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`a, 20x^3y^5 : 5x^2y^2`

`= (20:5)x^(3-2) . y^(5-2)`

`= 4xy^3`

`b, 18x^3y^5 : (3(-x^3)y^2)`

`= -(18:3)y^(5-3)`

`= -6y^2`

`a, (4x^3y^2 - 8x^2y + 10xy) : 2xy`

`= 2x^2y - 4x + 5`.

`b, 7x^4y^2 - 2x^2y^2 - 5x^3y^4 : 3x^2y`

`= 7/3 x^2y - 3/2y - 5/3xy^3`

`a, (5ab - 2a^2):a`

`= 5b - 2a`

`b, (6x^2y^2-xy^2+3x^2y) : (-3xy)`

`= -2xy + y/3 - x`.

7 tháng 10 2021

\(a,=5^3:5^2=5\\ b,=\left(\dfrac{3}{4}\right)^{5-3}=\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\\ c,=1728-512=1216\\ d,=x^{10}:x^8=x^2\\ e,=\left(-x\right)^{5-3}=\left(-x\right)^2=x^2\\ f,=\left(-y\right)^{5-4}=-y\)

22 tháng 7 2023

a) \(\left(-5a^4\right)\cdot\left(a^2b-ab^2\right)\)

\(=\left(-5a^4\cdot a^2b\right)-\left(-5a^4\cdot ab^2\right)\)

\(=-5a^6b+5a^5b^2\)

b) \(\left(x+2y\right)\left(xy^2-2y^3\right)\)

\(=x^2y^2-2xy^3+2xy^3-4y^4\)

\(=x^2y^2-4y^4\)

`a, (-5a^4)(a^2b - ab^2)`

`= -5(a^(4+2) . b) + 5a^(4+1) . b^2`

`= -5a^6b + 5a^5b^2`

`b, (x+2y)(xy^2-2y^3)`

`= x^2y^2 + 2xy^3 - 2xy^3 - 4y^4`

29 tháng 9 2019

a) =(x-y)5+(x-y)3=(x-y)3[(x-y)2+1]

b) =33(y-2x)3:-9(y-2x)=-3(y-2x)2

c) =(x-y)2 [3(x-y)3-2(x-y)2+3]:5(x-y)2=[3(x-y)3-2(x-y)2+3]/5

23 tháng 7 2023

a) \(18x^4y^3:12\left(-x\right)^3y\)

\(=\left(18:-12\right)\left(x^4:x^3\right)\left(y^3:y\right)\)

\(=-\dfrac{3}{2}xy^2\)

b) \(x^2y^2-2xy^3:\dfrac{1}{2}xy^2\)

\(=\dfrac{xy^2\left(x-2y\right)}{\dfrac{1}{2}xy^2}\)

\(=\dfrac{x-2y}{\dfrac{1}{2}}\)

\(=2x-4y\)

`a)`

`3x(2xy - 5x^2y)`

`= 3x*2xy + 3x* (-5x^2y)`

`= 6x^2y - 15x^3y`

`b)`

`2x^2y (xy - 4xy^2 + 7y)`

`= 2x^2y * xy + 2x^2y * (-4xy^2) + 2x^2y * 7y`

`= 2x^3y^2 - 8x^3y^3 + 14x^2y^2`

`c)`

`(-2/3xy^2 + 6yz^2)*(-1/2xy)`

`= (-2/3xy^2)*(-1/2xy) + 6yz^2 * (-1/2xy)`

`= 1/3x^2y^3 - 3xy^2z^2`

`a, 3x(2xy-5x^2y)`

`= 6x^2y - 15x^3y`

`b, 2x^2y(xy-4xy^2+7y)`

`= 2x^3y^2 - 8x^3y^3 + 14x^2y^2`

`c, (-2/3xy^2 + 6yz^2).(-1/2xy)`

`= 1/3x^2y^3 - 3xy^2z^2`

8 tháng 7 2023

\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)

\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)

\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)

\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)

3 tháng 11 2018

\(=\frac{15\left(x-y\right)^5}{5\left(x-y\right)^3}-\frac{10\left(x-y\right)^4}{5\left(x-y\right)^3}+\frac{20\left(x-y\right)^3}{5\left(x-y\right)^3}\)

\(=3\left(x-y\right)^5-2\left(x-y\right)^4+4\left(x-y\right)^3\)

3 tháng 11 2018

sửa dòng cuối

\(=3\left(x-y\right)^2-2\left(x-y\right)+4\)