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NV
6 tháng 10 2021

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

\(\left(y-x\right)^2=\left(x-y\right)^2\)

\(\Rightarrow\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]:\left(y-x\right)^2=5\left(x-y\right)^2-3\left(x-y\right)+4\)

b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)

\(\Leftrightarrow-4x+3+5x+2=0\)

\(\Leftrightarrow x=-5\)

7 tháng 10 2021

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

5 tháng 10 2017

a)\(\left(x+y\right)^2:\left(x+y\right)=\left(x+y\right)^{2-1}=x+y\)

b)\(\left(x-y\right)^5:\left(y-x\right)^4=\left(x-y\right)^5:\left(-\left(x-y\right)^4\right)=-\left(x-y\right)^{5-4}=-\left(x-y\right)\)

c)\(\left(x-y+z\right)^4:\left(x-y+z\right)^3=\left(x-y+z\right)^{4-3}=x-y+z\)

8 tháng 10 2017

a) (x+y)^2:(x+y)=x+y

b) (x−y)^5:(y−x)^4=(x-y)^5:[-(x-y)]^4=x-y

c) (x−y+z)^4:(x−y+z)^3=x-y+z

20 tháng 4 2017

Bài giải:

[3(x – y)4 + 2(x – y)3 – 5(x – y)2] : (y – x)2

= [3(x – y)4 + 2(x – y)3 – 5(x – y)2] : [-(x – y)]2

= [3(x – y)4 + 2(x – y)3 – 5(x – y)2] : (x – y)2

= 3(x – y)4 : (x – y)2 + 2(x – y)3 : (x – y)2 + [– 5(x – y)2 : (x – y)2]

= 3(x – y)2 + 2(x – y) – 5

17 tháng 10 2017

Bài 65: (SGK/29):

Cách 1:

[ 3(x-y)4 + 2(x-y)3 - 5(x-y)2] : (y-x)2

= [ 3(x-y)4 + 2(x-y)3 - 5(x-y)2] : (x-y)2

= 3.(x-y)4 : (x-y)2 + 2.(x-y)3 : (x-y)2 - 5.(x-y)2 : (x-y)2

= 3.(x-y)2 + 2.(x-y) - 5

Cách theo SGK:

[ 3(x-y)4 + 2(x-y)3 - 5(x-y)2] : (y-x)2

Đặt (x-y) = z => (y-x) = z

=> (x-y)2 = z2 = (y-x)2 = (-z2) = z2

Ta có: ( 3.z4 + 2.z3 - 5.z2) : z2

= (3z4 : z2) + (2z3 : z2) - (5z2 : z2)

= 3z2 + 2z - 5

Cách 2:

[ 3(x-y)4 + 2(x-y)3 - 5(x-y)2] : (y-x)2

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

= 3(x-y)2 + 2(x-y) - 5

a: \(x^3-2y^2=2^3-2\cdot\left(-2\right)^2=8-2\cdot4=0\)

=>\(C=x\left(x^2-y\right)\left(x^3-2y^2\right)\left(x^4-3y^3\right)\left(x^5-4y^4\right)=0\)

b: x+y+1=0

=>x+y=-1

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

\(=x^2\cdot\left(-1\right)-y^2\left(-1\right)+\left(x^2-y^2\right)+2\cdot\left(-1\right)+3\)

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

=1

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

6 tháng 6 2017

\(a,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)

\(=2x^2+2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2+y^2\right)\)\(b,\left(5x-1\right)+2\left(1-5x\right)\left(4x+5\right)+\left(5x+4\right)\)\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2=25\)

6 tháng 6 2017

c)\(Q=\left(x-y\right)^3+\left(x+y\right)^3+\left(x-y\right)^3-3xy\left(x+y\right)\)

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

\(=x^3+y^3\)

d)\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(2P=5^{32}-1\Rightarrow P=\dfrac{5^{32}-1}{2}\)