a:Mg +2HCL---->MGCL2+ H2O

a. Mg + 2HCl ---> MgCl₂ + H₂↑

b. Fe₂O₃ + 6HCl ---> 2FeCl₃ + 3H₂O

c. 2Al + 6HCl ---> 2AlCl₃ + 3H₂↑

d. 2C₂H₆ + 7O₂ ---t^o---> 4CO₂↑ + 6H₂O

e. BaCl₂ + 2AgNO₃ ---> Ba(NO₃)₂ + 2AgCl↓

f. Al₂(SO₄)₃ + 3Ba(OH)₂ ---> 2Al(OH)₃ + 3BaSO₄↓

Trong đó:

t^o: nhiệt độ

↑: bay hơi

↓: kết tủa

(Câu f sai nên mik sửa từ Al₂(SO₄)₃ thành Al(OH)₃)

nãy mik lm r bn xem lại nhé

`#Hưng`

\(a,3\sqrt{8\sqrt{5}}-2\sqrt{9\sqrt{20}}\\ =\sqrt{9.8\sqrt{5}}-\sqrt{4.9\sqrt{20}}\\ =\sqrt{72\sqrt{5}}-\sqrt{36\sqrt{20}}\\ =\sqrt{\sqrt{5184.5}}-\sqrt{\sqrt{1296.20}}\\ =\sqrt{\sqrt{25920}}-\sqrt{\sqrt{25920}}\\ =0\)

\(b,ĐKXĐ:x\sqrt{x}-\sqrt{x}+x-1\ne0\\ \Rightarrow\sqrt{x}\left(x-1\right)+\left(x-1\right)\ne0\\ \Rightarrow\left(x-1\right)\left(\sqrt{x}+1\right)\ne0\\ \Rightarrow x-1\ne0\left(vì.\sqrt{x}+1>0\right)\\ \Rightarrow x\ne1\)

a , 3 √ 8 √ 5 − 2 √ 9 √ 20 = √ 9.8 √ 5 − √ 4.9 √ 20 = √ 72 √ 5 − √ 36 √ 20 = √ √ 5184.5 − √ √ 1296.20 = √ √ 25920 − √ √ 25920 = 0 b , Đ K X Đ : x √ x − √ x + x − 1 ≠ 0 ⇒ √ x ( x − 1 ) + ( x − 1 ) ≠ 0 ⇒ ( x − 1 ) ( √ x + 1 ) ≠ 0 ⇒ x − 1 ≠ 0 ( v ì . √ x + 1 > 0 ) ⇒ x ≠ 1

a: \(B=\dfrac{x+\sqrt{x}-1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)

\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

b: \(B-\dfrac{1}{3}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}\)

\(=\dfrac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{-\left(\sqrt{x}+1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)

=>B<1/3

hộ e ik mà ;-;

6) ĐKXĐ: \(x\le-7\)

9) ĐKXĐ: \(x\ne1\)

Câu 1.

\(CH_3COOH+NaOH\rightarrow CH_4\uparrow+Na_2CO_3\)

\(2CH_4\xrightarrow[LLN]{1500^oC}C_2H_2+2H_2\uparrow\)

\(C_2H_2+H_2\underrightarrow{t^o,xtPd}C_2H_4\)

a)\(C_2H_4+HBr\rightarrow C_2H_5Br\)

b)\(C_2H_4+HCl\underrightarrow{as}C_2H_5Cl\)

B

nghĩ là D vì có trạng ngữ .-.

BT A - B hay A.B vậy bn ?

\(\)với \(x\ge0,x\ne4\) 

\(=>B=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{3}{\sqrt{x}+1}\)

\(B=\left[\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right].\dfrac{\sqrt{x}+1}{3}\)

\(B=\dfrac{3.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)3}=\dfrac{1}{\sqrt{x}-2}\)

\(=>P=\dfrac{18}{A.B}=\dfrac{18}{\dfrac{x-4}{\sqrt{x}-1}.\dfrac{1}{\sqrt{x}-2}}=\dfrac{18}{\dfrac{\sqrt{x}+2}{\sqrt{x}-1}}\)

\(=\dfrac{18\left(\sqrt{x}-1\right)}{\sqrt{x}+2}\)\(=\dfrac{18\sqrt{x}-18}{\sqrt{x}+2}=\dfrac{18\sqrt{x}}{\sqrt{x}+2}-\dfrac{18}{\sqrt{x}+2}\ge-9\)

dấu"=" xảy ra<=>x=0(tm)

\(\)