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ừ thì giải theo cách lớp 5
\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\times x=\frac{1}{9}+\frac{2}{8}+...+\frac{9}{1}\)
Gọi vế phải là A
\(A=\frac{10}{9}+\frac{10}{8}+...+\frac{10}{1}-9\)
\(A=10\times\left(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+...+\frac{1}{2}\right)+1\)
\(A=10\times\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+...+\frac{1}{2}-\frac{1}{10}\right)+1\)
\(A=10\times\left(\frac{1}{10}+\frac{1}{9}+...+\frac{1}{2}\right)-1+1\)
\(\Rightarrow VP=10\times A\)
\(\Rightarrow x=10\)
\(a.\frac{19}{5}\cdot\frac{4}{7}+\frac{3}{7}\cdot\frac{19}{5}-\frac{4}{5}\)
\(=\frac{19}{5}\cdot\left(\frac{4}{7}+\frac{3}{7}\right)-\frac{4}{5}\)
\(=\frac{19}{5}\cdot1-\frac{4}{5}\)
\(=\frac{19}{5}-\frac{4}{5}=\frac{15}{5}=3\)
\(b.2\frac{2}{7}\cdot5\frac{2}{5}+\frac{16}{7}\cdot1\frac{3}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\frac{27}{5}+\frac{16}{7}\cdot\frac{8}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\left(\frac{27}{5}+\frac{8}{5}\right)+\frac{1}{2}\)
\(=\frac{16}{7}\cdot7+\frac{1}{2}\)
\(=16+\frac{1}{2}=\frac{33}{2}\)
\(c.\frac{3}{7}\cdot3\frac{3}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{15}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\left(\frac{15}{4}-\frac{5}{4}\right)-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{5}{2}-\frac{1}{4}\)
\(=\frac{15}{14}-\frac{1}{4}=\frac{23}{28}\)
Chú ý: \(\cdot:\times\)
2/3+5/8=16/24+15/24=31/24
4 và 2/5-2 và 2/3=22/5-7/3=66/15-35/15=31/15
x + 2 và 1/5 =4 và 2/3
x + 11/5 = 14/3
x = 14/3-11/5
x = 37/15
x : 3/2 = 4/15
x = 4/15*3/2
x = 2/5
a) ( 1/2-1/3-1/6).(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10
0.(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10
0+3/4.x = 9/10
3/4.x = 9/10
x = 9/10: 3/4
x = 6/5
b) x + ( 3/1.3+3/3.5+...+3/13.15) = 11/5
x + 3/2. ( 1-1/3 + 1/3 - 1/5 + ...+ 1/13 - 1/15) = 11/5
x + 3/2. ( 1-1/15) = 11/5
x + 3/2.14/15 = 11/5
x + 7/5 = 11/5
x = 11/5 - 7/5
x = 4/5
(y - 2/5) x 1/2 =1
y - 2/5 =1:1/2
y-2/5 = 1x2/1
y-2/5 =2
y =2+2/5
y =2/1 +2/5
y =10/5+2/5
y = 12/5
\(x+12=\frac{1}{4}+\frac{2}{4}\)
\(=x+12=\frac{3}{4}\)
\(\Rightarrow x=\frac{3}{4}-12\)
\(\Rightarrow x=-\frac{45}{4}\)
\(x+12=\frac{1}{4}+\frac{2}{4}\)
\(\Rightarrow x+12=\frac{3}{4}\)
\(\Rightarrow x=-\frac{45}{4}\)