Cho \(x\ne0\) thỏa mãn \(x+\dfrac{1}{x}=a\) là một hằng số. Tính giá trị biểu thức sau theo a:
C\(=x^7+\dfrac{1}{x^7}\)
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Sửa đề: \(x+\dfrac{1}{x}=a\)
\(A=x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=a^3-3a\\ B=x^6+\dfrac{1}{x^6}=\left(x^3+\dfrac{1}{x^3}\right)^2-2=\left(a^3-3a\right)^2-2=a^6-6a^4+9a^2-2\\ C=x^7+\dfrac{1}{x^7}=\left(x^3+\dfrac{1}{x^3}\right)\left(x^4+\dfrac{1}{x^4}\right)-\left(x+\dfrac{1}{x}\right)\)
Mà \(x^4+\dfrac{1}{x^4}=\left(x^2+\dfrac{1}{x^2}\right)^2-2=\left[\left(x+\dfrac{1}{x}\right)^2-2\right]^2-2=\left(a^2-2\right)^2-2=a^4-4a^2+2\)
\(\Leftrightarrow C=\left(a^3-3a\right)\left(a^4-4a^2+2\right)-a=...\)
a) A = \(\dfrac{1}{x-1}-\dfrac{4}{x+1}+\dfrac{8x}{\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{x+1-4x+4+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x+5}{\left(x-1\right)\left(x+1\right)}=\dfrac{5}{x-1}\) => đpcm
b) \(\left|x-2\right|=3=>\left[{}\begin{matrix}x-2=3< =>x=5\left(C\right)\\x-2=-3< =>x=-1\left(L\right)\end{matrix}\right.\)
Thay x = 5 vào A, ta có:
A = \(\dfrac{5}{5-1}=\dfrac{5}{4}\)
c) Để A nguyên <=> \(5⋮x-1\)
x-1 | -5 | -1 | 1 | 5 |
x | -4(C) | 0(C) | 2(C) | 6(C) |
P(x)=\(ax^2+bx+c\) (1)(a\(\ne0\) )
Ta có :
\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}\)\(\Rightarrow\left\{{}\begin{matrix}b=2a\\c=3a\end{matrix}\right.\)(2)
Thay(2) vào (1)\(\Rightarrow P\left(x\right)=ax^2+2ax+3a\)
\(\Rightarrow\dfrac{P\left(-2\right)-3P\left(-1\right)}{a}=\dfrac{4a-4a+3a-3\left(a-2a+3a\right)}{a}\)=\(\dfrac{3a-3a+6a-9a}{a}=\dfrac{-3a}{a}=-3\)
\(A=\left(\dfrac{1}{x^2-1}+\dfrac{1}{x+1}\right):\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)\)
\(\Rightarrow A=\left(\dfrac{1}{\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x}{x\left(x-1\right)}-\dfrac{x-1}{x\left(x-1\right)}\right)\)
\(\Rightarrow A=\dfrac{1+x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-x+1}{x\left(x-1\right)}\)
\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{x\left(x-1\right)}\)
\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}.x\left(x-1\right)\)
\(\Rightarrow A=\dfrac{x^2}{x+1}\)
đk : xkhác -1 ; 1
\(A=\left(\dfrac{1+x-1}{\left(x+1\right)\left(x-1\right)}\right):\left(\dfrac{x-x+1}{x\left(x-1\right)}\right)=\dfrac{x}{\left(x+1\right)\left(x-1\right)}:\dfrac{1}{x\left(x-1\right)}=\dfrac{x^2}{x+1}\)
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{-2\left(-4k\right)-7k+5.3k}{2.\left(-4k\right)-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-5k}=-\dfrac{16}{5}\)