a) - Dùng quỳ tím:

+ Hóa đỏ -> dd H2SO4

+ Hóa xanh -> dd NaOH

+ Không đổi màu -> dd NaCl

b) - Dùng quỳ tím:

+ Hóa xanh -> dd Ca(OH)2

+ Hóa đỏ -> dd H2SO4, dd HCl

- Dùng dd BaCl2:

+ Có kết tủa trắng BaSO4 -> dd H2SO4 

+ Không có kt trắng -> dd HCl

PTHH: H2SO4 + BaCl2 -> BaSO4 (kt trắng) + 2 HCl

c) - Dùng quỳ tím:

+ Hóa xanh : dd KOH 

+ Hóa đỏ : dd HNO3, dd HCl

- Dùng dd AgNO3:

+ Có kt trắng AgCl -> dd HCl

PTHH: HCl + AgNO3 -> AgCl (trắng) + HNO3

+ Còn lại ko ht -> dd AgNO3

refer

1/ I like table tennis and that’s why I spend a lot of time on that game.
=>I spend a lot of time on table tennis because I like it.
2/ Because the weather was cold, we had to cancel our picnic.
=> The weather was cold, so we had to cancel our picnic.
3/ I won’t forget to phone her. You know, I like her so much.
=> You know I like her so much so I won't forget to phone her
4/ They tried their best to complete the course and that was why they passed.
=> they tried their best to complete the course, so they passed
5/ I love my city so much, so I think I will stay and work here in my city.
=> I think I will stay and word in my city because I love it so much.

1 Because I like table tennis, I spend a lot of time on that game

2 The weather was so cold that we had to cancel our picnic

3 I won't forget to phone her because I like her so mich

4 They pass the course because they tried their best to complete the course

5 Because I love my city so much, I think I will stay and work here in my city

1 Because I like table tennis, I spend a lot of time on that game

2 The weather was so cold that we had to cancel our picnic

3 I won't forget to phone her because I like her so mich

4 They pass the course because they tried their best to complete the course

5 Because I love my city so much, I think I will stay and work here in my city

3.B (thì HTHT với dấu hiệu yet)

4. B

5. B (wish S Ved: ước trong iện tại)

6. D (thì QKĐ với dấu iệu Last night)

Câu 2:

1: \(y=\sqrt{3}+5\)

=>\(\left(\sqrt{3}-1\right)x+4=\sqrt{3}+5\)

=>\(\left(\sqrt{3}-1\right)\cdot x=\sqrt{3}+5-4=\sqrt{3}+1\)

=>\(x=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}=\dfrac{\left(\sqrt{3}+1\right)^2}{3-1}=\dfrac{4+2\sqrt{3}}{2}=2+\sqrt{3}\)

2: \(x^2-2\left(1-m\right)x-2m-5=0\)

=>\(x^2+\left(2m-2\right)x-2m-5=0\)

a: \(\Delta=\left(2m-2\right)^2-4\left(-2m-5\right)\)

\(=4m^2-8m+4+8m+20\)

\(=4m^2+24>=24>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

Câu 1:

2: Thay x=2 và y=-1 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}2a-\left(-1\right)=5\\b\cdot2+a\cdot\left(-1\right)=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2a=5+\left(-1\right)=4\\2b-a=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\2b=a+4=6\end{matrix}\right.\)

=>a=2 và b=3

2: Gọi phân số cần tìm có dạng là \(\dfrac{a}{b}\left(b\ne0\right)\)

Khi tăng mẫu số thêm 4 đơn vị thì phân số đó bằng 1/3 nên ta có:

\(\dfrac{a}{b+4}=\dfrac{1}{3}\)

=>3a=b+4

=>3a-b=4(1)

Khi giảm mẫu số đi 2 đơn vị thì phân số bằng với 2/3 nên ta có:

\(\dfrac{a}{b-2}=\dfrac{2}{3}\)

=>3a=2(b-2)

=>3a=2b-4

=>3a-2b=-4(2)

Từ (1) và (2) ta có hệ phương trình:

\(\left\{{}\begin{matrix}3a-b=4\\3a-2b=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=8\\3a-b=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=8\\3a=b+4=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=8\end{matrix}\right.\)(nhận)

Vậy: Phân số cần tìm là \(\dfrac{4}{8}\)

1) Vì a⊥d , b⊥d  ⇒ a // b

⇒\(\widehat{A_1}=\widehat{B}=80^o\) (ở vị trí so le trong)

⇒\(\widehat{A_3}=\widehat{B}=80^o\)(ở vị trí đồng vị)

Do  \(\widehat{A_2}+\widehat{B}=180^o\)

     (hai góc trong cùng phía)

Thay số:\(\widehat{A_2}+80^o=180^o\)

           ⇒\(\widehat{A_2}=100^o\)

2)a.Vì Ax//By⇒\(\widehat{A}=\widehat{ABy}=30^o\)

Mà \(\widehat{ABC}=70^o\)

\(\Rightarrow\widehat{yBC}=\widehat{ABC}-\widehat{CBy}=70^o-30^o=40^o\)

b. Xét Bx và Ct có :\(\widehat{CBy}=\widehat{C}=40^o\) là hai góc so le trong bằng nhau

⇒Bx//Ct  . Mà Ax//By

⇒Ax//Ct

5:

\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}>=3\cdot\sqrt[3]{\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}}=3\)

a^2+b^2>=2ab

b^2+c^2>=2bc

a^2+c^2>=2ac

=>a^2+b^2+c^2>=ab+bc+ac

=>(ab+bc+ac)/(a^2+b^2+c^2)>=1

=>a/b+b/c+c/a+(ab+ac+bc)/(a^2+b^2+c^2)>=4