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a) Ta có: f(2)-f(-1)=(m-1).2-[(m-1).(-1)]=7
<=> 2m-2+m-1=7 <=> 3m=10 => m=10/3
b) m=5 => f(x)=4x
=> f(3-2x)=4(3-2x)=20 <=> 3-2x=5 => 2x=-2 => x=-1
\(a.f\left(1\right)=f\left(-1\right)\Leftrightarrow a+b+c=a-b+c\Leftrightarrow2b=0\Leftrightarrow b=0\)
\(\Rightarrow f\left(x\right)=ax^2+c\)
Khi đó ta có:
\(\left\{{}\begin{matrix}f\left(m\right)=am^2+c\\f\left(-m\right)=am^2+c\end{matrix}\right.\Rightarrow f\left(m\right)=f\left(-m\right)\forall m\)
a, Ta có:
\(f\left(2\right)=\left(m-1\right).2=2m-2\)
\(f\left(-1\right)=\left(m-1\right).\left(-1\right)=-m+1\)
Theo bài ra ta có:
\(f\left(2\right)-f\left(-1\right)=7\)
\(\Rightarrow2m-2-\left(-m+1\right)=7\)
\(\Rightarrow2m-2+m-1=7\)
\(\Rightarrow3m=10\Rightarrow m=\dfrac{10}{3}\)
b, Ta có:
\(f\left(3-2x\right)=\left(5-1\right)\left(3-2x\right)=20\)
\(\Rightarrow4\left(3-2x\right)=20\)
\(\Rightarrow3-2x=5\Rightarrow2x=2\Rightarrow x=1\)
Chúc bạn học tốt!!!
a) \(f\left(2\right)-f\left(-1\right)=7\)
\(\Rightarrow\left(m-1\right)2-\left(m-1\right)\left(-1\right)=7\)
\(\Rightarrow2m-2+m-1=7\)
\(\Rightarrow3m=10\Rightarrow m=\dfrac{10}{3}\)
b) \(f\left(3-2x\right)=20\)
\(\Rightarrow\left(m-1\right)\left(3-2x\right)=20\)
\(\Rightarrow\left(5-1\right)\left(3-2x\right)=20\)
\(\Rightarrow4\left(3-2x\right)=20\)
\(\Rightarrow12-8x=20\)
\(\Rightarrow8x=12-20=-8\)
\(\Rightarrow x=-1\)
a) Ta có: f(2)-f(-1)=(m-1).2-[(m-1).(-1)]=7
<=> 2m-2+m-1=7 <=> 3m=10 => m=10/3
b) m=5 => f(x)=4x
=> f(3-2x)=4(3-2x)=20 <=> 3-2x=5 => 2x=-2 => x=-1