K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

Bài 2 ; 2 ; 3 ; 4 phần tự luận bạn tham khảo ở Câu hỏi của Phùng minh long - Toán lớp 7 - Học trực tuyến OLM nhé !

Bài 1 :

\(1)\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)

\(\Rightarrow\left(\frac{11}{24}+\frac{13}{24}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)+\frac{1}{2}\)

\(\Rightarrow1-1+\frac{1}{2}=\frac{1}{2}\)

\(2)\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1\frac{15}{17}+\frac{2}{3}\)

\(=\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1-\frac{15}{17}+\frac{14}{21}\)

\(=\left(\frac{15}{34}+\frac{19}{34}-1\right)+\left(\frac{7}{21}+\frac{14}{21}\right)-\frac{15}{17}\)

\(=1-\frac{15}{17}=\frac{2}{17}\)

\(3)1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)

\(=1+1+0,5=2,5\)

\(4)\left(-12\right):\left(\frac{3}{4}-\frac{5}{6}\right)^2\)

\(=\left(-12\right):\left(\frac{-1}{12}\right)^2\)

\(=\left(-12\right).12^2=-12^3\)

\(5)\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)

\(=\frac{-891}{25}:4=\frac{-891}{100}\)

\(6)\left(-6,5\right).5,7+5,7.\left(-3,5\right)\)

\(=\left[\left(-6,5\right)+\left(-3,5\right)\right].5,7\)

\(=-10.5,7=57\)

\(7)-3,75.\left(-7,2\right)+2,8.3,75\)

\(=3,75.7,2+2,8.3,75\)

\(=3,75.\left(7,2+2,8\right)\)

\(=3,75.10=37,5\)

\(8)23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)

\(=23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)

\(=\left(23\frac{1}{4}-13\frac{1}{4}\right).1,4\)

\(=10.1,4=14\)

\(9)16\frac{2}{7}:\left(\frac{-3}{5}\right)+28\frac{2}{7}:\frac{3}{5}\)

\(=-16\frac{2}{7}:\frac{3}{5}+28\frac{2}{7}:\frac{3}{5}\)

\(=-16\frac{2}{7}.\frac{5}{3}+28\frac{2}{7}.\frac{5}{3}\)

\(=\left(-16\frac{2}{7}+28\frac{2}{7}\right).\frac{5}{3}\)

\(=12.\frac{5}{3}=20\)

15 tháng 9 2021

    3456789123456789

18 tháng 3 2022

ảnh lỗi

14 tháng 3 2022

1.A
2.A
3.B
4.C
5.B
6.C
7.A
8.A
9.B
10.A
11.B
12.A
13.C
14.B
15.B
16.A
17.A
18.A
19.A
20.C

14 tháng 12 2021

\(a,PT\left(1\right)=\dfrac{75y^4}{42x^2y^5};PT\left(2\right)=\dfrac{28x}{42x^2y^5}\\ b,PT\left(1\right)=\dfrac{11y^2}{102x^4y^3};PT\left(2\right)=\dfrac{9x^3}{10x^4y^3}\\ c,PT\left(1\right)=\dfrac{3x\left(3x+1\right)}{36x^2y^4};PT\left(2\right)=\dfrac{4y\left(y-2\right)}{36x^2y^4}\\ d,PT\left(1\right)=\dfrac{6y^2}{36x^3y^4};PT\left(2\right)=\dfrac{4x\left(x+1\right)}{36x^3y^4};PT\left(3\right)=\dfrac{9x^2y\left(x-1\right)}{36x^3y^4}\)

\(e,PT\left(1\right)=\dfrac{12y^4\left(3+2x\right)}{120x^4y^5};PT\left(2\right)=\dfrac{75x^2y^3}{120x^4y^5};PT\left(3\right)=\dfrac{8x^3}{120x^4y^5}\\ f,PT\left(1\right)=\dfrac{3\left(x+1\right)\left(4x-4\right)}{6x\left(x+3\right)\left(x+1\right)};PT\left(2\right)=\dfrac{2\left(x+3\right)\left(x-3\right)}{6x\left(x+1\right)\left(x+3\right)}\)

14 tháng 12 2021

\(g,PT\left(1\right)=\dfrac{4x^2}{2x\left(x+2\right)^3};PT\left(2\right)=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)^3}\\ h,PT\left(1\right)=\dfrac{5}{3x\left(x-2\right)\left(x+2\right)}=\dfrac{10\left(x+3\right)}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\\ PT\left(2\right)=\dfrac{3}{2\left(x+2\right)\left(x+3\right)}=\dfrac{9x\left(x-2\right)}{6x\left(x+2\right)\left(x+3\right)\left(x-2\right)}\)

25 tháng 2 2022

\(=\dfrac{2x^2y^2}{3xy^2}-\dfrac{2ax+3x}{3a}=\dfrac{2x}{3}-\dfrac{2ax+3x}{3a}\)

\(=\dfrac{2xa-2xa-3x}{3a}=\dfrac{-3x}{3a}=-\dfrac{x}{a}\)

\(=\dfrac{5}{3}-\dfrac{5a-6}{3a}=\dfrac{5a-5a+6}{3a}=\dfrac{6}{3a}=\dfrac{2}{a}\)

\(=\dfrac{2x-3a}{2a}+\dfrac{3}{2}=\dfrac{2x-3a+3a}{2a}=\dfrac{2x}{2a}=\dfrac{x}{a}\)

\(=\dfrac{2-a}{2a}+\dfrac{1}{2x}=\dfrac{4x-2xa+2a}{4xa}=\dfrac{2x-xa+a}{xa}\)

1 tháng 3 2022

a) \(n_{Fe}=\dfrac{28}{56}=0.5\left(mol\right)\)

\(n_{O_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

     \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)

\(Bđ:0.5....0.2\)

\(Pư:0.3.....0.2........0.1\)

\(Kt:0.2.......0..........0.1\)

\(m_{Fe\left(dư\right)}=0.2\cdot56=11.2\left(g\right)\)

\(m_{Fe_3O_4}=0.1\cdot232=23.2\left(g\right)\)

1 tháng 3 2022

a. \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)

\(n_{O_2}=\dfrac{4.48}{22,4}=0,2\left(mol\right)\)

Ta thấy : 0,5 > 0,2 => Fe dư , O2 đủ

PTHH : 3Fe + 2O---to---> Fe3O4

             0,3      0,2                0,1

\(m_{Fe\left(dư\right)}=\left(0,5-0,3\right).56=11,2\left(g\right)\)

b. \(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

16 tháng 3 2022

bn đg thi hả

16 tháng 3 2022

đr bn ạ