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\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)
\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)
\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)
\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)
\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)
\(=\dfrac{-1}{4}x^3y^3z^2\)
Đa thức chỉ dương khi xy<0
\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)
\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)
a) (31 6/13 + 5 9/41) - 36 6/13
= 409/13 + 214/41 - 474/13
= (409/13 - 474/13) + 214/41
= -5 + 214/41
= 9/41
b) 5/3 + (-2/7) - (-1,2)
= 5/3 - 2/7 + 6/5
= 29/21 + 6/5
= 271/105
c) 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)
= 1/4 + 3/5 - 1/8 + 2/5 - 5/4
= (1/4 - 5/4) + (3/5 + 2/5) - 1/8
= -1 + 1 - 1/8
= -1/8
A= \(\dfrac{11}{9}\).\(\dfrac{-3}{2}\)- \(\dfrac{11}{9}\).\(\dfrac{15}{2}\)+(2021)0
= \(\dfrac{11}{9}\)(\(\dfrac{-3}{2}\)-\(\dfrac{15}{2}\)) + 1
= \(\dfrac{11}{9}\)(-9) + 1
= -11 + 1
= -10
Lần sau bn ko đc phép chuyển \(\left(-2021\right)^0\) sang \(\left(2021\right)^0\) đou nha
a: Sửa đề: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
Để A là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3+4⋮\sqrt{x}-3\)
=>\(4⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7\right\}\)
=>\(x\in\left\{16;4;25;1;49\right\}\)
b:
được
\(\dfrac{1}{2}\)\(x^2\) = \(\dfrac{1}{4}x^2\) + \(\dfrac{1}{4}\)\(x^2\)