Cho hàm số f(x)=2x+1/2x+3
a) Tìm x để f(x)=3
b)Tìm giá trị nguyên của x để f(x) có giá trị nguyên
c) Tìm giá trị của x để f(x)>1
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a: ĐKXĐ: x<>-2/3
b: F=0
=>8-2x=0
=>x=4
d: F<0
=>(2x-8)/(3x+2)>0
=>x>4 hoặc x<-2/3
a: f(-3)=10
f(0)=-8
f(1)=-6
f(2)=0
b: f(x)=0
=>(x-2)(x+2)=0
=>x=2 hoặc x=-2
a) x khác 1
b) f(7)=\(\frac{3}{2}\)
c)\(\frac{x+2}{x-1}\)=\(\frac{1}{4}\)<=> 4(x+2)=x-1<=>x=-3
d) f(x)=\(\frac{x+2}{x-1}\)=\(\frac{x-1+3}{x-1}\)= 1+\(\frac{3}{x-1}\)
f(x) có giá trị nguyên <=> x-1 thuộc Ư(3) <=> x-1 thuộc {+1;+3}
x-1 | -1 | 1 | 3 | -3 |
x | 0 | 2 | 4 | -2 |
e) f(x)>1 <=> 1+\(\frac{3}{x-1}\)> 1 <=> \(\frac{3}{x-1}\)> 0 <=> x-1 >0 <=> x>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 \(f\left(x\right)_{min}=4031\Leftrightarrow-2016\le x\le2015\)
a) \(f\left(x\right)=\frac{x+2}{x-1}\)
\(f\left(x\right)=\frac{1}{4}\Leftrightarrow\frac{x+2}{x-1}=\frac{1}{4}\)
\(\Leftrightarrow4\left(x+2\right)=x-1\)
\(\Leftrightarrow4x+8=x-1\)
\(\Leftrightarrow4x-x=-1-8\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\)
Vậy x = -3 thì hàm số y = f(x) = \(\frac{1}{4}\)
b) \(f\left(x\right)=\frac{x+2}{x-1}=\frac{x-1+3}{x-1}=1+\frac{3}{x-1}\)
Để f(x) nguyên thì \(\frac{3}{x-1}\)nguyên
hay \(3⋮\left(x-1\right)\Leftrightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Lập bảng:
\(x-1\) | \(1\) | \(-1\) | \(3\) | \(-3\) |
\(x\) | \(2\) | \(0\) | \(4\) | \(-2\) |
Vậy \(x\in\left\{2;0;4;-2\right\}\) thì f(x) nguyên
a) Ta có: f(x) = 1/4
=> \(\frac{x+2}{x-1}=\frac{1}{4}\)
=> \(4\left(x+2\right)=x-1\)
=> 4x + 8 = x - 1
=> 4x - x = -1 - 8
=> 3x = -9
=> x = -3
b) Ta có: \(f\left(x\right)=\frac{x+2}{x-1}=\frac{\left(x-1\right)+3}{x-1}=1+\frac{3}{x-1}\)
Để f(x) có giá trị nguyên <=> \(3⋮x-1\) <=> \(x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Lập bảng :
x - 1 | 1 | -1 | 3 | -3 |
x | 2 | 0 | 4 | -2 |
Vậy ...
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