1/2+2/3+3/4+4/5+5/6+6/7+7/8+8/9+9/10x9/10

=9/10x(1/2+2/3)+(3/4+4/5)+(5/6+6/7)+(7/8+8/9)

=9/10x(1/3+3/5+5/7+7/9)

9/10x(1/3+3/5)+(5/7+7/9)

=9/10x1/5+5/9

9/50+5/9

=10

Bn Long làm đúng rồi bn   nguyễn kim arica cứ làm theo cách đó là được .

Bn nào thấy đúng thì ủng hộ nha .

hình như đề bạn sai thì phải đúng ko biết ngỏ bạn làm

cái chỗ có 2 phân số 9/10 đấy

\(\frac{1}{2}+\frac{3}{4}=\frac{1.2}{2.2}+\frac{3}{4}=\frac{2}{4}+\frac{3}{4}=\frac{5}{4}\)

\(\frac{2}{-5}-\frac{-7}{9}=-\frac{2}{5}-\frac{-7}{9}=\frac{-2}{5}+\left(-\frac{7}{9}\right)=-\frac{18}{45}+-\frac{35}{45}=-\frac{53}{45}\)

cho kẹo~~~

Câu 1:5/4

nhớ nha gửi lượt kb cho tớ nha

\(\frac{ }{ }\)

\(1)\frac{1}{2}x-\frac{3}{5}=\frac{-4}{5}\)

\(\Rightarrow\frac{1}{2}x=\frac{-4}{5}+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}x=\frac{-1}{5}\)

\(\Rightarrow x=\frac{-1}{5}:\frac{1}{2}=\frac{-1}{5}\cdot\frac{2}{1}=\frac{-2}{5}\)

\(\Leftrightarrow x=\frac{-2}{5}\)

\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)

\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)

Đến đây tìm được rồi nhé

3,4, áp dụng bài 1,2 rồi làm :v

1. 1/2x - 3/5 = -4/5

=> 1/2x = -4/5 + 3/5

=> 1/2x = -1/5

=> x = -1/5 : 1/2

=> x = -2/5

9/2

k mình nha

= \(\frac{9}{2}\)

ket qua bang 9/2

g) Ta có: \(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)

\(=\frac{34}{5}-\left(\frac{5}{3}+\frac{19}{5}\right)\)

\(=\frac{34}{5}-\frac{5}{3}-\frac{19}{5}\)

\(=3-\frac{5}{3}=\frac{9}{3}-\frac{5}{3}=\frac{4}{3}\)

h) Ta có: \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)

\(=\frac{68}{9}-\frac{11}{4}-\frac{32}{9}\)

\(=4-\frac{11}{4}=\frac{16}{4}-\frac{11}{4}=\frac{5}{4}\)

i) Ta có: \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)

\(=6+\frac{5}{7}-1-\frac{3}{4}-2-\frac{5}{7}\)

\(=3-\frac{3}{4}=\frac{9}{4}\)

k) Ta có: \(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)\)

\(=7+\frac{5}{11}-2-\frac{3}{7}-3-\frac{5}{11}\)

\(=2-\frac{3}{7}=\frac{11}{7}\)

g) \(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)

\(=6\frac{4}{5}-1\frac{2}{3}-3\frac{4}{5}\)

\(=\left(6\frac{4}{5}-3\frac{4}{5}\right)-1\frac{2}{3}\)

\(=\left(6+\frac{4}{5}-3-\frac{4}{5}\right)-1\frac{2}{3}\)

\(=3-\frac{5}{3}=\frac{9}{3}-\frac{5}{3}=\frac{4}{3}\)

h) \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)

\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)

\(=\left(7\frac{5}{9}-3\frac{5}{9}\right)-2\frac{3}{4}\)

\(=\left(7+\frac{5}{9}-3-\frac{5}{9}\right)-2\frac{3}{4}\)

\(=4-\frac{11}{4}=\frac{16}{4}-\frac{11}{4}=\frac{5}{4}\)

i) \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)

\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)

\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)

\(=\left(6+\frac{5}{7}-2-\frac{5}{7}\right)-\frac{7}{4}\)

\(=4-\frac{7}{4}=\frac{16}{4}-\frac{7}{4}=\frac{9}{4}\)

k) \(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)\)

\(=7\frac{5}{11}-2\frac{3}{7}-3\frac{5}{11}\)

\(=\left(7\frac{5}{11}-3\frac{5}{11}\right)-2\frac{3}{7}\)

\(=4-\frac{17}{7}=\frac{28}{7}-\frac{17}{7}=\frac{11}{7}\)

\(a)x+30\%x=-1,31\)

\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)

\(\Leftrightarrow10x+3x=-13,1\)

\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)

\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)

\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)

\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)

\(\Leftrightarrow\frac{6x-3}{2}=9\)

\(\Leftrightarrow6x-3=18\)

\(\Leftrightarrow x=\frac{7}{2}\)

\(c)\frac{1}{2}x-\frac{3}{4}=\frac{14}{9}.\frac{3}{7}\)

\(\Leftrightarrow\frac{x}{2}-\frac{3}{4}=\frac{2}{3}\)

\(\Leftrightarrow\frac{x}{2}=\frac{17}{12}\)

\(\Leftrightarrow12x=34\Leftrightarrow x=\frac{17}{6}\)

D=115/132/103/132=115/103

\(D=\frac{\frac{88}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}\)

\(D=\frac{\frac{115}{132}}{\frac{103}{132}}\)

\(D=\frac{115}{103}\)

D = \(\frac{115}{103}\)

\(a)\frac{-3}{4}+\frac{3}{7}+-\frac{1}{4}+\frac{4}{9}+\frac{4}{7}\)

\(=\left(\frac{-3}{4}+-\frac{1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)

\(=-1+1+\frac{4}{9}\)

\(=\frac{4}{9}\)

\(\frac{-3}{4}+\frac{3}{7}+\frac{-1}{4}+\frac{4}{9}+\frac{4}{7}\)

\(=\left(\frac{-3}{4}+\frac{-1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)

\(=\left(-1\right)+1+\frac{4}{9}\)

\(=0+\frac{4}{9}\)

\(=\frac{4}{9}\)

GIÚP MIK IK XIN CÁC BN ĐÓ