thiếu r e, xét TH 2021 và -2021 nữa

\(a,A=\dfrac{3}{x^2-2x+2}=\dfrac{3}{\left(x-1\right)^2+1}\le\dfrac{3}{0+1}=3\\ A_{max}=3\Leftrightarrow x-1=0\Leftrightarrow x=1\\ b,A=\dfrac{3}{x^2-2x+2}\Leftrightarrow Ax^2-2Ax+2A-3=0\\ \Leftrightarrow\Delta'=A^2-A\left(2A-3\right)=-A\left(A-3\right)\ge0\\ \Leftrightarrow0\le A\le3\)

Do đó \(A\in Z\Leftrightarrow A\in\left\{0;1;2;3\right\}\)

Với \(A=0\Leftrightarrow3=0\left(\text{vô lí}\right)\)

Với \(A=1\Leftrightarrow x^2-2x+2=3\Leftrightarrow x^2-2x-1=0\Leftrightarrow x=1\pm\sqrt{2}\)

Với \(A=2\Leftrightarrow2x^2-4x+4=3\Leftrightarrow2x^2-4x+1=0\Leftrightarrow x=\dfrac{2\pm\sqrt{2}}{2}\)

Với \(A=3\Leftrightarrow x^2-2x+2=1\Leftrightarrow x^2-2x+1=0\Leftrightarrow x=1\)

Vậy \(A\in Z\Leftrightarrow x\in\left\{1\pm\sqrt{2};\dfrac{2\pm\sqrt{2}}{2};1\right\}\)

hảo chất này có thật à :)))??

\(\dfrac{x}{3}=\dfrac{y}{20}=4\\ \Rightarrow\left\{{}\begin{matrix}x=4\cdot3=12\\y=4\cdot20=80\end{matrix}\right.\)

\(\dfrac{\left(3x-1\right)\left(x+2\right)}{3}-\dfrac{2x^2+1}{2}=\dfrac{11}{2}\\ \Leftrightarrow\dfrac{3x^2+5x-2}{3}=\dfrac{2x^2+12}{2}=x^2+6\\ \Leftrightarrow3x^2+5x-2=3x^2+18\\ \Leftrightarrow5x=20\Leftrightarrow x=4\)

Vì \(\overline{17x8y}⋮2\text{ và }5\Rightarrow y=0\)

\(\Rightarrow\overline{17x80}⋮9\\ \Rightarrow1+7+8+x+0⋮9\\ \Rightarrow16+x⋮9\\ \Rightarrow x=2\)

Vậy số đó là \(17280\)

Cho 2 mẫu thử vào dung dịch NaOH

- Tạo kết tủa xanh lơ: CuCl₂

- Không hiện tượng: KCl

\(2NaOH+CuCl_2\to Cu(OH)_2\downarrow+2NaCl\)

\(a,2Al+3Cl_2\xrightarrow{t^o}2AlCl_3\\ b,Ca(OH)_2+N_2O_5\to Ca(NO_3)_2+H_2O\\ c,Cu+2H_2SO_{4(đ)}\to CuSO_4+2H_2O+SO_2\uparrow\\ d,3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ e,FeO+H_2SO_4\to FeSO_4+H_2O\\ f,2Al(OH)_3+3H_2SO_4\to Al_2(SO_4)_3+6H_2O\\ g,AgNO_3+HCl\to AgCl\downarrow+HNO_3\\ h,3Mg+2AlCl_3\to 3MgCl_2+2Al\\ i,CaCO_3+2HCl\to CaCl_2+H_2O+CO_2\uparrow\\ j,2Al+3S\xrightarrow{t^o}Al_2S_3\)

\(\dfrac{9x^2+3xy-2y^2}{9x^2+9xy+2y^2}=\dfrac{9x^2-3xy+6xy-2y^2}{9x^2+3xy+6xy+2y^2}\\ =\dfrac{3x\left(3x-y\right)+2y\left(3x-y\right)}{3x\left(3x+y\right)+2y\left(3x+y\right)}=\dfrac{\left(3x-y\right)\left(3x+2y\right)}{\left(3x+y\right)\left(3x+2y\right)}\\ =\dfrac{3x-y}{3x+y}\)

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