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a/ \(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{32}{17}+\frac{14}{21}=\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)-\frac{32}{17}=1+1-\frac{32}{17}=\frac{2}{17}\)
\(\left(-2\right)^3\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
\(=-8.\frac{1}{2}:\left(\frac{9}{4}-\frac{7}{6}\right)\)
\(=-4.\frac{12}{13}\)
\(=-\frac{48}{13}\)
\(=\left(-8\right).\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)
\(=\left(-8\right).\frac{1}{2}:\frac{13}{12}\)
\(=\left(-4\right):\frac{13}{12}\)
\(=-\frac{48}{13}\)
a)\(23\frac{1}{4}\cdot\frac{7}{5}+13\frac{1}{4}\)/\(\frac{-5}{7}\)
=\(23\frac{1}{4}\cdot\frac{7}{5}+13\frac{1}{4}\cdot\frac{-7}{5}\)
=\(23\frac{1}{4}\cdot\frac{7}{5}+\left(-13\frac{1}{4}\right)\cdot\frac{7}{5}\)
=\(\frac{7}{5}\cdot\left(23\frac{1}{4}-13\frac{1}{4}\right)\)
=\(\frac{7}{5}\cdot\left(23-13+\frac{1}{4}-\frac{1}{4}\right)\)
=\(\frac{7}{5}\cdot10\)
=\(14\)
b)\(\left(-3\right)^2\cdot\left(\frac{3}{4}-0.25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)
=\(9\cdot\left(\frac{3}{4}-\frac{1}{4}\right)-\left(3-1+\frac{1}{2}-\frac{1}{2}\right)\)
=\(9\cdot\frac{1}{2}-2\)
=\(\frac{9}{2}-2\)
=\(\frac{9}{2}-\frac{4}{2}\)
=\(\frac{5}{2}\)
Nếu mình còn sai hay thiếu chỗ nào thì ae bảo mình nha :))
1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
A) \(A=\left(-\frac{3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(-\frac{1}{4}+\frac{1}{3}\right):\frac{5}{11}\)
\(A=-11.\frac{1}{12}:5+\frac{1}{3}-\frac{1}{4}:\frac{5}{11}\)
\(A=-\frac{11.\frac{1}{12}}{5}+\frac{11.\frac{1}{12}}{5}\)
\(\Rightarrow A=0\)
b) \(B=\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)
\(B=\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-\left(\frac{7}{2}-\frac{4}{2}\right)\)
\(B=\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-2\)
\(B=3^2.\left(\frac{3}{4}-0,25\right)-2\)
\(B=4,5-2\)
\(\Rightarrow B=2\)
Lộn nha :v ở phần b) ấy, bạn sửa 4,5 - 2 = 2 thành 4,5 - 2 = 2,5 hộ mình nha
1, \(\left(xy\right)^2-\frac{1}{2}x^2y^2+3xy^2.\left(-\frac{1}{3}x\right)\)
\(=x^2y^2-\frac{1}{2}x^2y^2-x^2y^2\)
\(=-\frac{1}{2}x^2y^2\)
2, \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
\(=x^2+\frac{3}{2}x^2+\frac{1}{3}x^2\)
\(=\frac{17}{6}x^2\)
3, \(-4.\left(2x\right)^2y^3+\frac{1}{2}xy.\left(-2xy^2\right)+\frac{1}{4}x^2y^3\)
\(=-16x^2y^3-x^2y^3+\frac{1}{4}x^2y^3\)
\(=-\frac{67}{4}x^2y^3\)
4, \(\frac{1}{3}x^4y-\frac{5}{3}x^3.\left(\frac{5}{2}xy\right)+\frac{3}{4}x^4y\)
\(=\frac{1}{3}x^4y-\frac{25}{6}x^4y+\frac{3}{5}x^4y\)
\(=-\frac{97}{30}x^4y\)
5, \(\left(-2x^3y^4\right)^2-5x^2y.\left(\frac{3}{4}x^4y^7\right)-\frac{2}{3}x^6y^8\)
\(=4x^6y^8-\frac{15}{4}x^6y^8-\frac{2}{3}x^6y^8\)
\(=-\frac{5}{12}x^6y^8\)
1.
\((\frac{1}{3}xy)^2.x^3+\frac{3}{2}(2x)^3(-\frac{7}{4}x^2y^2)-\frac{2}{3}x^5y^2\)
\(=(\frac{1}{9}x^2y^2)x^3+\frac{3}{2}(8x^3)(-\frac{7}{4}x^2y^2)-\frac{2}{3}x^5y^2\)
\(=\frac{1}{9}(x^2.x^3)y^2+(\frac{3}{2}.8.\frac{-7}{4})(x^3.x^2).y^2-\frac{2}{3}x^5y^2\)
\(=\frac{1}{9}x^5y^2-21x^5y^2-\frac{2}{3}x^5y^2=\frac{-194}{9}x^5y^2\)
2.
\(\frac{-2}{5}x^2y(-y^6)+\frac{3}{2}xy(\frac{-1}{15}xy^6)+(-2xy)^2y^5\)
\(=\frac{2}{5}x^2(y.y^6)+(\frac{3}{2}.\frac{-1}{15})(x.x).(y.y^6)+4x^2(y^2.y^5)\)
\(=\frac{2}{5}x^2y^7-\frac{1}{10}x^2y^7+4x^2y^7=\frac{43}{10}x^2y^7\)
3.
\(\frac{3}{7}xy^2z+\frac{1}{2}x^3y^2+\frac{1}{3}x^3y^2-\frac{3}{7}xy^2z\)
\(=(\frac{3}{7}xy^2z-\frac{3}{7}xy^2z)+(\frac{1}{2}x^3y^2+\frac{1}{3}x^3y^2)\)
\(=\frac{5}{6}x^3y^2\)
4.
\(\frac{2}{3}xy^2-\frac{5}{2}yz+\frac{1}{2}xy^2-\frac{2}{3}yz\)
\(=(\frac{2}{3}xy^2+\frac{1}{2}xy^2)-(\frac{5}{2}yz+\frac{2}{3}yz)\)
\(=\frac{7}{6}xy^2+\frac{19}{6}yz\)
5.
\(\frac{3}{2}xy^2z^5-\frac{5}{4}xyz^2+\frac{4}{3}xy^2z^5+\frac{1}{2}xyz^2\)
\(=(\frac{3}{2}xy^2z^5+\frac{4}{3}xy^2z^5)+(\frac{-5}{4}xyz^2+\frac{1}{2}xyz^2)\)
\(=\frac{17}{6}xy^2z^5-\frac{3}{4}xyz^2\)
\(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
\(=\left(-8\right).\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)
\(=\left(-8\right).\frac{1}{2}:\frac{13}{12}\)
\(=\left(-4\right).\frac{12}{13}\)
\(=\frac{-48}{13}\)