\(2Cu\left(NO_3\right)_2\xrightarrow[]{t^0}2CuO+4NO_2+O_2\)

\(C_nH_{2n}O_2+\dfrac{3n-2}{2}O_2\)   \(\underrightarrow{to}\)  \(nCO_2+nH_2O\)

\(C_xH_yO_zN_t+\left(x+\dfrac{y}{4}-\dfrac{z}{2}\right)O_2\xrightarrow[]{t^o}xCO_2+\dfrac{y}{2}H_2O+\dfrac{t}{2}N_2\)

13. PTHH: Fe_xO_y + yH₂ ---t^o→ xFe + yH₂O

\(2P+5H_2SO_{4_{đặc}}\xrightarrow[]{t^o}2H_3PO_4+5SO_2+2H_2O\)

2P + 5H₂SO₄ → 2H₃PO₄ + 5SO₂ + 2H₂O

\(C_nH_{2n+2}+\dfrac{3n-1}{2}O_2\underrightarrow{^{to}}nCO_2+\left(n+1\right)H_2O\)

\(2C_nH_{2n+2}+\left(3n+1\right)O_2\rightarrow2nCO_2+2\left(n+1\right)H_2O\)

\(2Al+3ZnSO_4\rightarrow Al_2\left(SO_4\right)_3+3Zn\)

\(f(x)=ax^2+bx+6\)

Để \(f(x)\) là đa thức bậc \(1\) thì \(ax^2=0\)

\(→a=0\)

Thay \(x=1\) vào \(f(x)=ax^2+bx+6\)

\(f(1)=b.1+6=b+6\)

Mà \(f(1)=3\)

\(\Rightarrow b+6=3\Rightarrow b=3−6\Rightarrow b=−3\)

Vậy \(a=0;b=−3\)

Câu 1: 

a: \(2\sqrt{18}-\dfrac{1}{5}\sqrt{50}+3\sqrt{98}-\sqrt{288}\)

\(=6\sqrt{2}-\sqrt{2}+21\sqrt{2}-12\sqrt{2}\)

\(=14\sqrt{2}\)

b: \(\sqrt{\left(1-\sqrt{3}\right)^2}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{3}-1+2-\sqrt{3}\)

=1