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a) \(f\left(-1\right)=2.\left(-1\right)^2-9=2.1-9=2-9=-7\)
\(f\left(0\right)=2.0^2-9=0-9=-9\)
\(f\left(1\right)=2.1^2-9=2-9=-7\)
b) \(f\left(x\right)=-1\)\(\Leftrightarrow2x^2-9=-1\)
\(\Leftrightarrow2x^2=8\)\(\Leftrightarrow x^2=4\)\(\Leftrightarrow x=\pm2\)
Vậy với \(x=\pm2\)thì \(f\left(x\right)=-1\)
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\(f\left(x\right)=\left|x-2015\right|+\left|x+2016\right|\)
a) Ta có: \(\left|x\right|=\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
+) Với \(x=\frac{1}{2}\):
\(f\left(\frac{1}{2}\right)=\left|\frac{1}{2}-2015\right|+\left|\frac{1}{2}+2016\right|=2\)
+) Với \(x=-\frac{1}{2}\)
\(f\left(-\frac{1}{2}\right)=\left|-\frac{1}{2}-2015\right|+\left|-\frac{1}{2}+2016\right|=0\)
c) Áp dụng BĐT |x| + |y| \(\ge\)|x + y|, ta được:
\(f\left(x\right)=\left|x-2015\right|+\left|x+2016\right|=\left|2015-x\right|+\left|x+2016\right|\)
\(\ge\left|\left(2015-x\right)+\left(x+2016\right)\right|=\left|4031\right|=4031\)
(Dấu "="\(\Leftrightarrow\left(2015-x\right)\left(x+2016\right)\ge0\)
TH1: \(\hept{\begin{cases}2015-x\ge0\\x+2016\ge0\end{cases}}\Leftrightarrow-2016\le x\le2015\)
TH2: \(\hept{\begin{cases}2015-x\le0\\x+2016\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2015\\x\le-2016\end{cases}}\left(L\right)\))
Vậy \(f\left(x\right)_{min}=4031\Leftrightarrow-2016\le x\le2015\)
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a) Để \(f\left(x\right)=3\)
\(\Leftrightarrow\frac{2x+1}{2x+3}=3\)
\(\Leftrightarrow3.\left(2x+3\right)=2x+1\)
\(\Leftrightarrow6x+9=2x+1\)
\(\Leftrightarrow6x-2x=1-9\)
\(\Leftrightarrow4x=-8\)
\(\Leftrightarrow x=-2\)
Để f(x) nguyên
\(\Leftrightarrow2x+1⋮2x+3\)
\(\Leftrightarrow2x+3-2⋮2x+3\)
mà \(2x+3⋮2x+3\)
\(\Rightarrow2⋮2x+3\)
\(\Rightarrow2x+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Lập bảng rồi tìm x nguyên nhé
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a) Thay f(-3) vào hàm số ta có :
y=f(-3)=2.(-3)2-8=10
Thay f(0) vào hàm số ta có :
y=(f0)=2.02-8=-8
Thay f(1) vào hàm số ta có :
y=f(1)=2.12-8=-6
Thay f(2) vào hàm số ta có :
y=f(2)=2.22-8=0
b) y=f(x)=0 <=> 2x2-8=0
2x2=8
x2=8:2
x2=4
=> x=2
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bài 1:
a) y=f(0)=|1-0|+2=3
y=f(1)=|1-(-1)|+2=4
y=f(-1/2)=|1-(-1/2)|+2=7/2
b) f(x)=3 <=> |1-x|+2=3
|1-x|=3-2
|1-x|=1
=> \(\orbr{\begin{cases}1-x=1\\1-x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
f(x)=3-x <=> |1-x|+2=3-x
|1-x|=3-x-2
|1-x|=1-x
=> (1-x)-(1-x)=0
2.(1-x)=0
=> 1-x=0
=> x=1
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a) * f(-2)
=-2.(-2)+1
=2
* f(3)
=-2.3+1
=-5
b) hàm số y=-2x+1
với x=-1 thì y=3 không bằng 1
Vậy M(-1,1)ko thuộc đồ thị hàm số f(x)
c) ta có 1>0
=> -2x+1=1
-2x=1-1
-2x=0
x=0/(-2)
x=0
=> x=0
vậy x=0 thì f(x)>0
nhớ k giùm mình nha
a)\(F\left(-2\right)=-2.\left(-2\right)+1=5\)
\(F\left(\frac{1}{2}\right)=-2.\left(\frac{1}{2}\right)+1=0\)
\(F\left(3\right)=-2.3+1=-5\)
\(F\left(1\right)=-2.1+1=-1\)
Để f(x) = -1 khi 2x2-9 = -1
2x2 = 8
x2=4
x=2 hoặc x=-2
Vậy x thuộc {-2;2) thì (x) = -1
a) f(x) = -1 <=> 2x^2-9= -1
<=> 2x^2 = -1 + 9
<=> 2x^2 = 8
<=> x^2 = 8 : 2
<=> x^2 = 4
<=> x = 2