Điều kiện: \(x\ge2\)

\(x+2021\sqrt{x-2}=2\sqrt{x-1}\)

\(\Leftrightarrow x-2\sqrt{x-1}=-2021\sqrt{x-2}\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+1=-2021\sqrt{x-2}\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2-2021\sqrt{x-2}\)

Nhận xét:

\(VT=\left(\sqrt{x-1}-1\right)^2\ge0\forall x\ge2\)

\(VP=-2021\sqrt{x-2}\le0\forall x\ge2\)

\(VT=VP\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\-2021\sqrt{x-2}=0\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)

\(\Leftrightarrow x=2\left(TMĐK\right)\)

Vậy \(S=\left\{2\right\}\)

\(\text{M là khối lg nhé.}\)

\(m_C=1,9926.10^{-23}\left(g\right)\)

\(\Rightarrow1đvC=\dfrac{1,9926.10^{-23}}{12}\approx1,66.10^{-24}\)

\(m_{Mg}=1,66.10^{-24}.24=3,984.10^{-23}\left(g\right)\)

\(\text{Theo đề ra, ta có:}\)

\(\dfrac{1}{5}M_F-\dfrac{1}{3}M_K=0\Rightarrow\dfrac{1}{5}M_F=\dfrac{1}{3}M_K\Rightarrow\dfrac{1}{5}M_F=\dfrac{1}{3}39=13\Rightarrow M_F=65\)\(\text{Vậy}\) \(F\) \(\text{là kẽm và có nguyên tử khối là}\) \(65\)

\(\text{Câu 4:}\)

\(\text{Theo đề ra, ta có tổng:}\) \(2p+n=48\left(1\right)\)

\(\text{Số hạt mang điện gấp đôi số hạt không mang điện:}\) \(2p=2n\Rightarrow p-n=0\left(2\right)\)

\(\text{Từ}\) \(\left(1\right);\left(2\right):\) \(\Rightarrow\left\{{}\begin{matrix}p=e=16\\n=16\end{matrix}\right.\)

\(\text{Câu 3:}\)

\(\text{Tổng ba loại hạt là 34}\) \(\Rightarrow2p+n=34\left(1\right)\)

\(\text{Theo đề ra, ta có hệ phương trình:}\) \(\left\{{}\begin{matrix}2p+n=34\\p+n=23\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}p=11\\n=12\end{matrix}\right.\)

\(\text{Mà:}\) \(p=e\Rightarrow p=e=11\)

\(\text{Vậy}\) \(\left\{{}\begin{matrix}p=e=11\\n=12\end{matrix}\right.\)

\(\text{Câu 1:}\)

\(\text{Chất: }\)\(\text{Đường, rượu, nước cất, muối ăn, thủy ngân, sắt}\)

\(\text{Hỗn hợp:}\) \(\text{Nước đồng, nước tự nhiên, nước chanh, sữa tươi, gang, thép}\)

\(d\dfrac{M_X}{H_2}=33\Rightarrow M_X=33.2=66\left(đvc\right)\)

\(\text{Không có nguyên tố hh nào tương ứng}\)

\(d\dfrac{M_X}{H_2}=40\Rightarrow M_X=40.2=80\left(đvc\right)\)

\(\text{Nguyên tố}\) \(Brom;KHHH:Br\)

d) \(\left(2x-3y\right)^3=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^2-\left(3y\right)^3=8x^3-3.4x^2.3y+6x.9y^2-27y^3=8x^3-36x^2y+54xy^2-27y^3\)

e) \(\left(2x^2+\dfrac{3}{2}\right)^3=\left(2x^2\right)^3+3\left(2x^2\right)^2\dfrac{3}{2}+3.2x^2\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3=8x^6+3.4x^4.\dfrac{3}{2}+6x^2.\dfrac{9}{4}+\dfrac{27}{8}=8x^6+18x^4+\dfrac{27}{2}x^2+\dfrac{27}{8}\)

f) \(\left(-2xy^2+\dfrac{1}{2}x^3y\right)^3=\left(-2xy^2\right)+3\left(-2xy^2\right)^2\dfrac{1}{2}x^3y+3\left(-2xy^2\right)\left(\dfrac{1}{2}x^3y\right)^2+\left(\dfrac{1}{2}x^3y\right)^3=-8x^3y^6+3.4x^2y^4.\dfrac{1}{2}x^3y-6xy^2.\dfrac{1}{4}x^6y^2+\dfrac{1}{8}x^9y^3=-8x^3y^6+6x^5y^5-\dfrac{3}{2}x^7y^4+\dfrac{1}{8}x^9y^3\)

a) \(\left(3x-2\right)^2=\left(3x\right)^2-2.3x.2+2^2=9x^2-12x+4\)

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

c) \(9x^2-225=\left(3x\right)^2-\left(15\right)^2=\left(3x-15\right)\left(3x+15\right)\)