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thay b=2/3, ta có: A=1/3x(2/3-2/5):7/5=4/63
thay A= 3/7, ta có: 3/7=1/3x(b-2/5):7/5
1/3x(b-2/5):7/5=3/7
1/3x(b-2/5) =3/7x7/5
1/3x(b-2/5) = 3/5
b-2/5 = 3/5:1/3
b-2/5 = 9/5
b = 9/5+2/5
b = 11/5
Đặt \(A=25\times4-0,5\times40\times5\times0,2\times20\times0,25\)
\(\Leftrightarrow A=100-\left(0,5\times0,2\right)\times\left(40\times0,25\right)\times\left(5\times20\right)\)
\(\Leftrightarrow A=100-0,1\times10\times100\)
\(\Leftrightarrow A=100-100=0\)
Vậy theo đề bài, ta có :
\(\frac{25\times4-0,5\times40\times5\times0,2\times20\times0,25}{1+2+8+...+128+256}=\frac{0}{1+2+8+...+128+256}=0\)
Vậy ..................
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\(\frac{25\times4-0,5\times40\times5\times0,2\times20\times0,25}{1+2+4+8+...+128+256}\)
\(=\frac{100-\left(20\times5\right)\times\left(0,5\times0,2\times0,25\times40\right)}{1+2+4+8+...+128+256}\)
\(=\frac{100-100\times\left(\frac{1}{40}\times40\right)}{1+2+4+8+...+128+256}\)
\(=\frac{100\times\left(1-1\right)}{1+2+4+8+...+128+256}\)
\(=0\)
\(\frac{25\times4-0,5\times40\times5\times0,2\times20\times0,25}{1+2+4+8+...128+256}\)
\(=\frac{100-\left(0,5\times20\right)\times\left(40\times0,25\right)\times\left(5\times0,2\right)}{1+2+4+8+...+128+256}\)
\(=\frac{100-10\times10\times1}{1+2+4+8+...+128+256}\)
\(=\frac{100-100}{1+2+4+8+...+128+256}\)
\(=\frac{0}{1+2+4+8+...+128+256}=0\)
a) \(=\frac{1}{6}x\frac{5}{2}+\frac{3}{7}\)
\(=\frac{5}{12}+\frac{3}{7}\)
\(=\frac{71}{84}\)
b) \(=\frac{8}{5}-\frac{1}{2}x\frac{1}{3}\)
\(=\frac{8}{5}-\frac{1}{6}\)
\(=\frac{43}{30}\)
a, \(\frac{1}{6}:\frac{2}{5}+\frac{3}{7}\)
\(=\frac{1}{6}\cdot\frac{5}{2}+\frac{3}{7}\)
\(=\frac{5}{12}+\frac{3}{7}\)
\(=\frac{71}{84}\)
b, \(\frac{8}{5}-\frac{1}{2}\cdot\frac{1}{3}\)
\(=\frac{8}{5}-\frac{1}{6}\)
\(=\frac{43}{30}\)