Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2+\frac{1}{x^2}=7\Leftrightarrow x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=9\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2=9\Leftrightarrow x+\frac{1}{x}=3\)
\(P=x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3x.\frac{1}{x}\left(x+\frac{1}{x}\right)=3^3-3.3=18\)
\(Q=\left(x^3+\frac{1}{x^3}\right)\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)=7.18-3=...\)
7) Ta có : \(\frac{5x-2}{3}=\frac{5-3x}{3}\)
=> \(5x-2=5-3x\)
=> \(5x+3x=5+2\)
=> \(8x=7\)
=> \(x=\frac{8}{7}\)
8) Ta có : \(\left(6x+3\right)\left(5x-20\right)=0\)
=> \(\left[{}\begin{matrix}6x+3=0\\5x-20=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=4\end{matrix}\right.\)
10) ĐKXĐ : \(x\ne5\)
Ta có : \(\frac{2x-5}{x+5}=3\)
=> \(2x-5=3\left(x+5\right)\)
=> \(2x-5-3x-15=0\)
=> \(x=-20\) ( TM )
11) ĐKXĐ : \(x-2\ne0\)
=> \(x\ne2\)
Ta có : \(\frac{1}{x-2}+4=\frac{x-3}{2-x}\)
=> \(\frac{1}{x-2}+\frac{4\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\)
=> \(1+4\left(x-2\right)=3-x\)
=> \(1+4x-8-3+x=0\)
=> \(5x=10\)
=> x = 2 ( KTM )
Vậy phương trình trên vô nghiệm.
7) \(\frac{5x-2}{3}=\frac{5-3x}{3}\)
\(\Leftrightarrow\) 5x-2=5-3x
\(\Leftrightarrow\) 5x+3x=5+2
\(\Leftrightarrow\) 8x=7
\(\Leftrightarrow\) x=\(\frac{7}{8}\)
8) (6x+3)(5x-20)=0
\(\Rightarrow\) 6x+3=0 hoặc 5x-20=0
\(\Rightarrow\) 6x=-3
\(\Rightarrow\) x=\(\frac{-1}{2}\)
a) \(B=\left[\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x-4}{x-3}-\frac{\left(x-1\right)}{x+3}\right]:\left(\frac{x+3-1}{x+3}\right)\)
ĐK: \(\hept{\begin{cases}x\ne3\\x\ne-3\end{cases}}\)
\(=\left[\frac{21+x-4-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+2}{x+3}\right)\)
\(=\left[\frac{21+x-4-x^2+3x+x-3}{\left(x+3\right)\left(x-3\right)}\right]\times\left(\frac{x+3}{x+2}\right)\)
\(=\left(\frac{-x^2+5x+14}{x-3}\right)\left(\frac{1}{x+2}\right)\)
\(=\frac{-\left(x^2+2x-7x-14\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{-\left(x+2\right)\left(x-7\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{7-x}{x-3}\)
b) \(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Mà \(x\ne-3\)
\(\Rightarrow x=2\)
Thế \(x=2\)vào B ta được:
\(B=\frac{7-2}{2-3}=-5\)
c) \(B=\frac{7-x}{x-3}=\frac{-3}{5}\)
\(\Leftrightarrow5\left(7-x\right)=-3\left(x-3\right)\)
\(\Leftrightarrow35-5x+3x-9=0\)
\(\Leftrightarrow-2x=-26\)
\(\Leftrightarrow x=13\)
Vậy để \(B=\frac{-3}{5}\)thì \(x=13\)
d) B<0\(\Rightarrow\frac{7-x}{x-3}< 0\)
TH1: \(\hept{\begin{cases}7-x< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x>7\\x>3\end{cases}\Rightarrow}x>7}\)
TH2: \(\hept{\begin{cases}7-x>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 7\\x< 3\end{cases}\Rightarrow}x< 3}\)
Để B<0 thì x>7 hoặc x<3
a) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\) ĐKXĐ: x khác =-3; x khác -2
\(B=\frac{21+x^2-x-12-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)
\(B=\frac{3x+6}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)
\(B=\frac{3\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+2}\)
\(B=\frac{3}{x-3}\)
b) bước đầu tiên ta phải tìm x:
\(\left|2x+1\right|=5\)
TH1: 2x+1=5 TH2: 2x+1=-5
2x=4 2x=-6
x=2 (nhận) x=-3 (loại)
thay x=2 vào biểu thức B, ta được:
\(B=\frac{3}{2-3}=\frac{3}{-1}=-3\)
vậy B=-3 tại x=2
c) Để \(B=-\frac{3}{5}\)thì \(\frac{3}{x-3}=-\frac{3}{5}\)
\(\Leftrightarrow-3\left(x-3\right)=15\)
\(\Leftrightarrow x-3=-5\)
\(\Leftrightarrow x=-2\)
vậy \(x=-2\)thì \(B=-\frac{3}{5}\)
d) để B<0 thì \(\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)
vậy để B<0 thì x phải < 3 và x khác -3
Ta có: \(\left(x+\frac{1}{x}\right)^3=x^3+\frac{3x^2}{x}+\frac{3x}{x^2}+\frac{1}{x^3}\)
\(x^3+\frac{1}{x^3}+3x+\frac{3}{x}=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=27\)
\(x^3+\frac{1}{x^3}=18\)
Ta có:
\(\left(x+\frac{1}{x}\right)^5=x^5+\frac{5x^4}{x}+\frac{10x^3}{x^2}+\frac{10x^2}{x^3}+\frac{5x}{x^4}+\frac{1}{x^5}\)
\(x^5+\frac{1}{x^5}+5x^3+10x+\frac{10}{x}+\frac{5}{x^3}=243\)
\(x^5+\frac{1}{x^5}+5\left(x^3+\frac{1}{x^3}\right)+10\left(x+\frac{1}{x}\right)=243\)
\(x^5+\frac{1}{x^5}+5\left(x^3+\frac{1}{x^3}\right)=213\)
\(x^5+\frac{1}{x^5}=123\)
a,x4+1/x4=(x2)2+(1/x2)2=(x2+1/x2)2-2 , bình phương gt lên rồi tính x2+1/x2
b,x5+1/x5=x5+(1/x)5=(x2+1/x2)(x3+1/x3)-x2/x3-x3x2=(x2+1/x2)(x3+1/x3)-(x+1/x)..... tự làm tiếp
\(C=\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{2x\left(x-1\right)}{\left(x+3\right)\left(x-3\right)}\) ( x khác 3 ; -3 )
\(=\frac{x^2+4x+3+x^2-4x+3+2x^2-2x}{\left(x+3\right)\left(x-3\right)}=\frac{4x^2-2x+6}{\left(x+3\right)\left(x-3\right)}=\frac{2\left(2x^2-x+3\right)}{\left(x+3\right)\left(x-3\right)}\)
Tại x = 5=> \(C=\frac{2\left(2.5^2-5+3\right)}{\left(5+3\right)\left(5-3\right)}=\frac{48}{8}=6\)
ta có
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)
Vì x+ 1/x =3 =>(x + 1/x)2=9=x2+1/x2+2 =>x2+1/x2 = 7
Vì x+ 1/x =3 =>(x+ 1/x)3=27=x3+ 1/x3 +3(x +1/x) =>x3 + 1/x3=18 ==>(x3 + 1/x3)(x2+1/x2)=126=x5+ 1/x5 +x +1/x
từ đó => x5+ 1/x5=123