(0,37-0,3+3/11+3/12)/(0,625-0,5+5/11+5/12)
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0,275-0,3+3/11+3/12-0,625+0,5-5/11-5/12
=11/14-3/10+3/11+3/12-5/8+1/2-5/11-5/12
=(11/14+1/2)-3/10-5/8+(3/11+3/12)+(-5/11-5/12)
=9/7-37/40+[3(1/11+1/12)]+[5(-1/11-1/12)]
=1/280+3+5
=1/280+8
=2441/280
=
\(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{-5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}=\dfrac{-3}{5}\)
\(M=0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}\)
\(M=\left(0,375-0,625\right)+\left(0,5-0,3\right)+\left(\dfrac{3}{11}-\dfrac{5}{11}\right)+\left(\dfrac{3}{12}-\dfrac{5}{12}\right)\)
\(M=-0,25+0,2-\dfrac{2}{11}-\dfrac{1}{6}\)
\(M=-\dfrac{263}{660}\)
b) Ta có : \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)và a2 - b2 + 2c2 = 108
⇒ \(\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}=\frac{a^2-b^2+2c^2}{4-9+32}=\frac{108}{27}=4\)
⇒ a2 = 4.4 =16 ⇔ a = 4 hoặc -4
b2 = 4.9 = 36 ⇔ b= 6 hoặc -6
2c2 = 4 .32 ⇔ c2 = 64 ⇔ c = 8 hoặc -8
Vậy các cặp ( a ; b ; c ) thỏa mãn là : ( 4; 6; 8 ) ; ( -4 ; -6 ; -8 )
\(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\\ A=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\\ A=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}=\dfrac{3\left(0,125-0,1+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(0,125-0,1+\dfrac{5}{11}+\dfrac{5}{12}\right)}+\dfrac{\dfrac{3}{5}\left(2,5+\dfrac{5}{3}-1,25\right)}{2,5+\dfrac{5}{3}-1,25}=-\dfrac{3}{5}+\dfrac{3}{5}=0\)
A=\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625+0,5-\frac{5}{11}-\frac{5}{12}}\)
A=\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
A=\(\frac{3\cdot\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\cdot\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
A=\(\frac{-3}{5}\)
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\(=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)