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Giúp em bài toán này với :
Bài 3: Tìm x :
b) X x \(\frac{1}{2}\)+ \(\frac{3}{2}\)x X = \(\frac{4}{5}\)
\(x.\frac{1}{2}+\frac{3}{2}.x=\frac{4}{5}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)
\(\Rightarrow x.1=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}\)
\(\left(x+\frac{3}{4}\right)\times\frac{5}{7}=\frac{10}{9}\)
\(\Rightarrow x+\frac{3}{4}=\frac{10}{9}:\frac{5}{7}=\frac{10}{9}\times\frac{7}{5}=\frac{14}{9}\)
\(\Rightarrow x=\frac{14}{9}-\frac{3}{4}=\frac{56-27}{36}=\frac{29}{36}\)
\(\left(x+\frac{3}{4}\right)\times\frac{5}{7}=\frac{10}{9}\)
\(\Leftrightarrow x+\frac{3}{4}=\frac{14}{9}\)
\(\Rightarrow x=\frac{29}{36}\)
P/s tham khảo nha
Bài làm
\(\frac{2}{3\times x}=\frac{3}{9}\)
\(\Leftrightarrow3\times x\times3=2\times9\)
\(\Leftrightarrow9\times x=18\)
\(\Leftrightarrow x=2\)
Đáp số \(x=2\)
\(\frac{1}{x}:\frac{1}{3}\times\frac{1}{4}=\frac{5}{6}\)
\(\frac{3}{x}\times\frac{1}{4}=\frac{5}{6}\)
\(\frac{3}{x}=\frac{5}{6}:\frac{1}{4}\)
\(\frac{3}{x}=\frac{10}{3}\)
\(\Leftrightarrow3.3=10.x\)
\(\Leftrightarrow9=10.x\)
\(\Leftrightarrow x=\frac{9}{10}\)
\(\frac{1}{x}:\frac{1}{3}\times\frac{1}{4}=\frac{5}{6}\)
\(\frac{1}{x}:\frac{1}{3}=\frac{10}{3}\)
\(\frac{3}{x}=\frac{10}{3}\)
\(x=3.\frac{3}{10}=\frac{9}{10}\)
3/4 * x + 1/2 * x -15 = 35
3/4 * x +1/2 * x = 35 + 15
3/4 * x +1/2 * x = 50
x * ( 3/4 + 1/2 ) = 50
x * 5/4 = 50
x = 50 : 5/4
x = 40
phan b mik ko nhap dc nen bn tu lm nha
a, \(\frac{3}{4}\times x+\frac{1}{2}\times x-15=35\)
\(x+x-15=\frac{3}{4}-\frac{1}{2}\)
\(x+x-15=\frac{2}{8}=\frac{1}{4}\)
\(x=35-15\)
\(x=20\)
Vậy \(x=\frac{1}{4}\)và \(x=20\)
b, Chịu
#)Giải :
Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{5}-\frac{1}{10}\)
\(A=\frac{1}{10}\)
\(x\times\frac{1}{2}+\frac{3}{2}\times x=\frac{4}{5}\)
\(x\times\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)
\(x\times2=\frac{4}{5}\)
\(x=\frac{4}{5}\div2\)
\(x=\frac{4}{10}\)