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f(x)=ax-b
=> f(2)=2a-b=8(thay x=2)
f(-2)=-2a-b=0(Thay x=-2)
Cộng vế với vế => 2a-b-2a-b=8
=> -2b=8
=>b=-4
=> a=2
\(a,\left|2x-5\right|=1\)
\(\Rightarrow\orbr{\begin{cases}2x-5=1\\2x-5=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\2x=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
b, đề thiếu
X2-\(\frac{7}{9}\)X=0 <=> X(X-\(\frac{7}{9}\))=0
=> x=0 hoặc x-\(\frac{7}{9}\)=0
x-\(\frac{7}{9}\)=0 <=>X=0+\(\frac{7}{9}\)=\(\frac{7}{9}\)
=> X=0 hoặc \(\frac{7}{9}\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(=>\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(=>\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(=>\left(\frac{x+2016}{2015}+\frac{x+2016}{2014}\right)-\left(\frac{x+2016}{2013}+\frac{x+2016}{2012}\right)=0\)
\(=>\left(x+2016\right).\left[\left(\frac{1}{2015}+\frac{1}{2014}\right)-\left(\frac{1}{2013}+\frac{1}{2012}\right)\right]=0\)
\(=>\orbr{\begin{cases}x+2016=0\\\left(\frac{1}{2015}+\frac{1}{2014}\right)-\left(\frac{1}{2013}+\frac{1}{2012}\right)=0\end{cases}}\)
Do 1/2015 + 1/2014 < 1/2013 + 1/2012
=> (1/2015 + 1/2014) - (1/2013 + 1/2012) khác 0
=> x - 2016 = 0
=> x = 2016
Vậy x = 2016
Ủng hộ mk nha ^_-
\(\frac{x-2}{8}=\frac{x-3}{9}\)
\(9\left(x-2\right)=8\left(x-3\right)\)
\(9x-18=8x-24\)
\(9x-8x=-24+18\)
\(x=\text{-6}\)
Vậy \(x=-6\)
\(\frac{x-2}{8}=\frac{x-3}{9}\)
\(\left(x-2\right)\cdot9=\left(x-3\right)\cdot8\)
\(9x-18=8x-24\)
\(9x-8x=-24+18\)
\(x=-6\)