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hộ mik nha
a)\(2.x-\frac{3}{2}=\frac{1}{2}\)
\(2x=\frac{4}{2}\)
\(x=\frac{4}{2}:2\)
\(x=1\)
Vậy x=1
b)\(\frac{8}{13}.x+\frac{5}{13}.x=\frac{1}{10}\)
\(x.\left(\frac{8}{13}+\frac{5}{13}\right)=\frac{1}{10}\)\
\(x.1=\frac{1}{10}\)
\(x=\frac{1}{10}\)
Vậy x=\(\frac{1}{10}\)
c)\(\frac{2}{15}+\frac{7}{15}.x=\frac{1}{15}\)
\(\frac{7}{15}x=\frac{-1}{15}\)
x=\(\frac{-1}{7}\)
Vậy \(x=\frac{-1}{7}\)
\(\frac{3}{2}\cdot x-\frac{3}{2}=\frac{1}{2}\) \(\frac{8}{13}\cdot x+\frac{5}{13}\cdot x=\frac{1}{10}\)
\(\frac{3}{2}\cdot x=\frac{3}{2}+\frac{1}{2}=2\) \(\left[\frac{8}{13}+\frac{5}{13}\right]\cdot x=\frac{1}{10}\)
\(x=2:\frac{3}{2}\) \(1\cdot x=\frac{1}{10}\)
x=\(\frac{4}{3}\) \(x=\frac{1}{10}:1\)
Vậy x=\(\frac{4}{3}\) \(x=\frac{1}{10}\)
Vậy x=\(\frac{1}{10}\)
c,\(\frac{2}{5}+\frac{7}{15}\cdot x=\frac{1}{15}\)
\(\frac{7}{15}\cdot x=\frac{1}{15}-\frac{2}{5}=\frac{-1}{3}\)
\(x=\frac{-1}{3}:\frac{7}{15}\)
\(x=\frac{-5}{7}\)
Vậy x=\(\frac{-5}{7}\)
Bài giải :
a, ( 3 - y ) x 15 = 0
3 - y = 0 : 15
3 - y = 0
y = 3 - 0
y = 3
Các ý khác tương tự
b, 13 x ( y - 10 ) = 13
y - 10 = 13 : 13
y - 10 = 1
y = 1 + 10
y = 11
\(\left(1-\frac{3}{4}\right)x\left(1-\frac{3}{7}\right)x\left(1-\frac{3}{10}\right)x\left(1-\frac{3}{13}\right)x...x\left(1-\frac{3}{97}\right)x\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}x\frac{4}{7}x\frac{7}{10}x\frac{10}{13}x...x\frac{94}{97}x\frac{97}{100}\)
\(=\frac{1}{100}\)
\(\left(1-\frac{3}{4}\right)\times\left(1-\frac{3}{7}\right)\times\left(1-\frac{3}{10}\right)...\times\left(1-\frac{3}{97}\right)\times\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}\times\frac{4}{7}\times\frac{7}{10}\times...\times\frac{94}{97}\times\frac{97}{100}\)
\(=\frac{1\times4\times7\times10\times...\times97}{1\times4\times7\times10\times...\times97\times100}\)
\(=\frac{1}{100}\)
a)\(3:\left(2x+1\right)=72-69\)
\(3:\left(2x+1\right)=3\)
\(2x+1=3:3\)
\(2x+1=1\)
\(2x=1-1\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
các bài còn lại giống như câu a nha nếu ko biết thì comment lại minhf sẽ giải cho . Nhớ k cho mình nha
#)Giải :
\(\left(1-\frac{3}{4}\right)x\left(1-\frac{3}{7}\right)x\left(1-\frac{3}{10}\right)x\left(1-\frac{1}{13}\right)x...x\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}x\frac{4}{7}x\frac{7}{10}x...x\frac{94}{97}x\frac{97}{100}\)
\(=\frac{1x4x7x...x94x100}{4x7x10x...x97x100}\)
\(=\frac{1}{100}\)
#~Will~be~Pens~#
\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\left(1-\frac{3}{10}\right)\left(1-\frac{1}{13}\right)...\left(1-\frac{1}{97}\right)\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}.\frac{4}{7}.\frac{7}{10}.\frac{10}{13}...\frac{94}{97}.\frac{97}{100}\)
\(=\frac{1}{100}\)
a) x= \(\frac{-5}{12}\)
b) x = \(\frac{2}{5}\)
c) x =\(\frac{-87}{140}\)
d) x = \(\frac{109}{140}\)
e) x = \(\frac{13}{63}\)
13 - 3x = 10
3x = 13 - 10
3x = 3
x = 3 : 3
x = 1
Vậy: ..
1