\(-\dfrac{3}{14}+\dfrac{8}{28}\\ =\dfrac{-3}{14}+\dfrac{4}{14}\\ =\dfrac{-3+4}{14}\\ =\dfrac{1}{14}\)

\(\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{12}{11}-\dfrac{5}{7}.\dfrac{7}{11}\\ =\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{12}{11}-\dfrac{7}{11}\right)\\ =\dfrac{5}{7}.\dfrac{7}{11}\\ =\dfrac{5}{11}\)

`ĐKXĐ:{(x ne 2),(x ne 5):}`

\(\dfrac{x^2-25}{2x-4}.\dfrac{4}{5-x}\\ =\dfrac{\left(x-5\right)\left(x+5\right)}{2\left(x-2\right)}.\dfrac{4}{5-x}\\ =\dfrac{\left(5-x\right)\left(x+5\right)}{2\left(2-x\right)}.\dfrac{2.2}{5-x}\\ =\dfrac{2\left(x+5\right)}{2-x}\)

\(\dfrac{0,8\times0,4\times1,25\times25+0,725+0,275}{1,25\times4\times8\times25}\\ =\dfrac{\left(0,8\times1,25\right)\times\left(0,4\times25\right)+\left(0,725+0,275\right)}{\left(1,25\times8\right)\times\left(4\times25\right)}\\ =\dfrac{1\times10+1}{10\times100}\\ =\dfrac{10+1}{1000}\\ =\dfrac{11}{1000}\)

thanks thôi hoặc thank you k có thanks you :)

\(a,ĐKXĐ:x\ge5\\ \sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\\ \Leftrightarrow\sqrt{9\left(x-5\right)}-14.\dfrac{\sqrt{x-5}}{\sqrt{49}}+\dfrac{1}{4}\sqrt{4\left(x-5\right)}=3\\ \Leftrightarrow\sqrt{9}.\sqrt{x-5}-14.\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}.\sqrt{4}\sqrt{x-5}=3\\ \Leftrightarrow3\sqrt{x-5}-2\sqrt{x-5}+\dfrac{1}{4}.2\sqrt{x-5}=3\\ \Leftrightarrow\sqrt{x-5}+\dfrac{1}{2}\sqrt{x-5}=3\\ \Leftrightarrow\dfrac{3}{2}\sqrt{x-5}=3\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)

\(b,ĐKXĐ:x\ge\dfrac{1}{2}\\ 2x=\sqrt{2x-1}+7\\ \Leftrightarrow2x-7-\sqrt{2x-1}=0\\ \Leftrightarrow2x-1-\sqrt{2x-1}-6=0\\ \Leftrightarrow2x-1-3\sqrt{2x-1}+2\sqrt{2x-1}-6=0\\ \Leftrightarrow\sqrt{2x-1}\left(\sqrt{2x-1}-3\right)+2\left(\sqrt{2x-1}-3\right)=0\\ \Leftrightarrow\left(\sqrt{2x-1}-3\right)\left(\sqrt{2x-1}+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}-3=0\\\sqrt{2x-1}+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=3\\\sqrt{2x-1}=-2\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow2x-1=9\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\left(tm\right)\)

`3x^3 -12x^2 +12x`

`=3x(x^2 -4x+4)`

`=3x(x-2)^2`

`A=x^4 -6x^3 +10x^2 -6x +9`

`=x^4 -6x^3 +9x^2  +x^2 -6x+9`

`=x^2 (x^2 -6x+9)+(x^2 -6x+9)`

`=x^2 (x-3)^2 +(x-3)^2`

`=[x(x-3)]^2+(x-3)^2`

Dấu "=" xảy ra `<=>{(x(x-3)=0),(x-3=0):}`

`<=>x-3=0`

`<=>x=3`

Vậy `A_(min) =0<=>x=3`

\(\left\{{}\begin{matrix}x-30y=-30\\1,5x-20y=30\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=30y-30\\1,5\left(30y-30\right)-20y=30\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=30y-30\\45y-45-20y=30\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=30y-30\\25y=75\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=30.3-30\\y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=60\\y=3\end{matrix}\right.\)

\(x^2+x+1+\dfrac{x^3}{1-x}\\ =x^2+x+1-\dfrac{x^3}{x-1}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)-x^3}{x-1}\\ =\dfrac{x^3-1-x^3}{x-1}\\ =\dfrac{-1}{x-1}\\ =\dfrac{1}{1-x}\)