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.

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}

\(2^x\times8^{x-1}=32\Leftrightarrow2^x\times2^{3x-3}=32\)

\(\Leftrightarrow2^{4x-3}=32\Leftrightarrow2^{4x-3}=2^5\)

\(\Rightarrow4x-3=5\Leftrightarrow4x=8\Rightarrow x=2\)

vậy \(x=2\)

\(Ta có : 2 ^x . 8\)^{\(x - 1\)} \(= 32\)

\(\Rightarrow\)\(2 ^ x . 2\)^{\(3. ( x - 1 )\)} \(=\) \(32\)

\(\Rightarrow\)\(2 \)^{\(x + 3. ( x - 1 ) \)}\(= 32\)

\(\Rightarrow\)\(2\)^{\(x + 3x - 3 \)} \(= 32\)

\(\Rightarrow\)\(2\)^{\(4x - 3\)} \(= 32\)

\(\Rightarrow\)\(2\)^{\(4x - 3\)}\(= 2^5\)

\(\Rightarrow\)\(4x - 3 = 5\)

\(\Rightarrow\)\(4x = 8\)

\(\Rightarrow\)\(x = 2\)

\(Vậy : x = 2\)

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

\(a ) Ta có : a / b = 2 và b / c = 3\)

\(\Rightarrow\)\(a = 2b \)\(và \)\(b = 3c\)

\(A = ( a + b ) / ( b + c ) \)

\(A = ( 2b + b ) / ( 3c + c )\)

\(A = 3b / 4c\)

\(A = 3 / 4 . b / c \)

\(A = 3 / 4 . 3 \)

\(A = 9 / 4\)