\(\left\{{}\begin{matrix}6u_2+u_5=1\\3u_3+2u_4=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6u_1.q+u_1.q^4=1\\3u_1.q^2+2u_1.q^3=-1\end{matrix}\right.\)

\(\Rightarrow u_1\left(6q+q^4+3q^2+2q^3\right)=0\)

\(\Leftrightarrow q^3+2q^2+3q+6=0\)

\(\Leftrightarrow\left(q+2\right)\left(q^2+3\right)=0\)

\(\Leftrightarrow q=-\text{​​}2\)

\(\Rightarrow u_1=\dfrac{1}{4}\)

\(\Rightarrow u_n=u_1.q^{n-1}=\dfrac{1}{4}.\left(-2\right)^{n-1}=\left(-2\right)^{n-3}\)

\(x=\left(\dfrac{1}{2}\right)^3:\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^{3-1}=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)

1/4

\(\dfrac{1}{4}\)

\(n_{NaOH}=\dfrac{12}{40}=0.3\left(mol\right)\)

\(n_{HCl}=\dfrac{7.3}{36.5}=0.2\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

Ta có : 

\(n_{NaOH}>n_{HCl}\Rightarrow NaOHdư\)

\(n_{NaOH\left(pư\right)}=n_{HCl}=n_{NaCl}=0.2\left(mol\right)\)

\(n_{NaOH\left(dư\right)}=0.3-0.2=0.1\left(mol\right)\)

\(m_{cr}=m_{NaOH\left(dư\right)}+m_{NaCl}=0.1\cdot40+0.2\cdot58.5=15.7\left(g\right)\)

Ta có: \(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)

\(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

PTHH: NaOH + HCl ---> NaCl + H₂O

Ta thấy: \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\)

Vậy NaOH dư, HCl hết.

Theo PT: \(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)

\(\Rightarrow m_{NaCl}=0,2.58,5=11,7\left(g\right)\)

\(1,\\ a,=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\\ b,=4x^3+5x^2-8x^2-10x+12x+15\\ =4x^3-3x^2+2x+15\\ 2,\\ a,=7\left(x^2-6x+9\right)=7\left(x-3\right)^2\\ b,=\left(x-y\right)^2-36=\left(x-y-6\right)\left(x-y+6\right)\\ 3,\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow x\left(x-0,6\right)\left(x+0,6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,6\\x=-0,6\end{matrix}\right.\)

Ok

\(M=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\left(đk:x\ge0,x\ne9\right)\)

Để \(M=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}< 0\) thì 

\(\sqrt{x}-3< 0\) ( do \(\sqrt{x}+3\ge3>0\))

\(\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{0;1;2;3;4;5;6;7;8\right\}\)

Để M là số nguyên âm thì \(\sqrt{x}-3< 0\)

\(\Leftrightarrow\sqrt{x}< 3\)

hay \(x\in\left\{0;1;4\right\}\)

=4090506-\(\dfrac{3}{2023}\)*2021+2020

=4090506-\(\dfrac{6063}{2023}\)+2020

=\(\dfrac{8279174035}{2023}\) tối giản luôn rồi nhé