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\(\frac{7}{3}\)\(+\frac{1}{2}\)\(+\frac{-3}{70}\)\(=\frac{293}{105}\)
\(\frac{5}{12}\)\(+\frac{3}{-16}\)\(+\frac{3}{4}\)\(=\frac{47}{48}\)
\(\frac{5}{3}\)\(+\left(7+\frac{-5}{3}\right)=\frac{5}{3}\)\(+\frac{-5}{3}\)\(+7=0+7=7\)
\(\frac{-7}{31}\)\(+\left(\frac{24}{17}+\frac{7}{31}\right)=\left(\frac{-7}{31}+\frac{7}{31}\right)+\frac{24}{17}=0+\frac{24}{17}\)\(=\frac{24}{17}\)
\(\frac{3}{7}\)\(+\left(\frac{-1}{5}+\frac{-3}{7}\right)=\left(\frac{3}{7}+\frac{-3}{7}\right)+\frac{-1}{5}\)\(=0+\frac{-1}{5}\)\(=\frac{-1}{5}\)
Nếu được cho mình xin 1 k đúng ^_^
\(a,\frac{7}{25}\cdot\frac{39}{-14}\cdot\frac{50}{78}=\frac{7}{25}\cdot-\frac{39}{14}\cdot\frac{25}{39}=-\frac{7}{14}=-\frac{1}{2}\)
\(b,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}=\frac{1}{4}+\frac{1}{2}=\frac{6}{8}=\frac{3}{4}\)
a) \(5+\left(-12\right)-10=\left(-7\right)-10=-17\)
b) \(25-\left(-17\right)+24-12=42+24-12=66-12=54\)
c) \(\left(-5-3\right)\cdot\left(3-5\right)\div\left(-3+5\right)=\left(-8\right)\cdot\left(-2\right)\div2=8\)
d) \(5^6\div5^4+2^3\cdot2^2-225\div15^2=25+32-1=56\)
e) \(4\cdot\left(13-16\right)-\left(3-5\right)\cdot\left(-3\right)^2=4\cdot\left(-3\right)-\left(-2\right)\cdot9=6\)
f) \(235-\left(34+135\right)-100=235-169-100=-34\)
\(\frac{6}{x+27}=-\frac{7}{x+1}\)
\(\Rightarrow6\left(x+1\right)=-7\left(x+27\right)\)
\(6x+6=-7x+\left(-189\right)\)
\(6x+7x=-189-6\)
\(13x=195\)
\(x=195:13\)
\(x=15\)
Vậy \(x=15\)
Ta có: \(\frac{6}{x+27}=\frac{-7}{x+1}\)
\(\Leftrightarrow6\cdot\left(x+1\right)=-7\cdot\left(x+27\right)\)
\(\Leftrightarrow6x+6=-7x-189\)
\(\Leftrightarrow6x+7x=-189-6\)
\(\Leftrightarrow13x=-195\)
\(\Leftrightarrow x=-15\)
Vậy \(x=-15\)
\(\approx GOOD\)\(LUCK\approx\)
a) S=1+2+4+8+...+512
=(1+2)+(4+8)+...+(508+512)
=(3+12+....+1020) chia hết cho 3
b S=1+2+4+8+..+512
số số hạng là:
2+(512-4):4+1=2+129=131(số hạng)
tổng là :
3+(512+4):2.129=33285
Bạn vào:câu hỏi của :Vũ Ngân Hà -olm
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)
\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)
\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)