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\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

\(SO_3+2KOH\rightarrow K_2SO_4+H_2O\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

\(1,\\ a,=4-5y\\ b,=\dfrac{20x}{3y}\cdot\dfrac{6y}{4}=\dfrac{10x}{1}=10x\\ 2,\\ a,=\left(2x+2-4\right)\left(2x+2+4\right)=2\left(x+1-2\right)2\left(x+1+2\right)\\ =4\left(x-1\right)\left(x+3\right)\\ b,=\left(2xy-xz\right)+\left(6y-3z\right)\\ =2x\left(y-z\right)+3\left(y-z\right)=\left(2x+3\right)\left(y-z\right)\)

a.

Đặt \(x+2y+1=a\)

\(\Rightarrow P=a^2+\left(a+4\right)^2=2a^2+8a+16=2\left(a+2\right)^2+8\ge8\)

\(P_{min}=8\) khi \(a=-2\) hay \(x+2y+3=0\)

b.

\(\sqrt{x}-1=a\ge0\Rightarrow\sqrt{x}=a+1\Rightarrow x=a^2+2a+1\)

\(Q=\dfrac{\left(a^2+2a+1\right)+\left(a+1\right)+1}{a}=\dfrac{a^2+3a+3}{a}=a+\dfrac{3}{a}+3\ge2\sqrt{\dfrac{3a}{a}}+3=3+2\sqrt{3}\)

\(Q_{min}=3+2\sqrt{3}\) khi \(a=\sqrt{3}\) hay \(x=4+2\sqrt{3}\)

Bài 12: 

a: Xét ΔABM và ΔACN có

AB=AC

\(\widehat{ABM}=\widehat{ACN}\)

BM=CN

Do đó: ΔABM=ΔACN