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\(\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2=7+2=9\)
\(\Rightarrow x+\frac{1}{x}=3\) (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.
thay z = -(x+y) , y = -(z+x),... vao
=> Duoc bieu thuc trong do co 1/xy + 1/yz + 1/zx = (x+y+z)/xyz = 0
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 \(x^5+\frac{1}{x^5}=125\)
1, \(=\left[\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right]:\frac{1-x^2}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(=\left(1+x+x^2-x\right):\frac{1-x^2}{\left(1-x\right)\left(1-x^2\right)}\)\(=\left(x^2+1\right)\left(1-x\right)\)
2, để B<0 <=> (x2+1)(1-x)<0
vì x^2+1 > 0 với mọi x
=> \(\hept{\begin{cases}x^2+1>0\\1-x< 0\end{cases}\Leftrightarrow x>1}\)
3, \(\left|x-4\right|=5\Leftrightarrow\orbr{\begin{cases}x=9\\x=-1\left(loại\right)\end{cases}}\)
Thay x=9 vào B ta có: B=(92+1)(1-9)=82.(-8)=-656
a)\(A=\frac{x+1}{x^2+2x+1}:\left(\frac{1}{x^2-x}+\frac{1}{x-1}\right)\left(ĐK:x\ne0;x\ne1\right)\)
\(=\frac{x+1}{\left(x+1\right)^2}:\frac{1+x}{x\left(x-1\right)}\)
\(=\frac{1}{x+1}\cdot\frac{x\left(x+1\right)}{x+1}=\frac{x}{x+1}\)
b)Có: \(x^2+x-2=0\\ \Leftrightarrow x^2-x+2x-2=0\\ \Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\left(loại\right)\\x=-2\end{array}\right.\)
Thay x=-2 vào A ta có
\(A=\frac{-2}{-2+1}=\frac{-2}{-1}=2\)
\(ĐKXĐ:\)\(x\ne\left\{0;1;2;3;4;5\right\}\)
\(P=\frac{1}{x^2-x}+\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}\)
\(=\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-5}-\frac{1}{x-4}\)
\(=\frac{1}{x-5}-\frac{1}{x}\)
\(=\frac{5}{x\left(x-5\right)}\)
Ta có: \(x^3-x^2+2=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-2x+2\right)=0\)
Xét: \(x^2-2x+2=\left(x-1\right)^2+1\)\(>0\)
\(\Rightarrow\)\(x+1=0\)
\(\Leftrightarrow\)\(x=-1\)(t/m)
Vậy tại \(x=-1\) thì:
\(P=\frac{5}{-1\left(-1-5\right)}=\frac{5}{6}\)
ĐKXĐ \(x\ne0,1,2,3,4,5\)
\(P=\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(P=\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+...+\frac{1}{x-5}-\frac{1}{x-4}\)
\(P=\frac{1}{x-5}-\frac{1}{x}\)
\(P=\frac{5}{x\left(x-5\right)}\)
a)\(A=\frac{x+1}{x^2-2x+1}:\left(\frac{1}{x^2-x}+\frac{1}{x-1}\right)\left(ĐK:x\ne0;x\ne1\right)\)
\(=\frac{x+1}{\left(x-1\right)^2}:\left(\frac{1}{x\left(x-1\right)}+\frac{1}{x-1}\right)\)
\(=\frac{x+1}{\left(x-1\right)^2}:\frac{1+x}{x\left(x-1\right)}\)
\(=\frac{x+1}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{1+x}\)
\(=\frac{x}{x-1}\)
b)Có:\(x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\left(loại\right)\\x=2\end{array}\right.\)
Thay x=2 vào A ta được:
\(A=\frac{2}{2-1}=2\)
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...