Lí do = 0 nè:
\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-5}{39}+\frac{x-69}{49}\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}=0+\left(\frac{x-5}{39}+\frac{x-69}{49}\right)\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}-\left(\frac{x-5}{39}+\frac{x-69}{49}\right)=0\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}-\frac{x-5}{39}-\frac{x-69}{49}=0\)
Tìm x :
a) \(\frac{x}{8}=\frac{2}{x}\)
b) \(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)
Tìm \(x\) :
\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)
Chưa hiểu lắm nên làm kĩ càng nhé
7/19 x 29 + 7/29 x 39 + 7/39 x 49 + 7/49 x 59 + 7/59 x 69
Giúp em với, trước 3:00 nha><
Tìm x, biết :
a, \(60\%x+0,4x+x:3=2\)
b, \(\left|2x-5\right|-7=\left(\frac{1}{49}-\frac{1}{3^2}\right)\left(\frac{1}{49}-\frac{1}{4^2}\right)...\left(\frac{1}{49}-\frac{1}{2015^2}\right)\)
c, \(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+...+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+...+\frac{40}{39}\)
Tìm \(x\) biết:
a)\(\left|2x-5\right|-7=\left(\frac{1}{49}-\frac{1}{3^2}\right)\left(\frac{1}{49}-\frac{1}{4^2}\right).....\left(\frac{1}{49}-\frac{1}{2015^2}\right)\)
b)\(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+...+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+...+\frac{40}{39}\)
a) tính nhanh :
54 x 275 + 825 x 15 + 275 = ....
b) tích sau tận cùng bằng chữ số nào ?
19 x 29 x 39 x 49 x 59 x 69 x 79 x 89 x 99
\(\frac{25}{49}\)x \(\frac{21}{29}\)-\(\frac{25}{49}\)x\(\frac{7}{29}\)+ \(\frac{24}{49}\)x\(\frac{15}{29}\)
Timf x,y
1) x+\(\frac{7}{12}\)=\(\frac{17}{18}\)-\(\frac{1}{9}\)
2)\(\frac{29}{30}\)-(\(\frac{13}{23}\)+x)=\(\frac{7}{69}\)
3)\(\frac{x}{15}=\frac{4}{10}\)
4)\(\frac{x-3}{49}=\frac{-2}{7}\)
5)\(\frac{x}{3}=\frac{1}{y}\)
Chứng minh tổng sau lớn hơn 1 : \(N=\frac{19}{60}+\frac{29}{100}+\frac{39}{150}+\frac{49}{300}\)
