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Ta có :
\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-1+\frac{1}{x^2}\right)\)
\(=\left(x+\frac{1}{x}\right)\left(7-1\right)\)(vì \(x^2+\frac{1}{x^2}=7\))
\(=6\left(x+\frac{1}{x}\right)\)
Đặt \(x+\frac{1}{x}=a\)thì \(\left(x+\frac{1}{x}\right)=a^2\). Suy ra \(a^2-2=x^2+\frac{1}{x^2}\)
\(\Rightarrow a^2-2=7\)(vì \(x^2+\frac{1}{x^2}=7\))
\(\Rightarrow a^2=9\)\(\Rightarrow\left(x+\frac{1}{x}\right)^2=9\)
Vì \(x\inℝ,x>0\)nên \(x+\frac{1}{x}>0\)
\(\Rightarrow\) \(\left(x+\frac{1}{x}\right)^2=3^2\Rightarrow x+\frac{1}{x}=3\)
Do đó \(x^3+\frac{1}{x^3}=6.3=18\)
Ta có:
\(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=x^5+\frac{1}{x^5}+1\)
Mà \(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=7.18=126\)
\(\Rightarrow x^5+\frac{1}{x^5}+1=126\)
\(\Rightarrow x^5+\frac{1}{x^5}=125\)
Vậy với \(x\inℝ,x>0\)và \(x^2+\frac{1}{x^2}=7\)thì \(x^5+\frac{1}{x^5}=125\)
2/ \(\frac{1}{2}x2y5z3=\left(\frac{1}{2}.2.5.3\right)xyz\)\(=15xyz\)
\(\Rightarrow\frac{1}{2}x2y5z3\)có bậc là 3
3/ \(\frac{x}{4}=\frac{9}{x}\Leftrightarrow x^2=9.4\Rightarrow x^2=36\) mà \(x>0\Rightarrow x=6\)
4/ \(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\Rightarrow\left|2x+\frac{1}{2}\right|=\frac{35}{7}=5\Rightarrow\hept{\begin{cases}2x+\frac{1}{2}=5\Rightarrow2x=\frac{9}{2}\Rightarrow x=\frac{9}{4}\\2x+\frac{1}{2}=-5\Rightarrow2x=\frac{-11}{2}\Rightarrow x=\frac{-11}{4}\end{cases}}\)
Ta có: \(x^2+\frac{1}{x^2}=7\)
\(\Rightarrow x^2+2+\frac{1}{x^2}=9\)
\(\Rightarrow\left(x+\frac{1}{x}\right)^2=9\)
Mà x>0
\(\Rightarrow x+\frac{1}{x}=3\)
Lại có: \(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-1+\frac{1}{x^2}\right)=3\left(7-1\right)=18\)
\(\Rightarrow\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=x^5+\frac{1}{x^5}+x+\frac{1}{x}\)
\(\Rightarrow x^5+\frac{1}{x^5}=7.18-3=123\)
Cho số x khác 0 thỏa mãn \(x^2-5x+1=0\).Tính giá trị của \(Q=x^7-x^5+\frac{1}{x^7}-\frac{1}{x^5}+1\)
\(\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2=7+2=9\)
\(\Rightarrow x+\frac{1}{x}=3\) (vì x > 0)
Mặt khác, \(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=3^3-3.3=18\)
Ta có: \(B=x^5+\frac{1}{x^5}=\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)\)
\(=7.18-3=123\)
Vậy B = 123
Chúc bạn học tốt.
Ta có : \(x^2+\dfrac{1}{x^2}=7\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}+2=9\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2=9\)
\(\Leftrightarrow x+\dfrac{1}{x}=3\left(x>0\right)\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=27\)
\(\Leftrightarrow x^3+3x^2.\dfrac{1}{x}+3x.\dfrac{1}{x^2}+\dfrac{1}{x^3}=27\)
\(\Leftrightarrow x^3+3x+\dfrac{3}{x}+\dfrac{1}{x^3}=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3.3=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)
Lại có : \(\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)\)
\(=x^5+x+\dfrac{1}{x}+\dfrac{1}{x^5}\)
\(=x^5+\dfrac{1}{x^5}+3\left(1\right)\)
Mặt khác : \(\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=7.18=126\left(2\right)\)
Từ ( 1 ) ; ( 2 ) \(\Rightarrow x^5+\dfrac{1}{x^5}+3=126\)
\(\Rightarrow x^5+\dfrac{1}{x^5}=123\in Z\)
\(\left(đpcm\right)\)
\(x^2+\frac{1}{x^2}=7\Leftrightarrow x^2+2+\frac{1}{x^2}=9\Leftrightarrow\left(x+\frac{1}{x}\right)^2=3^2.\)Do x > 0 nên \(x+\frac{1}{x}\)>0 và \(x+\frac{1}{x}=3\)
\(\Rightarrow\left(x+\frac{1}{x}\right)^3=27\Rightarrow x^3+\frac{1}{x^3}+3\cdot x\cdot\frac{1}{x}\left(x+\frac{1}{x}\right)=27\Rightarrow x^3+\frac{1}{x^3}+3\cdot3=27\Rightarrow x^3+\frac{1}{x^3}=18\)
\(\Rightarrow\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=7\cdot18\Rightarrow x^5+\frac{1}{x^5}+x+\frac{1}{x}=126\Rightarrow x^5+\frac{1}{x^5}+3=126\Rightarrow x^5+\frac{1}{x^5}=123.\)
Vậy \(x^5+\frac{1}{x^5}\)là 1 số nguyên và bằng: 123