Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{3.6+2.9.5+18.\left(-4\right)}{7.\left(-7\right)+12.\left(-7\right)+7}\)=\(\frac{18+18.5+18.\left(-4\right)}{-7.\left(7+12-1\right)}\)=\(\frac{18.\left(1+5-4\right)}{-7.18}\)=\(\frac{18.2}{18.\left(-7\right)}=\frac{-2}{7}\)
b) \(\frac{2.3.4.5.6-3.4.5.6.7}{2.4.6-4.6.8}=\frac{3.4.5.6.\left(2-7\right)}{4.6.\left(2-8\right)}\)=\(\frac{3.5.\left(-5\right)}{\left(-6\right)}=\frac{-25}{-2}=\frac{25}{2}\)
a)\(\frac{3.6+2.9.5+18.\left(-4\right)}{7.\left(-7\right)+12.\left(-7\right)+7}\)
\(=\frac{18+18.5+18.\left(-4\right)}{7.\left(-7\right)+\left(-12\right).7+7}\)
\(=\frac{18.\left(1+5+-4\right)}{7.\left(-7+-12+1\right)}\)
\(=\frac{15.2}{7.\left(-18\right)}\)
\(=\frac{3.5.2}{7.2.3.-3}\)
\(=\frac{5}{-21}\)
b) \(\frac{2.3.4.5.6-3.4.5.6.7}{2.4.6-4.6.8}\)
\(=\frac{3.4.5.6.\left(2-7\right)}{4.6.\left(2-8\right)}\)
\(=\frac{3.4.5.6.\left(-5\right)}{4.6.\left(-3\right)}\)
\(=\frac{3.5.\left(-5\right)}{\left(-3\right)}\)
\(=\frac{-75}{\left(-3\right)}\)
\(=25\)
Ta có : |x + 5| - (x + 5) = 0
<=> |x + 5| = (x + 5)
<=> x + 5 = x + 5 ( x bằng bất kì)
-x + 5 = x + 5
<=> -x - x = 5 - 5
=> -2x = 0
=> x = 0
ai có thể trả lời cho mình phần b ko rồi mình sẽ k
a/b= (1+1/6) + (1/2+1/5) + (1/3+1/4)
a/b= 7/6 + 7/10 + 7/12
a/b= 7(1/6+1/10+1/12)
Vì 6x10x12 khong la boi so cua 7 => a/b chia het cho 7 <=> a chia het cho 7 (dpcm)
A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
= \(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{30}{31}\)
=\(\frac{-671}{4805}\)