lần đổ 1

\(\left(mC+m'C'\right).\left(38-20\right)=mC.\left(60-38\right)\)

\(\Leftrightarrow\left(mC+m'C'\right)18=mC.22\)

\(\Leftrightarrow2mC=9m'C'\)

lần 2 \(\left(2mC+m'C'\right)\left(t_x-38\right)=mC.\left(60-t_x\right)\)

\(11m'C'\left(t_x-38\right)=\dfrac{9}{2}.m'C'\left(60-t_x\right)\)

\(\Rightarrow t_x=...\)

 1 had better eat more fruits and vegetables.

2 likes painting very much.

3 play  card.

*Le's -> Let's

4 had better not eat canned  food.

5 don't we go camping for some days?

cảm ơn cậu nhìu ạ

Chụp rõ đề hơn đi bạn.

Bài 1:

a: \(\frac17-\left(-\frac35\right)+\frac67=\frac77+\frac35=1+\frac35=\frac85\)

b: \(\frac{20}{13}\cdot\frac{-8}{9}+\frac{7}{18}\cdot\frac{20}{13}\)

\(=\frac{20}{13}\left(-\frac89+\frac{7}{18}\right)\)

\(=\frac{20}{13}\left(-\frac{16}{18}+\frac{7}{18}\right)=\frac{20}{13}\cdot\frac{-9}{18}=\frac{20}{13}\cdot\frac{-1}{2}=-\frac{20}{26}=-\frac{10}{13}\)

c: \(0,75+\frac{15}{6}:5-\frac{1}{36}\cdot\left(-3\right)^2+\left(-\frac{2020}{2021}\right)^0\)

\(=\frac34+\frac{15}{6\cdot5}-\frac{1}{36}\cdot9+1\)

\(=\frac34-\frac14+\frac12+1=\frac12+\frac12+1=2\)

Bài 2:

a: \(3x-\frac23=-\frac59\)

=>\(3x=-\frac59+\frac23=-\frac59+\frac69=\frac19\)

=>\(x=\frac19:3=\frac{1}{27}\)

b: \(\frac45-\left|x+\frac12\right|=\frac14\)

=>\(\left|x+\frac12\right|=\frac45-\frac14=\frac{16-5}{20}=\frac{11}{20}\)

=>\(\left[\begin{array}{l}x+\frac12=\frac{11}{20}\\ x+\frac12=-\frac{11}{20}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{20}-\frac12=\frac{1}{20}\\ x=-\frac{11}{20}-\frac12=-\frac{21}{20}\end{array}\right.\)

c: \(\left(\frac32x-\frac15\right)^2\cdot\left(\frac{51}{67}x^2+\frac12\right)=0\)

=>\(\left(\frac32x-\frac15\right)^2=0\)

=>\(\frac32x-\frac15=0\)

=>\(\frac32x=\frac15\)

=>\(x=\frac15:\frac32=\frac15\cdot\frac23=\frac{2}{15}\)

a)

ta có \(n_{N_2}=\dfrac{4,2}{28}=0,15\left(mol\right)\)

\(\Rightarrow V_{N_2}=0,15.22,4=3,36l\)

ta có \(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,1.22,4=2,24l\)

b) 

ta có \(n_{SO_2}=\dfrac{32}{64}=0,5\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,5.22,4=11,2l\)

ta có \(n_{O_2}=\dfrac{32}{32}=1\left(mol\right)\)

\(\Rightarrow V_{O2}=1.22,4=22,4l\)

c) 

ta có \(n_{H_2}=\dfrac{2}{2}=1\left(mol\right)\)

\(\Rightarrow V_{H_2}=1.22,4=22,4l\)

ta có \(n_{O_2}=\dfrac{8}{32}=0,25\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,25.22,4=5,6l\)

d) 

ta có \(n_{N_2}=\dfrac{8,4}{28}=0,3\left(mol\right)\)

\(\Rightarrow V_{N_2}=0,3.22,4=6,72l\)

ta có \(n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\)

\(\Rightarrow V_{N_2}=0,2.22,4=4,48l\)

Cảm mơn bạn nha !

Bài 6: 

a: Xét tứ giác DEBF có 

DE//BF

DE=BF

Do đó: DEBF là hình bình hành

Bài 5 Mn ơi

Giúp em bài số 4 ạ😓

1 Having slept

2 not being invited

3 Having had

4 having

5 talking

6 succeeded - launching

7 Having travelled

8 Have - considered - trying

9 Having seen - had - to go

10 Being invited

11 Being found

12 having

13 taken - being photographed 

14 to fix

15 living

16 Having waited - to deliver - decided to cancel

17 Having photocopied 

18 to have happen

19 to give

20 spoiling

Bài 12:

Chiều rộng mảnh vườn:

25 x 3/5 = 15(m)

Diện tich phần trồng cây có chiều dài:

25 - 2=23(m)

Diện tích phần trồng cây có chiều rộng:

15 - 2 = 13(m)

Diện tích phần trồng cây:

23 x 13= 299(m²)

Đ.số: 299m²

12: Chiều rộng là 25*3/5=15m

Chiều dài mảnh đất trồng cây là 25-2=23m

Chiều rộng mảnh đất trồng cây là 15-2=13m

Diện tích mảnh đất trồng cây là:

23*13=299m²