\(\frac{0,25xq+0,5}{2}=0,25\)
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a)
\(\begin{array}{l}\left( {0,25 - \frac{5}{6}} \right).1,6 + \frac{{ - 1}}{3}\\ =(\frac{25}{100}-\frac{5}{6}).\frac{16}{10}+\frac{-1}{3}\\= \left( {\frac{1}{4} - \frac{5}{6}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \left( {\frac{6}{{24}} - \frac{{20}}{{24}}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{24}}.\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 5}}{{15}}\\ = \frac{{ - 19}}{{15}}\end{array}\)
b)
\(\begin{array}{l}3 - 2.\left[ {0,5 + \left( {0,25 - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left[ {\frac{1}{2} + \left( {\frac{1}{4} - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left( {\frac{1}{2} + \frac{1}{{12}}} \right)\\ =3-2.(\frac{6}{12}+\frac{1}{12})\\= 3 - 2.\frac{7}{{12}}\\ = 3 - \frac{7}{6}\\=\frac{18}{6}-\frac{7}{6}\\ = \frac{{11}}{6}\end{array}\)
#)Giải :
a) \(\frac{2xX-4,36}{0,125}=\)0,25 x 42,9 - 11,7 x 0,25 + 0,25 x 0,8
\(\frac{2xX-4,36}{0,125}=\)0,25 x ( 42,9 - 11,7 + 0,8 )
\(\frac{2xX-4,36}{0,125}=\)0,25 x 32
\(\frac{2xX-4,36}{0,125}=8\)
\(2xX-4,36=8x0,125\)
\(2xX-4,36=1\)
\(2xX=1+4,36\)
\(2xX=5,36\)
\(X=5,36:2\)
\(X=2,68\)
#~Will~be~Pens~#
Lời giải:
Gọi phân số trên là $A$. Ta có:
$A=\frac{12,48:6,48\times 2\times 8\times 6,25\times 3,25\times 4\times 4\times 2}{3,12\times 21,6\times 1,25\times 2,6\times 4\times 4\times 10\times 10}$
$=\frac{12,48:6,48\times (2\times 6,25)\times (8\times 3,25)\times 2}{3,12\times 21,6\times (1,25\times 10)\times (2,6\times 10)}$
$=\frac{12,48:6,48\times 12,5\times 26\times 2}{3,12\times 21,6\times 12,5\times 26}$
$=\frac{12,48:6,48\times 2}{3,12\times 21,6}=\frac{125}{2187}$
\(=\frac{\frac{1}{3}-\frac{1}{4}+1}{\frac{1}{2}-\frac{3}{7}}\)
\(=\frac{\frac{4}{12}-\frac{3}{12}+\frac{12}{12}}{\frac{7}{14}-\frac{6}{14}}\)
\(=\frac{\frac{13}{12}}{\frac{1}{14}}\)
\(=\frac{13}{12}\times\frac{14}{1}\)
\(=\frac{13}{6}\times7\)
\(=\frac{91}{6}\)
\(C=\left(0,5\right)^{-4}-625^{0,25}-\left(2\frac{1}{4}\right)^{-1\frac{1}{2}}+19\left(-3\right)^{-3}=\left(2^{-1}\right)^{-4}-\left(5^4\right)^{\frac{1}{4}}-\left[\left(\frac{3}{2}\right)^2\right]^{-\frac{3}{2}}+19.\frac{1}{\left(-3\right)^3}\)
\(=2^4-5-\left(\frac{3}{2}\right)^{-3}-\frac{19}{27}\)
\(=11-\left(\frac{2}{3}\right)^3-\frac{19}{27}=10\)
\(C=\left(0,5\right)^{-4}-625^{0,25}-\left(2\frac{1}{4}\right)^{-1\frac{1}{2}}+19.\left(-3\right)^{-3}\)
\(=\left(\frac{1}{2}\right)^{-4}-625^{\frac{1}{4}}-\left(\frac{9}{4}\right)^{-\frac{3}{2}}+19.\left(-3\right)^{-3}\)
\(=2^4-\sqrt[4]{625}-\left(\frac{4}{9}\right)^{\frac{3}{2}}+19.\left(\frac{1}{\left(-3\right)^3}\right)\)
=\(16-5-\sqrt[2]{\left(\frac{4}{9}\right)^3}+19.\frac{1}{-27}=11-\frac{8}{27}-\frac{19}{27}=10\)
Đề chắc thế này đúng ko bạn
\(\frac{0,25\cdot q+0,5}{2}=0,25\)
=> 0,25 . q + 0,5 = 0,5
=> 0,25 . q = 0
=> q = 0
Vậy,........