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`#3107.101107`
a)
\(x+x+\dfrac{1}{2}\times\dfrac{2}{5}+x+\dfrac{8}{10}=121\\3x+\dfrac{1}{5}+\dfrac{4}{5}=121\\ 3x+1=121\\ 3x=121-1\\ 3x=120\\ x=40 \)
Vậy, `x = 40`
b)
\(\dfrac{12+x}{42}=\dfrac{5}{6}\\ \dfrac{12+x}{42}=\dfrac{35}{42}\\ \dfrac{12+x}{42}-\dfrac{35}{42}=0\\ \dfrac{12+x-35}{42}=0\\ \dfrac{x-\left(35-12\right)}{42}=0\\ \dfrac{x-23}{42}=0\\ x-23=0\\ x=23\)
Vậy,` x = 23.`
a: \(x+x+\dfrac{1}{2}\cdot\dfrac{2}{5}+x+\dfrac{8}{10}=121\)
=>\(3x+\dfrac{1}{5}+\dfrac{4}{5}=121\)
=>3x+1=121
=>3x=120
=>x=40
b: \(\dfrac{x+12}{42}=\dfrac{5}{6}\)
=>\(x+12=42\cdot\dfrac{5}{6}=35\)
=>x=35-12=23
\(3\dfrac{7}{8}\times x-2\dfrac{3}{4}=3\dfrac{6}{12}\times\dfrac{10}{8}-\dfrac{1}{3}\)
\(\dfrac{31}{8}\times x-\dfrac{11}{4}=\dfrac{7}{2}\times\dfrac{10}{8}-\dfrac{1}{3}\)
\(\dfrac{31}{8}\times x-\dfrac{11}{4}=\dfrac{35}{8}-\dfrac{1}{3}\)
\(\dfrac{31}{8}\times x=\dfrac{97}{24}+\dfrac{11}{4}\)
\(\dfrac{31}{8}\times x=6\dfrac{19}{24}\)
\(x=1\dfrac{70}{93}\)
42/12x10/8-1/3= 84/24x30/24-8/24= 315/72-8/24= 291/72
xong bạn rút gọn nha
1)\(y\times7:5+4\times8=134\)
\(\Leftrightarrow y\times7:5+32=134\)
\(\Leftrightarrow y\times7:5=102\)
\(\Leftrightarrow y\times7=510\)
\(\Leftrightarrow y=72,86\)
2) \(\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+\dfrac{1}{2}:0,5-\dfrac{1}{10}\)
\(=0,25:0,25-0,125:0,125+0,5:0,5-\dfrac{1}{10}\)
\(=1-1+1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
Lời giải:
$8\frac{8}{10}< x< 8\frac{2}{10}$
$\Rightarrow 8,8< x< 8,2$ (vô lý vì $8,8> 8,2$)
Bạn xem lại đề.