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\(a,\frac{5}{6}+x\)x \(\frac{3}{4}=3\)
\(x\)x \(\frac{3}{4}=3-\frac{5}{6}\)
\(x\)x \(\frac{3}{4}=\frac{13}{6}\)
\(x=\frac{13}{6}:\frac{3}{4}\)
\(x=\frac{26}{9}\)
b, \(5\frac{1}{6}-x:\frac{3}{4}=\frac{2}{3}\)
\(\frac{31}{6}-x:\frac{3}{4}=\frac{2}{3}\)
\(x:\frac{3}{4}=\frac{31}{6}-\frac{2}{3}\)
\(x:\frac{3}{4}=\frac{9}{2}\)
\(x=\frac{9}{2}\)x \(\frac{3}{4}\)
\(x=\frac{27}{8}\)
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\(a.\frac{4}{3}-\frac{3}{2}:X=\frac{1}{6}\)
\(\frac{3}{2}:X=\frac{4}{3}-\frac{1}{6}\)
\(\frac{3}{2}:X=\frac{8}{6}-\frac{1}{6}\)
\(\frac{3}{2}:X=\frac{7}{6}\)
\(X=\frac{3}{2}:\frac{7}{6}\)
\(X=\frac{3}{2}\times\frac{6}{7}\)
\(X=\frac{9}{7}\)
\(b.\left(X+\frac{2}{3}\right):\frac{1}{3}=\frac{41}{3}\)
\(X-\frac{2}{3}=\frac{41}{3}.\frac{1}{3}\)
\(X-\frac{2}{3}=\frac{41}{9}\)
\(X=\frac{41}{9}+\frac{2}{3}\)
\(X=\frac{41}{9}+\frac{6}{9}\)
\(X=\frac{47}{9}\)
\(\frac{4}{13}+\frac{9}{5}=\frac{137}{65}\)
\(3-\frac{2}{16}=\frac{23}{8}\)
\(\frac{12}{31}\times\frac{1}{19}=\frac{12}{589}\)
\(\frac{3}{8}\div\frac{9}{12}=\frac{1}{2}\)
ab . 9 = a0b
( 10a + b ) x 9 = 100a + b
90a + 9b = 100a + b
8b = 10a
=> a/b = 4/5 = 8/10
=> \(\orbr{\begin{cases}a=4;b=5\\a=8;b=10\end{cases}}\)
Ta có:\(\overline{ab}\cdot9=\overline{a0b}\)
\(\left(10a+b\right)\cdot9=100a+b\)
\(90a+9b=100a+b\)
\(8b=10a\)(giảm mỗi bên 90a+b)
\(4b=5a\)
Mà chỉ có a=4;b=5 là thỏa mãn
Vậy a=4;b=5
\(\left(2.8x-32\right):\frac{2}{3}=90\)
\(2.8\cdot x-32=90\cdot\frac{2}{3}\)
\(\frac{14}{5}x-32=60\)
\(\frac{14}{5}x=60+32\)
\(\frac{14}{5}x=92\)
\(x=\frac{230}{7}\)
B , c , d tương tự
Ko hiểu đề bài cho lắm.
a11 a - x3 = [ - b4 ] b