Vì \(\left|4-x\right|\ge0\forall x\)

\(\Rightarrow11+\left|4-x\right|\ge11+0=11\)

\(\Rightarrow A\ge11\)

\(\Rightarrow\) GTNN của Alà 11\(\Leftrightarrow\left|4-x\right|=0\)

\(\left|4-x\right|=0\)

\(4-x=0\)

\(x=0+4=4\)

Vậy GTNN của A là 11 khi x = 4

\(A=\left[x+\left(x+7\right)\right]-\left[\left(x+4\right)-\left(x-4\right)\right]\)

\(A=\left(x+x+7\right)-\left(x+4-x+4\right)\)

\(A=\left(2x+7\right)-\left(x-x+4+4\right)\)

\(A=2x+7-8\)

\(A=2x-1\)

\(B=x\left\{\left(x-3\right)-\left[\left(x+3\right)-\left(-x-2\right)\right]\right\}\)

\(B=x\left\{\left(x-3\right)-\left[x+3-\left(-x\right)+2\right]\right\}\)

\(B=x\left[\left(x-3\right)-\left(x+3+x+2\right)\right]\)

\(B=x\left(x-3-2x+5\right)\)

\(B=x\left(x-2x-3+5\right)\)

\(B=x\left(-x+2\right)\)

\(B=-x^2+2x\)

\(B=2x-x^2\)

Nhưng mà mình k bt ss kiểu j

\(\dfrac{121\times75\times130\times169}{39\times60\times11\times198}\)

\(=\dfrac{25\times169}{6\times18}\)

\(=\dfrac{4225}{108}\)

\(\dfrac{7}{9}+\dfrac{1}{3}< x< \dfrac{43}{8}+\dfrac{1}{10}\)

\(\dfrac{10}{9}< x< \dfrac{219}{40}\)

Mà \(x\inℕ\)

\(\Rightarrow\dfrac{10}{9}< 2\le x\le5< \dfrac{219}{40}\)

\(\Rightarrow2\le x\le5\)

\(\Rightarrow x\in\left\{2;3;4;5\right\}\)

Vậy: \(x\in\left\{2;3;4;5\right\}\)

\(7,75-\left(0,5\times y\div5-6,2\right)=5\)

\(0,5\times y\div5-6,2=7,75-5=2,75\)

\(0,5\div5\times y-6,2=2,75\)

\(0,1\times y=2,75+6,2=8,95\)

\(\dfrac{1}{10}y=8,95\)

\(y=8,95\times10=89,5\)

\(y\div6\times7,2+1,3\times y+y\div2+15=19,95\)

\(1,2\times y+1,3\times y+0,5y=19,95-15=4,95\)

\(y\left(1,2+1,3+0,5\right)=4,95\)

\(2y=4,95\)

\(y=4,95\div2=2,475\)

\(\dfrac{\left(x+1\right)}{3}=\dfrac{3}{\left(x+1\right)}\)

\(\left(x+1\right)^2=3\cdot3=9\)

\(\left(x+1\right)^2=\left(\pm3\right)^2\)

\(\Rightarrow x+1=\pm3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3-1=2\\x=-3-1=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{-4;2\right\}\)

\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{3}\)

\(\dfrac{x}{2}=\dfrac{2y}{8}=\dfrac{z}{3}\)

ADTCDTSBN, ta có:

\(\dfrac{x}{2}=\dfrac{2y}{8}=\dfrac{z}{3}=\dfrac{x-2y+z}{2-8+3}=\dfrac{6}{-3}=-2\)

\(\dfrac{x}{2}=-2\Rightarrow x=-2\cdot2=-4\)

\(\dfrac{y}{4}=-2\Rightarrow x=-2\cdot4=-8\)

\(\dfrac{z}{3}=-2\Rightarrow x=-2\cdot3=-6\)\

Vậy x=-4

y=-8

z=-6

\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)

\(\dfrac{x}{2}=\dfrac{2y}{8}=\dfrac{z}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{2}=\dfrac{2y}{8}=\dfrac{z}{6}=\dfrac{x-2y+z}{2-8+6}=\dfrac{6}{0}\)(vô lí)

=> Không có x, y, z thỏa mãn đề bài

\(\left(\left|x\right|-2011\right)^{\left(2+2008\right)}\cdot\left(2+2009\right)=-\left(2^3-3^2\right)^{2009}\)

\(\left(\left|x\right|-2011\right)^{2010}\cdot2011=-\left(8-9\right)^{2009}\)

\(\left(\left|x\right|-2011\right)^{2010}\cdot2011=-\left(-1\right)^{2009}\)

\(\left(\left|x\right|-2011\right)^{2010}\cdot2011=-\left(-1\right)\)

\(\left(\left|x\right|-2011\right)^{2010}\cdot2011=1\)

\(\left(\left|x\right|-2011\right)^{2010}=\dfrac{1}{2011}\)

???