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Ta có:
\(\left(x^2+\frac{1}{x^2}\right)^4=x^8+4x^6.\frac{1}{x^2}+6x^4.\frac{1}{x^4}+4x^2.\frac{1}{x^6}+\frac{1}{x^8}=7^4\)
\(\Leftrightarrow x^8+4x^4+6+\frac{4}{x^4}+\frac{1}{x^8}=2401\)(1)
Ta thấy x=0 không phải là nghiệm của phương trình nên ta có
\(\left(1\right)\Leftrightarrow\left(x^8+\frac{1}{x^8}\right)+\left(4x^4+\frac{4}{x^4}\right)+6=2401\)
\(\Leftrightarrow\left(x^4+\frac{1}{x^4}\right)^2-2.x^4.\frac{1}{x^4}+4\left(x^4+\frac{1}{x^4}\right)+6=2401\)
\(\Leftrightarrow\left(x^4+\frac{1}{x^4}\right)^2+4\left(x^4+\frac{1}{x^4}\right)=2397\)(2)
Đặt \(x^4+\frac{1}{x^4}=t\)ta có:
\(\left(2\right)\Leftrightarrow t^2+4t=2397\)
\(\Leftrightarrow t^2+4t-2397=0\)
\(\Leftrightarrow\left(t^2-47t\right)+\left(51t-2397\right)=0\)
\(\Leftrightarrow t\left(t-47\right)+51\left(t-47\right)=0\)
\(\Leftrightarrow\left(t-47\right)\left(t+51\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t-47=0\\t+51=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=47\\t=-51\end{cases}}}\)
Vì \(t=x^4+\frac{1}{x^4}\ge0\)nên \(t\ne-51\Rightarrow t=47\)
Ta lại có:
\(x^4+\frac{1}{x^4}=47\)
\(\Leftrightarrow\left(x^4+\frac{1}{x^4}\right)^2-2.x^4.\frac{1}{x^4}=47^2\)
\(\Leftrightarrow x^4+\frac{1}{x^8}=2209\)
Ta có:
\(\left(x^2+\frac{1}{x^2}\right)^2=x^4+\frac{1}{x^4}+2.x^4.\frac{1}{x^4}=7^2.\)
\(\Leftrightarrow x^4+\frac{1}{x^4}+2=49.\)
\(\Leftrightarrow x^4+\frac{1}{x^4}=47\)
\(\Leftrightarrow\left(x^4+\frac{1}{x^4}\right)^2=47^2\)
\(\Leftrightarrow x^8+\frac{1}{x^8}+2.x^4.\frac{1}{x^4}=2209\)
\(\Leftrightarrow x^8+\frac{1}{x^8}+2=2209.\)
\(\Leftrightarrow x^8+\frac{1}{x^8}=2207\)
\(\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2=7+2=9\)
\(\Rightarrow x+\frac{1}{x}=3\) (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 :
\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-1+\frac{1}{x^2}\right)\)
\(=\left(x+\frac{1}{x}\right)\left(7-1\right)\)(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 \(x^5+\frac{1}{x^5}=125\)
a) \(A=\frac{3x^2+6x+10}{x^2+2x+3}\)
\(A=\frac{3x^2+6x+9+1}{x^2+2x+3}\)
\(A=\frac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}\)
\(A=\frac{3\left(x^2+2x+3\right)}{x^2+2x+3}+\frac{1}{x^2+2x+1+2}\)
\(A=3+\frac{1}{^{\left(x+1\right)^2+2}}\le3+\frac{1}{2}=\frac{7}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-1\)
`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`
`b)` Với `x ne -1;x ne -5` có:
`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`
`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`
`A=[x^2-3x-4]/[(x+1)(x+5)]`
`A=[(x+1)(x-4)]/[(x+1)(x+5)]`
`A=[x-4]/[x+5]`
`c)` Với `x ne -5; x ne -1; x ne 4` có:
`P=A.B=[x-4]/[x+5].[-10]/[x-4]`
`=[-10]/[x+5]`
Để `P` nguyên `<=>[-10]/[x+5] in ZZ`
`=>x+5 in Ư_{-10}`
Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`
`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)
1) \(a^2+\frac{1}{a^2}=14\Leftrightarrow a^2+\frac{1}{a^2}+2a.\frac{1}{a}=16\Leftrightarrow\left(a+\frac{1}{a}\right)^2=16\Rightarrow a+\frac{1}{a}=4\)
\(\Rightarrow\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}\right)=a^3+\frac{1}{a}+a+\frac{1}{a^3}=a^3+4+\frac{1}{a^3}=4.14=56\)
\(\Rightarrow a^3+\frac{1}{a^3}=52\)
Ta có : \(\left(a^2+\frac{1}{a^2}\right)\left(a^3+\frac{1}{a^3}\right)=a^5+\frac{1}{a}+a+\frac{1}{a^5}=a^5+4+\frac{1}{a^5}=14.52\)
\(\Rightarrow a^5+\frac{1}{a^5}=14.52-4=724\)
2) \(A=2xy-x^2-4y^2+2x+10y-2000\)
\(=\left(-x^2+2xy-y^2\right)+\left(2x-2y\right)+\left(-3y^2+12y-12\right)-1988\)
\(=-\left(x-y\right)^2+2\left(x-y\right)-1-3\left(y^2-4y+4\right)-1987\)
\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2-1987\le-1987\forall x;y\) có GTLN là 2013
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}}\)
Vậy \(A_{max}=-1987\) tại \(x=3;y=2\)
ta có \(x^2+\frac{1}{x^2}\)
=\(\left(x+\frac{1}{x}\right)^2-2x\frac{1}{x}=\left(x+\frac{1}{x}\right)^2-2\)
=> \(\left(x+\frac{1}{x}\right)^2=25.vì\)\(x>0\Rightarrow x+\frac{1}{x}>0\Rightarrow x+\frac{1}{x}=5\)
\(\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3x+\frac{3}{x}=x^3+\frac{1}{x^3}+15\)
\(\Rightarrow x^3+\frac{1}{x^3}=5^3+15=110\)
\(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=x^5+\frac{1}{x^5}+x+\frac{1}{x}=x^5+\frac{1}{x^5}+5\)
\(\Rightarrow x^5+\frac{1}{x^5}=23\cdot110-5=2525\)
Vậy...
ta có : \(x^2+\dfrac{1}{x^2}=7\Leftrightarrow\dfrac{x^4+1}{x^2}=7\Leftrightarrow x^4+1=7x^2\)
\(\Leftrightarrow x^4-7x^2+1=0\)
đặc \(x^2=t\) \(\left(t\ge0\right)\)
khi đó : \(x^4-7x^2+1=0\Leftrightarrow t^2-7t+1=0\)
\(\Delta=\left(-7\right)^2-4.1.1=49-4=45>0\)
vậy phương trình có 2 nghiệm phân biệt
\(t_1=\dfrac{7+\sqrt{45}}{2}\left(tmđk\right)\) ; \(t_2=\dfrac{7-\sqrt{45}}{2}\left(tmđk\right)\)
ta có : \(t=x^2=\dfrac{7+\sqrt{45}}{2}\Leftrightarrow x=\pm\sqrt{\dfrac{7+\sqrt{45}}{2}}\)
\(t=x^2=\dfrac{7-\sqrt{45}}{2}\Leftrightarrow x=\pm\sqrt{\dfrac{7-\sqrt{45}}{2}}\)
tìm được giá trị của \(x\) thế vào \(x^5+\dfrac{1}{x^5}\) tìm giá trị là xong
ko hiểu gì hết