a) \(\left|x+2\right|+\left|2y-1\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}\left|x+2\right|=0\\\left|2y-1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+2=0\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=0,5\end{cases}}\)

Vậy (x; y) = (-2; 0,5)

b) \(\left|x-y\right|+\left|2x+3\right|=0\Leftrightarrow\hept{\begin{cases}\left|x-y\right|=0\\\left|2x+3\right|=0\end{cases}}\)

+) |2x + 3| = 0

2x + 3 = 0

2x = -3

x = -1,5

+) |x - y| = 0

x - y = 0

-1,5 - y = 0

y = -1,5

Vậy (x; y) = (-1,5; -1,5)

c, \(\left|2x+y\right|+\left|y+\left(1:4\right)\right|=0\)

\(\left|2x+y\right|+\left|y+\frac{1}{4}\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}\left|2x+y\right|=0\\\left|y+\frac{1}{4}\right|=0\end{cases}}\)

\(\left|y+\frac{1}{4}\right|=0\Leftrightarrow y+\frac{1}{4}=0\Leftrightarrow y=-\frac{1}{4}\)

\(\left|2x+y\right|=0\Leftrightarrow2x+y=0\Leftrightarrow2x-\frac{1}{4}=0\Leftrightarrow2x=\frac{1}{4}\Leftrightarrow x=\frac{1}{8}\)

Vậy \(\left(x;y\right)=\left(\frac{1}{8};-\frac{1}{4}\right)\)

a) |x+2|+|2y-1|=0

=> |x+2| =0=> x+2=0 => x= -2

     |2y-1|=0=> 2y-1=0 => 2y=1 => y = 1:2 =0,5

b) |x-y|+|2x+3|=0

=> |2x+3|=0 => 2x+3 => 2x= -3 => x = -3:2 = -1,5

     |x-y|=0 => x-y =0 => y = -1,5 - 0 = -1,5

Ta thấy :

\(\left|x-2012\right|+\left|x-2013\right|=\left|x-2012\right|+\left|2013-x\right|\ge\left|x-2012+2013-x\right|=1\)

\(\Rightarrow VT\ge VP\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-2012\right)\left(2013-x\right)\ge0\Leftrightarrow2012\le x\le2013\)

Vậy \(2012\le x\le2013\)

x-2013+x-2015=1

x-(2013+2015)=1

x-2025=1

x

​=1+2025=2026

\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c};c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)

 \(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)

\(\Rightarrow k^3=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\) (TC DTSBN) (1)

Ta lại có : \(k^3=\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\) (2)

Từ (1) ; (2) \(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\) (đpcm)

thanks bạn