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\(\text{ Bài giải}\)
\(\text{ Ta có :}\)
\(68< x< 92\) \(\text{ Mà }x\text{ là một số tròn chục}\)
\(\Rightarrow\text{ }x=70\text{ ; }80\text{ ; }90\)
\(\text{ Vậy }x=70\text{ ; }80\text{ ; }90\)
a) \(\frac{2}{5}+\frac{9}{15}=\frac{2}{5}+\frac{3}{5}=\frac{5}{5}=1\)
b) \(\frac{15}{45}+\frac{25}{30}=\frac{1}{3}+\frac{5}{6}=\frac{2}{6}+\frac{5}{6}=\frac{7}{6}\)
c) \(\frac{28}{32}+\frac{45}{72}=\frac{7}{8}+\frac{5}{8}=\frac{12}{8}=\frac{3}{2}\)
d) \(\frac{8}{28}+\frac{5}{30}=\frac{2}{7}+\frac{1}{6}=\frac{12}{42}+\frac{7}{42}=\frac{19}{42}\)
a) \(\frac{2}{5}+\frac{9}{15}\)=\(\frac{2}{5}+\frac{3}{5}=\frac{2+3}{5}=1\)
b)\(\frac{15}{45}+\frac{25}{30}=\frac{1}{3}+\frac{5}{6}=\frac{2}{6}+\frac{5}{6}=\frac{2+5}{6}=\frac{7}{6}\)
c)\(\frac{28}{32}+\frac{45}{72}=\frac{7}{8}+\frac{5}{8}=\frac{7+5}{8}=\frac{12}{8}=\frac{3}{2}\)
d)\(\frac{8}{28}+\frac{5}{30}=\frac{2}{7}+\frac{1}{6}=\frac{12}{42}+\frac{7}{42}=\frac{12+7}{42}=\frac{19}{42}\)
(a+b) x 2=88 => a+b=44
(b+c) x 2= 148 => b+c=74
(a+c) x 2=108 => a+c=54
(c+d) x 2=208 => c+d=104
(b+d) x 2=188 => b+d=94
(a+d) x 2=148 => a+d= 74
=> a+b+b+c+a+c+c+d+b+d+a+d = 44+74+54+104+94+74
=> 3 x a + 3 x b + 3 x c + 3 x d = 444
3 x (a+b+c+d) = 444
a+b+c+d = 444 : 3
a+b+c+d = 148
- a B,b D
- a \(\frac{1}{2}\)b \(\frac{2}{5}\)
- \(\frac{2}{3};\frac{10}{17};\frac{5}{11};\frac{4}{9}\)
- a\(\frac{5}{12}\)b\(\frac{97}{36}\)
a)\(\frac{7\cdot15:25\cdot102}{7\cdot4\cdot15\cdot3:25\cdot4\cdot102\cdot3}=\frac{1}{4\cdot3:4\cdot3}=\frac{1}{1}=1\)
b)\(\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}=\frac{2+3+4+5+6+7+8}{10}=\frac{44}{10}=\frac{22}{5}\)
Đáp án C