I = |x+\(\frac{1}{2}\)| + |x+\(\frac{1}{3}\)| + |x+\(\frac{1}{4}\)| tìm min hoặc max của I
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nếu x>0
=> 2x=x+5
<=> 2x-x=5
<=> x=5
nếu x\(\le\)0
=>-2x=x+5
<=> -2x-x=5
<=> -3x=5
<=> x=-5/3
vậy pt có tập nghiệm x={5;-5/3}
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\(2^x\times8^{x-1}=32\Leftrightarrow2^x\times2^{3x-3}=32\)
\(\Leftrightarrow2^{4x-3}=32\Leftrightarrow2^{4x-3}=2^5\)
\(\Rightarrow4x-3=5\Leftrightarrow4x=8\Rightarrow x=2\)
vậy \(x=2\)
\(Ta có : 2 ^x . 8\)\(x - 1\) \(= 32\)
\(\Rightarrow\)\(2 ^ x . 2\)\(3. ( x - 1 )\) \(=\) \(32\)
\(\Rightarrow\)\(2 \)\(x + 3. ( x - 1 ) \)\(= 32\)
\(\Rightarrow\)\(2\)\(x + 3x - 3 \) \(= 32\)
\(\Rightarrow\)\(2\)\(4x - 3\) \(= 32\)
\(\Rightarrow\)\(2\)\(4x - 3\)\(= 2^5\)
\(\Rightarrow\)\(4x - 3 = 5\)
\(\Rightarrow\)\(4x = 8\)
\(\Rightarrow\)\(x = 2\)
\(Vậy : x = 2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{x}{2015}=\frac{y}{2016}=\frac{z}{2017}=k\)
\(\Rightarrow x=2015k;y=2016k;z=2017k\)
Ta có:
\(\left(x-z\right)^3=\left(2015k-2017k\right)^3=-8k^3\left(1\right)\)
Mặt khác:
\(-8\left(x-y\right)^2\left(z-y\right)=-8\left(2015k-2016k\right)^2\left(2017k-2016k\right)\)
\(=-8k^2\cdot k=-8k^3\left(2\right)\)
Từ ( 1 );( 2 ) suy ra đpcm
Ta có:
\(I=\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{4}\right|=\left(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{4}\right|\right)+\left|x+\frac{1}{3}\right|\)
\(=\left(\left|x+\frac{1}{2}\right|+\left|-x-\frac{1}{4}\right|\right)+\left|x+\frac{1}{3}\right|\ge\left|x+\frac{1}{2}-x-\frac{1}{4}\right|+\left|x+\frac{1}{3}\right|=\frac{1}{4}+\left|x+\frac{1}{3}\right|\ge\frac{1}{4}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x+\frac{1}{2}\right)\left(-x-\frac{1}{4}\right)\ge0\\x+\frac{1}{3}=0\end{cases}}\Leftrightarrow x=-\frac{1}{3}\)
Vậy min I = 1/4 đạt tại x = -1/3.