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

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

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

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

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

a: \(x^2+y^2=\left(x-y\right)^2+2xy=15^2+2\cdot50=325\)

b: \(x+y=\sqrt{\left(x-y\right)^2+4xy}=\sqrt{15^2+4\cdot50}=5\sqrt{17}\)

\(x^2-y^2=\left(x-y\right)\left(x+y\right)=5\sqrt{17}\cdot15=75\sqrt{17}\)

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

\(=75\sqrt{17}\cdot325=24375\sqrt{17}\)

x1 = 3; y1 = 6 nên hệ số tỉ lệ của y đối với x là 6 : 3 = 2

chắc đề cho x,y chứ x+y=6,x-y=4,xy=5

(làm ra bạn tự thay số vào tính)

a,\(=>A=\left(x+y\right)^2-2xy=.....\)

b,\(=>B=\left(x+y\right)^3-3xy\left(x+y\right)+xy=....\)

c,\(=>C=\left(x-y\right)\left(x+y\right)=....\)

d,\(=>D=\dfrac{x+y}{xy}=.....\)

e,\(=>E=\dfrac{x^2+y^2}{xy}=\dfrac{\left(x+y\right)^2-2xy}{xy}=...\)

a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=6^2-2\cdot5=26\)

b: \(B=x^3+y^3+xy\)

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

\(=6^3-3\cdot5\cdot6+5\)

\(=216-90+5=131\)

c: \(C=x^2-y^2=\left(x-y\right)\left(x+y\right)=4\cdot6=24\)

d: \(D=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{6}{5}\)

e: \(E=\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x^2+y^2}{xy}=\dfrac{\left(x+y\right)^2-2xy}{xy}=\dfrac{6^2-2\cdot5}{5}=\dfrac{26}{5}\)

Lời giải:
a.

$x^3+y^3=(x+y)^3-3xy(x+y)=9^3-3.9.18=243$

$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$

$=[9^2-2.18]^2-2.18^2=1377$

Nếu $x\geq y$ thì:

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

$=|x-y|[(x+y)^2-xy]=\sqrt{(x+y)^2-4xy}[(x+y)^2-xy]$

$=\sqrt{9^2-4.18}(9^2-18)=189$

Nếu $x< y$ thì $x^3-y^3=-189$

b.

$A=(x+y)^2-6(x+y)+y-5$

$=(-9)^2-6(-9)+y-5=130+y$

Chưa đủ cơ sở để tính biểu thức.

a) \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=9^3-3\cdot18\cdot9=243\)

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

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

\(=\left(9^2-2\cdot18\right)^2-2\cdot18^2\)

\(=45^2-2\cdot324\)

=1377

\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz\) Thay x+y+z=0 vào

\(\Rightarrow0=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)

\(\Leftrightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\) (1)

Ta có

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

Bình phương 2 vế của (1)

\(\left(x^2+y^2+z^2\right)^2=4\left(xy+yz+xz\right)^2\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2=4\left(x^2y^2+y^2z^2+x^2z^2+2xy^2z+2xyz^2+2x^2yz\right)\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2=4\left[x^2y^2+y^2z^2+x^2z^2+2xyz\left(x+y+z\right)\right]\)

Do x+y+z=0 nên

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

\(\Rightarrow\dfrac{\left(x^2+y^2+z^2\right)^2}{2}=2x^2y^2+2y^2z^2+2x^2z^2\) (3)

Thay (3) vào (2)

\(\left(x^2+y^2+z^2\right)^2=x^4+y^4+z^4+\dfrac{\left(x^2+y^2+z^2\right)^2}{2}\)

\(\Rightarrow2\left(x^4+y^4+z^4\right)=\left(x^2+y^2+z^2\right)^2\) (đpcm)