1. Những cây sẵn trong tự nhiên, tự bản thân nó được dùng để trang trí: cây hoa (hoa hồng, hoa cẩm chướng..), cây tùng, cây sanh. 
2. Phương pháp sinh sản vô tính: giâm cành bằng cát, ghép, chiết cành, nuôi cấy mô tế bào. 
phương pháp sinh sản hữu tính: thụ phấn trong tự nhiên. 
3. chọn chậu cây cảnh dựa trên các yếu tố: chất liệu, kích thước, 

4. tránh hư hỏng do va đập cơ học

5. Sử dụng axit abxixic để ức chế sinh trưởng. 
6. kỹ thuật sản xuất, an toàn thực phẩm, môi trường làm việc đảm bảo, nguồn gốc sản phẩm rõ ràng. 

\(\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{AC}{BC}:\dfrac{AB}{BC}=\dfrac{AC}{AB}=\tan\alpha\)

\(\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{AB}{BC}:\dfrac{AC}{BC}=\dfrac{AB}{AC}=\cot\alpha\)

\(\tan\alpha\cot\alpha=\dfrac{AC}{AB}\cdot\dfrac{AB}{AC}=1\)

\(\sin^2\alpha+\cos^2\alpha=\dfrac{AC^2}{BC^2}+\dfrac{AB^2}{BC^2}=\dfrac{AB^2+AC^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\left(pytago\right)\)

556667576

Câu 4:

a) nC2H6O=0,3(mol)

PTHH: C2H6O + 3 O2 -to-> 2 CO2 + 3 H2O

0,3___________0,9_____0,6(mol)

=>V(CO2,đktc)=0,6 x 22,4= 13,44(l)

b) V(kk,dktc)=V(O2,dktc) . 100/20 = (0,9.22,4).5=100,8(l)

Câu 5:

C2H6O + 3 O2 -to-> 2 CO2 + 3 H2O

nH2O=0,9(mol)

=> nCO2= 2/3. 0,9=0,6(mol)

a) V(CO2,đktc)=0,6.22,4=13,44(l)

b) Vkk=5.V(O2,dktc)= 5.(0,9.22,4)= 100,8(l)

Em cần hỗ trợ cụ thể bài nào em?

cái thứ 2 em tải hình xuống đề phòng hình 1 mất ạ

Bài 1: 

1: \(\sqrt{3+2\sqrt{2}}=\sqrt{2}+1\)

2: \(\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}\)

3: \(\sqrt{11-2\sqrt{30}}=\sqrt{6}-\sqrt{5}\)

4: \(\sqrt{7-2\sqrt{10}}=\sqrt{5}-\sqrt{2}\)

g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

1) \(\sqrt{2x-5}=7\)

\(\left(\sqrt{2x-5}\right)^2=7^2\)

\(2x-5=49\)

\(2x=54\)

\(x=27\)

2) \(3+\sqrt{x-2}=4\)

\(\sqrt{x-2}=1\)

\(\left(\sqrt{x-2}\right)^2=1^2\)

\(x-2=1\)

\(x=3\)

1) \(\sqrt{2x-5}=7\left(đk:x\ge\dfrac{5}{2}\right)\)

\(\Leftrightarrow2x-5=49\Leftrightarrow2x=54\Leftrightarrow x=27\left(tm\right)\)

2) \(3+\sqrt{x-2}=4\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)

3) \(\Leftrightarrow\sqrt{\left(x-1\right)^2}=1\Leftrightarrow\left|x-1\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

4) \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\Leftrightarrow\left|x-2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

5) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x+4\right)^2}\)

\(\Leftrightarrow\left|2x-1\right|=\left|x+4\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

6) \(ĐK:x\ge-2\)

 \(\Leftrightarrow5\sqrt{x+2}-3\sqrt{x+2}-\sqrt{x+2}=\sqrt{x+7}\)

\(\Leftrightarrow\sqrt{x+2}=\sqrt{x+7}\)

\(\Leftrightarrow x+2=x+7\Leftrightarrow2=7\left(VLý\right)\)

Vậy \(S=\varnothing\)

7) \(ĐK:x\ge-1\)

\(\Leftrightarrow5\sqrt{2x+1}+3\sqrt{x+1}=4\sqrt{x+1}+4\sqrt{2x+1}\)

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

\(\Leftrightarrow2x+1=x+1\Leftrightarrow x=0\left(tm\right)\)

\(3,\sqrt{x^2-2x+1}=1\left(x\in R\right)\\ \Leftrightarrow\sqrt{\left(x-1\right)^2}=1\\ \Leftrightarrow\left|x-1\right|=1\Leftrightarrow\left[{}\begin{matrix}x-1=1\left(x\ge1\right)\\x-1=-1\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(4,ĐK:x\in R\\ PT\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\\ \Leftrightarrow\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x-2=1\left(x\ge2\right)\\x-2=-1\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

\(5,ĐK:x\in R\\ PT\Leftrightarrow\left|2x-1\right|=\left|x+4\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=x+4\\1-2x=x+4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

\(6,ĐK:x\ge-2\\ PT\Leftrightarrow5\sqrt{x+2}-3\sqrt{x+2}-\sqrt{x+2}=\sqrt{x+7}\\ \Leftrightarrow\sqrt{x+2}=\sqrt{x+7}\Leftrightarrow x+2=x+7\Leftrightarrow0x=5\Leftrightarrow x\in\varnothing\)

\(7,ĐK:x\ge-1\\ PT\Leftrightarrow5\sqrt{x+2}+3\sqrt{x+1}=4\sqrt{x+1}+4\sqrt{x+2}\\ \Leftrightarrow\sqrt{x+2}=\sqrt{x+1}\\ \Leftrightarrow x+2=x+1\\ \Leftrightarrow0x=-1\Leftrightarrow x\in\varnothing\)

Xet tam giac BDC va tam giac CEB ta co 

^BDC = ^CEB = 90⁰

BC _ chung 

^BCD = ^CBE ( gt ) 

=> tam giac BDC = tam giac CEB ( ch - gn ) 

=> ^DBC = ^ECB ( 2 goc tuong ung ) 

Ta co ^B - ^DBC = ^ABD 

^C - ^ECB = ^ACE 

=> ^ABD = ^ACE 

Xet tam giac IBE va tam giac ICD 

^ABD = ^ACE ( cmt )

^BIE = ^CID ( doi dinh ) 

^BEI = ^IDC = 90⁰

Vay tam giac IBE = tam giac ICD (g.g.g) 

c, Do BD vuong AC => BD la duong cao 

CE vuong BA => CE la duong cao 

ma BD giao CE = I => I la truc tam 

=> AI la duong cao thu 3 

=> AI vuong BC