-5a^3+5a\)