Ta có \(28=2^2.7;41=41\)

\(\RightarrowƯCLN\left(28,41\right)=1\)

Vậy chia nhiều nhất thành 1 tổ

\(a,M=\dfrac{x^2-5x+5x+25+10x}{\left(x+5\right)\left(x-5\right)}\cdot\dfrac{x-5}{x}=\dfrac{\left(x+5\right)^2\left(x-5\right)}{x\left(x+5\right)\left(x-5\right)}=\dfrac{x+5}{x}\\ b,M=3\Leftrightarrow3x=x+5\Leftrightarrow2x=5\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

\(=\sqrt{\left(\sqrt{\dfrac{3}{4}}+\sqrt{\dfrac{1}{4}}\right)^2}-\dfrac{\sqrt{3}}{2}=\sqrt{\dfrac{3}{4}}+\sqrt{\dfrac{1}{4}}-\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}=\dfrac{1}{2}\)

PT có 2 nghiệm phân biệt \(\Leftrightarrow\Delta=\left(2m-3\right)^2-4\left(m-3\right)=9>0\)

Vậy PT có 2 nghiệm phân biệt với mọi m

Ta có \(\left[{}\begin{matrix}x_1=\dfrac{2m-3+3}{2}=m\\x_2=\dfrac{2m-3-3}{2}=m-3\end{matrix}\right.\)

Ta thấy \(m>m-3\) nên \(1< m-3< m< 6\Leftrightarrow4< m< 6\)

Vậy \(4< m< 6\)  thỏa yêu cầu đề

\(=\dfrac{\left(\sqrt{a}-2\right)\left(a+2\sqrt{a}+4\right)}{\sqrt{a}-2}+2\sqrt{a}=a+2\sqrt{a}+4+2\sqrt{a}=\left(\sqrt{a}+2\right)^2\)

\(a,=3x^3y^3-3x^2y^3+3x^2y^4+3xy^5\\ b,=\left(2x^3-6x^2+10x-3x^2+9x-15\right):\left(x^2-3x+5\right)\\ =\left[2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)\right]:\left(x^2-3x+5\right)\\ =2x-3\\ c,=\left[x^2\left(x-3\right)+\left(x-3\right)\right]:\left(x-3\right)=x^2+1\)

Viết đoạn văn kể lại giây phút gặp lại người thân sau bao ngày xa cách sử dụng lời dẫn trực tiếp

\(x^2+y^2-2x+2y+2=0\\ \Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(M=\left(x+y\right)^{2018}+\left(x-2\right)^{2019}+\left(y+1\right)^{2020}\\ M=\left(1-1\right)^{2018}+\left(1-2\right)^{2019}+\left(-1+1\right)^{2020}\\ M=0^{2018}+\left(-1\right)^{2019}+0^{2020}=-1\)

Ta có \(VT=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\)

Mà \(VP=\dfrac{8}{3\left(x+1\right)^2+2}\le\dfrac{8}{3\cdot0+2}=4\)

\(\Rightarrow VT\ge4\ge VP\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3\right)\left(1-2x\right)\ge0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{2}\le x\le\dfrac{1}{2}\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\)