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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) \(\frac{2.3.4.5.6-3.4.5.6.7}{2.4.6-4.6.8}\)
\(=\frac{3.4.5.6.\left(2-7\right)}{4.6.\left(2-8\right)}\)
\(=\frac{3.4.5.6.\left(-5\right)}{4.6.\left(-3\right)}\)
\(=\frac{3.5.\left(-5\right)}{\left(-3\right)}\)
\(=\frac{-75}{\left(-3\right)}\)
\(=25\)
to noi that: nguoi ta chi thich giup lam bai kho thoi chu bai nay ai di hoc cung lam dc
tai sao ban phai hoi
a) − 7 .3 + 4. − 6 − 5 .3 + 2.3 = − 45 = 9 = 6
b) 3.6 + 2.9.5 − 18. − 4 7. − 7 + 12. − 7 + 7 = 18. 1 + 5 + 4 − 7 . 7 + 12 − 1 = 18.10 − 7.18 = − 10 7
\(\frac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right)\left(2^4-4^2\right)}{15^2+5^3+67^7}=\frac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right).0}{15^2+5^3+67^7}=0\)
a, 26/x + 3 nguyên
=> 26 ⋮ x + 3
=> x + 3 thuộc Ư(26)
=> x + 3 thuộc {-1; 1; -2; 2; -13; 13; -26; 26}
=> x thuộc {-4; -2; -5; -1; -16; 10; -29; 23}
vậy_
b, x+6/x+1 nguyên
=> x + 6 ⋮ x + 1
=> x + 1 + 5 ⋮ x + 1
=> 5 ⋮ x + 1
=> x + 1 thuộc Ư(5)
=> x + 1 thuộc {-1; 1; -5; 5}
=> x thuộc {-2; 0; -6; 4}
vậy_
c, x-2/x+3 nguyên
=> x - 2 ⋮ x + 3
=> x + 3 - 5 ⋮ x + 3
=> 5 ⋮ x + 3
=> x + 3 thuộc Ư(5)
=> x + 3 thuộc {-1; 1; -5; 5}
=> x thuộc {-4; -2; -8; 2}
vậy_
\(a,\frac{26}{x+3}\in Z\Leftrightarrow26\)\(⋮\)\(x+3\)\(\Rightarrow x+3\inƯ_{26}\)
Mà \(Ư_{26}=\left\{\pm1;\pm2;\pm13;\pm26\right\}\)\(\Rightarrow...\)
\(b,\frac{x+6}{x+1}=\frac{x+1+5}{x+1}=1+\frac{5}{x+1}\)
\(\frac{5}{x+1}\in Z\Leftrightarrow5\)\(⋮\)\(x+1\Rightarrow x+1\inƯ_5\)
MÀ \(Ư_5=\left\{\pm1;\pm5\right\}\)\(\Rightarrow...\)
\(c,\frac{x-2}{x+3}=\frac{x+3-3-2}{x+3}=1-\frac{5}{x+3}\)
\(\frac{5}{x+3}\in Z\Leftrightarrow\)\(5\)\(⋮\)\(x+3\Rightarrow x+3\inƯ_5\)
Mà \(Ư_5=\left\{\pm1;\pm5\right\}\)\(\Rightarrow...\)
\(\frac{3.6+2.9.5-18.\left(-4\right)}{7.\left(-7\right)+12.\left(-7\right)+7}=\frac{18+18.5-18.\left(-4\right)}{\left(-7\right).\left(7+12\right)+7}\)
\(=\frac{18\left[1+5-\left(-4\right)\right]}{\left(-7\right).19+7}=\frac{18.10}{7.\left(-19\right)+7}\)
\(=\frac{180}{7\left[\left(-19\right)+1\right]}=\frac{180}{-126}=\frac{=10}{7}\)