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\(\frac{x^9-1}{x^9+1}=7\)=>x9-1=7x9+1
=>x9=\(\frac{-8}{6}\)
=>(x9)2=(\(\frac{-8}{6}\))2
=>x18=\(\frac{16}{9}\)=>..................................
Đặt \(a=x;2b=y;3c=z\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow Q=\frac{\frac{y}{2}.\frac{z}{3}}{x^2}+\frac{x.\frac{c}{3}}{2y^2}+\frac{x.\frac{y}{2}}{3z^2}\)
\(=\frac{x^3y^3+y^3z^3+z^3x^3}{6x^2y^2z^2}\)
\(=\frac{x^3y^3+y^3z^3+z^3x^3-3x^2y^2z^2+3x^2y^2z^2}{6x^2y^2z^2}\)
\(=\frac{\left(xy+yz+zx\right)\left(x^2y^2+y^2z^2+z^2x^2-x^2yz-y^2zx+z^2xy\right)+3x^2y^2z^2}{6x^2y^2z^2}\)
\(=\frac{3x^2y^2z^2}{6x^2y^2z^2}=\frac{1}{2}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1-\frac{1}{100}\)
\(B=\frac{99}{100}\)
a,\(A=\frac{6x+12}{\left(x+2\right)\left(2x-6\right)}=\frac{6\left(x+2\right)}{2\left(x+2\right)\left(x-3\right)}=\frac{3}{x-3}\)
b, Giá trị của x để phân thức có giá trị bằng (-2) :
\(\frac{3}{x-3}=-2\Rightarrow x=1,5\)
a)Ta có : \(4x^2=1\)
\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
mà \(x\ne-\frac{1}{2}\Rightarrow x=\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào B , ta được:
\(B=\frac{\left(\frac{1}{2}\right)^2-\frac{1}{2}}{2.\frac{1}{2}+1}=\frac{\frac{1}{4}-\frac{1}{2}}{1+1}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}\)
Vậy \(B=-\frac{1}{8}\)khi \(4x^2=1\)
b)Ta có : \(A=\frac{1}{x-1}-\frac{x}{1-x^2}\)
\(=\frac{1}{x-1}+\frac{x}{x^2-1}\)
\(=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow M=A.B=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x^2-x}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x\left(x-1\right)}{2x+1}\)
\(=\frac{x}{x+1}\)
Vậy \(M=\frac{x}{x+1}\)
c)Ta có: \(x< x+1\forall x\)
\(\Rightarrow M=\frac{x}{x+1}< \frac{x+1}{x+1}=1\forall x\ne-1\)
Vậy với mọi \(x\ne-1\)thì \(M< 1\)
B=\(\frac{1}{3\cdot5\cdot7}+\frac{1}{5\cdot7\cdot9}+........+\frac{1}{2009+2011+2013}\)
\(\Leftrightarrow4B=\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+.....+\frac{4}{2009+2011+2013}\)
4B=\(\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+\frac{1}{5\cdot7}-\frac{1}{7\cdot9}+......-\frac{1}{2009\cdot2011}+\frac{1}{2009\cdot2011}-\frac{1}{2011\cdot2013}\)
4B=\(\frac{1}{3\cdot5}-\frac{1}{2011\cdot2013}\)=>B=\(\left(\frac{1}{3\cdot5}-\frac{1}{2011\cdot2013}\right)\div4\)
thanks Nguyễn Tấn Tài nạk :))