Lời giải:
$2014|x-12|+(x-12)^2=2013|12-x|$

$\Rightarrow 2014|x-12|+|x-12|^2=2013|x-12|$

$\Rightarrow |x-12|+|x-12|^2=0$

$\Rightarrow |x-12|(1+|x-12|)=0$

Hiển nhiên $1+|x-12|\geq 1>0$ với mọi $x$

$\Rightarrow |x-12|=0$

$\Rightarrow x=12$

Lời giải:

Áp dụng TCDTSBN:

$\frac{x}{2}=\frac{y}{3}=\frac{x-y}{2-3}=\frac{4}{-1}=-4$

$\Rightarrow x=2(-4)=-8; y=3(-4)=-12$

a. Nếu $x\geq 4$ thì:

$x-1+x-4=7$

$\Rightarrow 2x=12$

$\Rightarrow x=6$ (tm) 

Nếu $1\leq x< 4$ thì:

$x-1+4-x=7$

$\Rightarrow 3=7$ (vô lý - loại) 

Nếu $x<1$ thì:

$1-x+4-x=7$

$5-2x=7$

$2x=-2$

$x=-1$ (tm)

b. Nếu $x\geq 3$ thì:

$x-3+x+10=13$

$\Rightarrow 2x+7=13$

$\Rightarrow 2x=6$

$\Rightarrow x=3$ (tm)

Nếu $-10\leq x< 3$ thì:

$3-x+x+10=13$

$\Rightarrow 13=13$ (luôn đúng) 

Nếu $x<-10$ thì:

$3-x-x-10=13$

$\Rightarrow -7-2x=13$

$\Rightarrow 2x=-20$

$\Rightarrow x=-10$ (loại vì $x<-10$)

Vậy $x=3$ hoặc $-10\leq x< 3$

Lời giải:

Gọi biểu thức là $A$.

\(A=\frac{(2^4)^3.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\\ =\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}(2.3+1)}\\ =\frac{2^{12}.3^{10}(1+5)}{7.2^{11}.3^{11}}=\frac{2^{12}.3^{10}.2.3}{7.2^{11}.3^{11}}\\ =\frac{2^{13}.3^{11}}{7.2^{11}.3^{11}}=\frac{2^2}{7}=\frac{4}{7}\)

\(\left(x-2\right)^8=\left(x-2\right)^6\)

\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^6=0\)

\(\Leftrightarrow\left(x-2\right)^6.\left[\left(x-2\right)^2-1\right]=0\)

Trường hợp 1: \(\left(x-2\right)^6=0\)

\(\Leftrightarrow\left(x-2\right)^6=0^6\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Trường hợp 2: \(\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2\right)^2=1^2\)

\(\Leftrightarrow x-2=1\)

\(\Leftrightarrow x=3\)

\(3x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=-\dfrac{16}{4}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{7}=-4\Rightarrow x=-4.7=-28\\\dfrac{y}{3}=-4\Rightarrow y=-4.3=-12\end{matrix}\right.\)

\(\Rightarrow x+y=\left(-28\right)+\left(-12\right)=-40\)

Theo t/c dãy tỉ số bằng nhau, ta có \(\frac{2}{3x}\)\(=\frac{1}{2y}\)\(=\frac{2}{z}\)\(=\frac{2+1+2}{3x+2y+z}=\frac{5}{1}=5\)

\(\to\) \(\frac{2}{3x}\)=5 \(\to\)x=2/15. Tương tự, tính dk y, z

Có:

\(\left|2x-27\right|^{2011}\ge0\forall x\)

\(\left(3y+10\right)^{2012}\ge0\forall y\)

\(\Rightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\forall x;y\)

\(\Rightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=-\dfrac{10}{3}\end{matrix}\right.\)