Bài 10:

\(=4,2\times100,21-0,21\times\left(2,3+1,9\right)\\ =4,2\times100,21-0,21\times4,2\\ =4,2\times\left(100,21-0,21\right)=4,2\times100=420\)

Tick mình nha <3

Câu 5:

\(1,\Leftrightarrow4x^2-20x-4x^2+7x-3=5\\ \Leftrightarrow-13x=8\Leftrightarrow x=-\dfrac{8}{13}\\ 2,\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\\ \Leftrightarrow4\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ 3,\Leftrightarrow\left(x+3\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ 4,\Leftrightarrow x^3-x-6x-6=0\\ \Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-x-6\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)

\(5,\Leftrightarrow3x^2-10x+6=3x^2-12x-3\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=\dfrac{9}{2}\\ 6,\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\\ 7,\Leftrightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\\ \Leftrightarrow-2\left(2x-5\right)=0\\ \Leftrightarrow x=\dfrac{5}{2}\)

Tick mình nha <3

Gọi d=ƯCLN(2n+9,4n+19)

\(\Rightarrow\left\{{}\begin{matrix}2n+9⋮d\\4n+19⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4n+18⋮d\\4n+19⋮d\end{matrix}\right.\\ \Rightarrow4n+19-4n-18⋮d\\ \Rightarrow1⋮d\Rightarrow d=1\)

Vậy ƯCLN(2n+9,4n+19)=1 hay 2 số trên ntcn

\(m_A=\dfrac{3,82\cdot10^{-23}}{1,66\cdot10^{-24}}\approx23\left(đvC\right)\)

Vậy A là Natri (Na)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{H_2SO_4}=n_{CuO}=\dfrac{19,6}{98}=0,2\left(mol\right)\\ \Rightarrow a=m_{CuO}=0,2\cdot80=16\left(g\right)\)

\(a,A=\dfrac{1}{4+1}=\dfrac{1}{5}\\ b,B=\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}-2}\\ c,P=\dfrac{A}{B}=\dfrac{1}{\sqrt{x}+1}\cdot\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\\ P-1=\dfrac{\sqrt{x}-2+\sqrt{x}+1}{\sqrt{x}+1}=\dfrac{-1}{\sqrt{x}+1}< 0\\ \Leftrightarrow P< 1\)

\(=\dfrac{\left(\sqrt{5}-a\right)\left(\sqrt{5}+a\right)}{a-\sqrt{5}}=-a-\sqrt{5}\left(A\right)\)

Ta có: \(\left\{{}\begin{matrix}m_S=64\cdot50\%=32\left(g\right)\\m_O=64\cdot50\%=32\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_S=\dfrac{32}{32}=1\left(mol\right)\\n_O=\dfrac{32}{16}=2\left(mol\right)\end{matrix}\right.\)

Vậy CTHH là \(SO_2\)