a) x = 4

b) x = 3

c) x = 2

d) x = 1

e) x = 3

f) x = 2

g) x = 4

h) x = 3

a, A =1 + 2 -  3 - 4 + 5 + 6 - 7 - 8 + 9 +...+ 82 - 83 - 84 + 85

A =(1+2-3-4)+(5+6-7-8)+...+(81+82-83-84) + 85

Xét dãy số: 1; 5;...;81

Dãy số trên là dãy số cách đều với khoảng cách là: 5-1 = 4

Số số hạng của dãy số trên là: (81-1):4+1 = 21 (số hạng)

Vậy tổng A là tổng của 21 nhóm và 85, trong đó mỗi nhóm có giá trị bằng 1+2-3-4 =  - 4

A = -4. 21 + 85 = 1

yêu cầu của câu c là gì thế em

\(\left\{5.\left[42:\left(5.3+50:10\right).3:2+4\right]\right\}+\left\{35.\left[63:9.\left(5^2\right)\right]\right\}\\ =\left\{5.\left[42:\left(15+5\right).1,5+4\right]\right\}+\left\{35.\left(9.25\right)\right\}\\ =\left\{5.\left[42:20.1,5+4\right]\right\}+\left\{35.225\right\}\\ =\left\{5.\left[2,1.1,5+4\right]\right\}+7875\\ =\left\{5.\left[3,15+4\right]\right\}+7875\\ =\left\{5.7,15\right\}+7875\\ =35,75+7875\\ =7910,75\)

{5.[42:(5.3+50:10).3:2+4]}+{35.[63:9.(52)]}={5.[42:(15+5).1,5+4]}+{35.(9.25)}={5.[42:20.1,5+4]}+{35.225}={5.[2,1.1,5+4]}+7875={5.[3,15+4]}+7875={5.7,15}+7875=35,75+7875=7910,75

\(a,x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x-\frac{61}{8}=\frac{5}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)

\(b,x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x+\frac{43}{5}=\frac{37}{4}\)

=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)

\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)

=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)

=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)

d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)

=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)

=> \(\frac{98x}{198}=99\)

=>  98x = 99 . 198

=> 98x = 19602

=> x = 19602 : 98 = 9801/49

a) \(x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{71}{8}\)

b) \(x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x=\frac{37}{4}-\frac{61}{8}\)

=> \(x=\frac{13}{8}\)

c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)

=> \(x-\frac{61}{8}=3.\frac{1}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}\)

=> \(x=\frac{73}{8}\)

d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)

=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)

=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)

=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)

=> \(x.\frac{98}{99}=198\)

=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)

1. a\()\) \(\frac{9}{4}\)\(-\) \(\frac{3}{4}\) \(\times\) \(\frac{1}{5}\)

= \(\frac{9}{4}\) \(-\) \(\frac{3}{20}\)

= \(\frac{45}{20}\) \(-\) \(\frac{3}{20}\)

= \(\frac{21}{10}\)

b\()\) \(\frac{2}{5}\) \(\times\) \(\frac{1}{3}\) \(\times\) \(\frac{3}{5}\)

= \(\frac{2}{15}\) \(\times\) \(\frac{3}{5}\)

= \(\frac{2}{25}\)

2.

a\()\) câu a mik vẫn chưa hiểu đề là gì

b\()\) \(\frac{7}{5}\) \(\div\) x \(=\) \(\frac{2}{3}\) \(\times\) \(\frac{1}{4}\)

\(\frac{7}{5}\) \(\div\) x \(=\) \(\frac{1}{6}\)

x \(=\) \(\frac{7}{5}\) \(\div\) \(\frac{1}{6}\)

x \(=\) \(\frac{7\times6}{5\times1}\)

x \(=\) \(\frac{42}{5}\)