$n_{H_2}=\dfrac{0,2}{2}=0,1(mol)$

$AO+H_2\xrightarrow{t^o}A+H_2O$

Theo PT: $n_A=n_{H_2}=0,1(mol)$

$\to M_A=\dfrac{6,4}{0,1}=64(g/mol)$

$\to A:Cu$

Vậy CT oxit là $CuO$

Giải PT nghiệm nguyên: $(x-y+2)^3=20+x^3-y^3$

Cho các số $a,b,c$ dương, tìm GTLN của:

\(M=\dfrac{ab}{bc+2a^2+3ab}+\dfrac{bc}{ca+2b^2+3bc}+\dfrac{ca}{ab+2c^2+3ca}\)

$n_{Al}=\dfrac{2,7}{27}=0,1(mol)$

$2Al+6HCl\to 2AlCl_3+3H_2$

Theo PT: $n_{HCl}=3n_{Al}=0,3(mol)$

$\to a=0,3$

$a\big)A_{Fe}=0,75.6.10^{23}=4,5.10^{23}$ (nguyên tử)

$b\big)A_{CaCO_3}=0,25.6.10^{23}=1,5.10^{23}$ (phân tử)

$c\big)A_{O_2}=0,05.6.10^{23}=3.10^{22}$ (phân tử)

$a\big)CH_3COOH+NaOH\to CH_3COONa+H_2O$

$b\big)$

Theo PT: $n_{CH_3COOH}=n_{CH_3COONa}=\dfrac{8,2}{82}=0,1(mol)$

$\to m_{CH_3COOH}=0,1.60=6(g)$

$\to m_{C_2H_5OH}=10,6-6=4,6(g)$

$\to \begin{cases}\%m_{CH_3COOH}\approx 56,6\%\\ \%m_{C_2H_5OH}=43,4\% \end{cases}$

\(f\left(2\right)=4a+2b+c\)

Mà \(4a+c=-2b+2022\Rightarrow4a+2b+c=2022\)

Vậy \(f\left(2\right)=2022\)

$S_{NaCl}=\dfrac{m_{NaCl}}{m_{H_2O}}.100=25$

$\to m_{NaCl}=0,25m_{H_2O}$

Mà $m_{NaCl}+m_{H_2O}=200(g)$

$\to m_{NaCl}=40(g);m_{H_2O}=160(g)$

đang nghĩ

Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)