:v

Làm hết các bài hả?

1 asked me what time it was and he had to go then

2 He told me not to shut the door because it was very hot

3 He told me to take my seat and I thanked him

4 He said that climate didn't suit his health so he had to go away as soon as he could

5 Mr Bill told me to be quiet because the child was sleeping

6 She told me to bring an umbrella because it was raining then

câu 1:có....từ phức.từ láy là : lẫm liệt,hoảng hốt,vươn vai.

câu 2.

câu 3:các danh từ ;vươn vai,tráng sĩ,oai phong,lẫm liệt

câu 4:lẫm liệt là oai phong,trông đáng sợ

1. A

2. B

3. D

4. B

5. C

6. B

7. D

8. B

1) \(\left(2x+3\right)^2=4x^2+12x+9\)

\(\left(3x+2\right)^2=9x^2+12x+4\)

\(\left(2x+5\right)^2=4x^2+20x+25\)

\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)

\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)

2)  \(\left(2x-3\right)^2=4x^2-12x+9\)

\(\left(3x-2\right)^2=9x^2-12x+4\)

\(\left(2x-5\right)^2=4x^2-20x+25\)

\(\left(2x-\dfrac{1}{3}\right)^2=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)

\(\left(3x-\dfrac{1}{3}\right)^2=9x^2-2x+\dfrac{1}{9}\)

3) \(\left(2x-3\right)\left(2x+3\right)=4x^2-9\)

\(\left(3x-4\right)\left(3x+4\right)=9x^2-16\)

\(\left(2x-5\right)\left(2x+5\right)=4x^2-25\)

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

\(\left(2x-\dfrac{1}{3}\right)\left(2x+\dfrac{1}{3}\right)=4x^2-\dfrac{1}{9}\)

1: \(\left(2x+3\right)^2=4x^2+12x+9\)

\(\left(3x+2\right)^2=9x^2+12x+4\)

\(\left(2x+5\right)^2=4x^2+20x+25\)

\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)

\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)

ta có Cu ko tác dụng được vs H2SO4 

-> 2,24 lít khí H2 đc tạo ra là do phản ứng HH của Fe va H2SO4 

\(PTHH:\) \(Fe+H_2SO_4->FeSO_4+H2\)

n H2 = 2,24 :22,4=0,1 mol 

n H2 = n Fe =0,1 mol 

m Fe = 0,1.56=5,6 g 

m Cu = 10-5,6=4,4 g 

vậy giá trị của x là 4,4 g

26 

PTHH : \(Zn+2HCl->ZnCl_2+H_2\)

n Zn = 6,5:65=0,1 mol

n ZnCl2 = n Zn =0,1 mol

m ZnCl2 = 0,1.(65+35,5.2)=13,6 g

ta có Cu ko tác dụng được vs H2SO4 

-> 2,24 lít khí H2 đc tạo ra là do phản ứng HH của Fe va H2SO4 

n H2 = 2,24 :22,4=0,1 mol 

n H2 = n Fe =0,1 mol 

m Fe = 0,1.56=5,6 g 

m Cu = 10-5,6=4,4 g 

vậy giá trị của x là 4,4 g

a)

%O=100-43.4-11.3=45.3%

m_Na=43.4*m_A/100=0.434m_A

n_Na=0.434m_A/23=0.018m_A

n_C=0.009m_A

n_O=0.028m_A

A có công thức: Na_xC_yO_z

x : y : z = n_Na: n_C : n_O =0.018: 0.009: 0.028 = 2:1:3

A là Na₂CO₃

câu b mình ko bik lm nha bạn

1. 5

2. 43/31

3. >

4. 35/24

5. 16/33

\(\frac{28}{15}\times\frac{1}{4}\times3+\left(\frac{8}{15}-\frac{79}{60}\right)\times\frac{24}{47}\)

=\(\frac{7}{15}\times3+\left(\frac{32}{60}-\frac{79}{60}\right)\times\frac{24}{47}\)

=\(\frac{7}{5}+\frac{47}{60}\times\frac{24}{47}\)

=\(\frac{7}{5}+\frac{2}{5}\)

=\(\frac{5}{5}\)

=\(1\)

\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+(\(\frac{8}{15}\)+\(\frac{-79}{60}\))x\(\frac{24}{47}\)=

=\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+(\(\frac{32}{60}\)+\(\frac{-79}{60}\))x\(\frac{24}{47}\)

=\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+\(\frac{-47}{60}\)x\(\frac{24}{47}\)

=\(\frac{7}{15}\)x3+\(\frac{-47}{60}\)x\(\frac{24}{47}\)

=\(\frac{7}{5}\)+\(\frac{-2}{5}\)

=\(\frac{5}{5}\)=1