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a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=15^2-2\cdot50=115\)

c: \(x-y=\sqrt{\left(x+y\right)^2-4xy}=\sqrt{15^2-4\cdot50}=5\)

\(C=x^2-y^2=\left(x+y\right)\left(x-y\right)=15\cdot5=75\)

AH
Akai Haruma
Giáo viên
29 tháng 12 2023

Lời giải:

$x^3+y^3=(x+y)^3-3xy(x+y)=2^3-3xy.2=8-6xy$

$=8-3.2xy=8-3[(x+y)^2-(x^2+y^2)]=8-3(2^2-34)=98$

----------------

$x^4+y^4=(x^2+y^2)^2-2x^2y^2=34^2-\frac{1}{2}(2xy)^2$

$=34^2-\frac{1}{2}[(x+y)^2-(x^2+y^2)]^2=34^2-\frac{1}{2}(2^2-34)^2=706$

a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=15^2-2\cdot50=125\)

b:\(B=x^4+y^4\)

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

\(=125^2-2\cdot2500\)

=10625

c:  \(x-y=\sqrt{\left(x+y\right)^2-4xy}=\sqrt{15^2-4\cdot50}=5\)

\(C=x^2-y^2=\left(x-y\right)\left(x+y\right)=15\cdot5=75\)

NV
15 tháng 12 2020

a.

\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)

e.

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

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

1 tháng 7 2021

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x2-4x-5

=(x-5)(x+1)

i.x4-y= (x2-y2)(x2+y2)

 

 

27 tháng 1 2019

x4 + y4 = (x2 + y2)2-2x2 y2 = 182-2.52 = 274

25 tháng 8 2021

a) \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)

b) \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x=x^3-3x^2+3x-1-x^3-x^2-x+x^2+x+1-3x+3x^2=0\)

a: Ta có: \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)

\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)

=0

b: Ta có: \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)

\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)

=0

Bài 3: 

\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15\)

\(=\left(x^2-9\right)\left(x^2-1\right)+15\)

\(=x^4-10x^2+9+15\)

\(=x^4-10x^2+24\)

\(=\left(x^2-4\right)\left(x^2-6\right)\)

\(=\left(x-2\right)\left(x+2\right)\left(x^2-6\right)\)

 

6) Ta có: \(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

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

7) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

8) Ta có: \(4x^4-32x^2+1\)

\(=4x^4+12x^3+2x^2-12x^3-36x^2-6x+2x^2+6x+1\)

\(=2x^2\left(2x^2+6x+1\right)-6x\left(2x^2+6x+1\right)+\left(2x^2+6x+1\right)\)

\(=\left(2x^2+6x+1\right)\left(2x^2-6x+1\right)\)

9) Ta có: \(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)

\(=3\left[x^4+2x^2+1-x^2\right]-\left(x^2+x+1\right)^2\)

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

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

\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)\)

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

19 tháng 10 2021

\(g,=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)=\left(x^4-y^4\right)\left(x^4+y^4\right)=x^8-y^8\)

\(b,=\left(x^2-9\right)\left(x-4\right)-\left(x^3+3x^2+3x+1\right)\\ =x^3-4x^2-9x+36-x^3-3x^2-3x-1\\ =-7x^2-12x+36\)