\(a,=3x^2-6x-3x^2+1=1-6x\\ b,=4x^2-4xy+y^2-4x^2+4xy=y^2\\ c,=x^3+3x^2y+3xy^2-x^3-3x^2y-3xy^2-y^3=-y^3\\ d,=\dfrac{x^2-x+3x^2-3x}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x}{x+1}\)

\(a,\Leftrightarrow2x^2-2x-2x^2=3\Leftrightarrow-2x=3\Leftrightarrow x=-\dfrac{3}{2}\\ b,\Leftrightarrow x\left(4x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-1\end{matrix}\right.\)

Chọn D

\(\Leftrightarrow n+1+4⋮n+1\\ \Leftrightarrow4⋮n+1\\ \Leftrightarrow n+1\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Leftrightarrow n\in\left\{-5;-3;-2;0;1;3\right\}\)

\(=x^2\left(x-y\right)-9\left(x-y\right)=\left(x-y\right)\left(x^2-9\right)=\left(x-y\right)\left(x-3\right)\left(x+3\right)\)

Thay \(x=0;y=-\sqrt{2}\Leftrightarrow-\sqrt{2}=1\left(loại\right)\)

Thay \(x=\sqrt{2};y=\sqrt{2}+1\Leftrightarrow\sqrt{2}-2+1=\sqrt{2}+1\left(loại\right)\)

Thay \(x=1-\sqrt{2};y=3-2\sqrt{2}\Leftrightarrow\left(1-\sqrt{2}\right)^2+1=4-2\sqrt{2}=3-2\sqrt{2}\left(loại\right)\)

Thay \(x=1+\sqrt{2};y=0\Leftrightarrow\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)+1=1-2+1=0\left(đúng\right)\)

Chọn D

\(a,\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\\dfrac{c}{a}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}25-12\left(2m+1\right)\ge0\\\dfrac{2m+1}{3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le\dfrac{13}{24}\\m< -\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow m< -\dfrac{1}{2}\)

\(b,\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\dfrac{c}{a}>0\\-\dfrac{b}{a}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{13}{24}\\m>-\dfrac{1}{2}\\-\dfrac{5}{3}< 0\left(đúng\right)\end{matrix}\right.\Leftrightarrow-\dfrac{1}{2}< m< \dfrac{13}{24}\)

\(c,\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\dfrac{c}{a}>0\\-\dfrac{b}{a}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{13}{24}\\m>-\dfrac{1}{2}\\-\dfrac{5}{3}>0\left(\text{vô lí}\right)\end{matrix}\right.\Leftrightarrow m\in\varnothing\)

\(d,\Leftrightarrow\Delta\ge0\Leftrightarrow m\le\dfrac{13}{24}\\ \text{Viét: }\left\{{}\begin{matrix}x_1+x_2=-\dfrac{5}{3}\\x_1x_2=\dfrac{2m+1}{3}\end{matrix}\right.\\ x_1^3+x_2^3+10\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=10\\ \Leftrightarrow-\dfrac{125}{27}-\left(2m+1\right)\left(-\dfrac{5}{3}\right)=10\\ \Leftrightarrow\dfrac{5}{3}\left(2m+1\right)=\dfrac{395}{27}\\ \Leftrightarrow2m+1=\dfrac{79}{9}\Leftrightarrow m=\dfrac{35}{9}\left(ktm\right)\\ \Leftrightarrow m\in\varnothing\)

Vậy ko có m thỏa đề

\(=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\)

Đề thiếu

Áp dụng BĐT cauchy ta có:

\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge2\sqrt{ab}\cdot2\sqrt{\dfrac{1}{ab}}=4\sqrt{ab\cdot\dfrac{1}{ab}}=4\)

Dấu \("="\Leftrightarrow a=b\)