có mấy cái e mới biết luôn á anh:"(

Muối bị nhiệt phân: FeCO₃;NaHCO₃;CaCO₃;Ba(HCO₃)₂;KNO₃;AgNO₃;NH₄Cl;NH₄NO₃

PTHH:

\(FeCO_3\rightarrow\left(t^o\right)FeO+CO_2\)

\(2NaHCO_3\rightarrow\left(t^o\right)Na_2CO_3+CO_2+H_2O\)

\(CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\)

\(Ba\left(HCO_3\right)_2\rightarrow\left(t^o\right)BaCO_3+CO_2+H_2O\)

\(2KNO_3\rightarrow\left(t^o\right)2KNO_2+O_2\)

\(2AgNO_3\rightarrow\left(t^o\right)2Ag+2NO_2+O_2\)

\(NH_4Cl\rightarrow\left(t^o\right)NH_3+HCl\)

\(NH_4NO_3\rightarrow\left(t^o\right)N_2O+2H_2O\)

\(A=\dfrac{3n+1}{n-4}\) ; \(\left(n\ne4\right)\)

\(A=\dfrac{3\left(n-4\right)+13}{n-4}\)

\(A=3+\dfrac{13}{n-4}\)

Để A nguyên thì \(\dfrac{13}{n-4}\in Z\Rightarrow n-4\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

`@n-4=1->n=5`

`@n-4=-1->n=3`

`@n-4=13->n=17`

`@n-4=-13->n=-9`

Vậy \(n\in\left\{5;3;17;-9\right\}\) thì A nguyên

Ta có: \(BC^2=\left(10\sqrt{2}\right)^2=200\)

          \(AB^2+AC^2=10^2+10^2=200\)

`=>` Tam giác ABC vuông cân tại A

Xét tam giác ABC vuông cân tại A, đcao AH:

`@`\(AB.AC=AH.BC\)

\(\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{10.10}{10\sqrt{2}}=5\sqrt{2}\left(cm\right)\)

`@`\(AB^2=BH.BC\)

\(\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{10^2}{10\sqrt{2}}=5\sqrt{2}\left(cm\right)\)

\(\Rightarrow CH=BC-BH=10\sqrt{2}-5\sqrt{2}=5\sqrt{2}\left(cm\right)\)

\(\dfrac{x+7}{2020}+\dfrac{x+6}{2021}+\dfrac{x+5}{2022}+3=0\)

\(\Rightarrow\dfrac{x+7}{2020}+1+\dfrac{x+6}{2021}+1+\dfrac{x+5}{2022}+1=0\)

\(\Rightarrow\dfrac{x+7+2020}{2020}+\dfrac{x+6+2021}{2021}+\dfrac{x+5+2022}{2022}=0\)

\(\Rightarrow\dfrac{x+2027}{2020}+\dfrac{x+2027}{2021}+\dfrac{x+2027}{2022}=0\)

\(\Rightarrow\left(x+2027\right)\left(\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}\right)=0\)

`=>x+2027=0`

`=>x=-2027`

Khi điều kiện của đề cho có từ " đốt, nung nóng " thì thêm t^o vào PTHH

Ví dụ: Đốt cháy hoàn toàn 3,1 gam Photpho trong không khí

khi đó PTHH sẽ là: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

`a.` Ta có: \(BC^2=10^2=100\)

                  \(AB^2+AC^2=6^2+8^2=100\)

`=>` Tam giác ABC vuông tại A

`b.` Xét tam giác ABC vuông A, đcao AH:

`@`\(AH.BC=AB.AC\)

\(\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{6.8}{10}=4,8\left(cm\right)\)

`@`\(AB^2=BH.BC\)

\(\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{6^2}{10}=3,6\left(cm\right)\)

`@`\(CH=BC-BH=10-3,6=6,4\left(cm\right)\)

`c.` Áp dụng t/c đường phân giác, ta có:

\(\dfrac{AB}{BD}=\dfrac{AC}{CD}=\dfrac{AB+AC}{BD+CD}=\dfrac{AB+AC}{BC}=\dfrac{6+8}{10}=\dfrac{14}{10}\)

\(\Rightarrow BD=6:\dfrac{14}{10}=\dfrac{30}{7}\left(cm\right)\)

\(\Rightarrow CD=8:\dfrac{14}{10}=\dfrac{40}{7}\left(cm\right)\)

\(HD=HC-CD=6,4-\dfrac{40}{7}=\dfrac{24}{35}\left(cm\right)\)

`a.`\(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\) ; \(\left(x\ge0;x\ne1\right)\)

\(A=\dfrac{\left(x\sqrt{x}+1\right)-\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

`b.`\(x=2021+2\sqrt{2020}\)

\(\Leftrightarrow x=2020+2\sqrt{2020}+1\)

`<=>`\(x=\left(\sqrt{2020}+1\right)^2\)

\(\Rightarrow A-1=\dfrac{\sqrt{\left(\sqrt{2020}+1\right)^2}}{\sqrt{\left(\sqrt{2020}+1\right)^2}-1}\)

\(\Leftrightarrow A-1=\dfrac{\sqrt{2020}+1}{\sqrt{2020}}\)

\(\Leftrightarrow A=\dfrac{2\sqrt{2020}+1}{\sqrt{2020}}\)