Bài 1.

Ta có: $x+y+z=0$

$\Rightarrow x+y=-z,\ y+z=-x,\ x+z=-y$.

Suy ra: $N=(x+y)(y+z)(x+z)$$=(-z)(-x)(-y)$$=-xyz$.

Mà: $xyz=2$.

Nên: $N=-2$.

Vậy: $N=-2$.

Bài 2.

$\frac{a}{b}=\frac{10}{3}\Rightarrow a=10k,\; b=3k$

$\frac{3a-2b}{a-3b}=\frac{3\cdot10k-2\cdot3k}{10k-3\cdot3k}$

$=\frac{30k-6k}{10k-9k}$

$=\frac{24k}{k}$

$=24$

Ta co : x+y=2

(x+y)^2=4

x^2+2xy+y^2=4

x^2+y^2+2xy=4

10+2xy=4

2xy=-6

xy=-3

Ta lai co : x^3+y^3 =(x+y)(x^2+xy+y^2)

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

=2.[10-(-3)]

=26

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\(A=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)

\(A=\left(x^3+y^3\right)-2\left(x^2+y^2\right)+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)

\(A=\left(x+y\right)^3-3xy\left(x+y\right)-2\left(\left(x+y\right)^2-2xy\right)+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)

\(A=\left(x+y\right)^3-3xy\left(x+y\right)-2\left(x+y\right)^2+4xy+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)

\(A=\left(5\right)^3-3xy\left(5\right)-2\left(5\right)^2+4xy+3xy\left(5\right)-4xy+3\left(5\right)+10\)

\(A=125-15xy-50+4xy+15xy-4xy+15+10\)

\(A=100\)

a: Khi x=2 và y=-3 thì \(x^2+2y=2^2+2\cdot\left(-3\right)=4-6=-2\)

b: \(A=x^2+2xy+y^2=\left(x+y\right)^2\)

Khi x=4 và y=6 thì \(A=\left(4+6\right)^2=10^2=100\)

c: \(P=x^2-4xy+4y^2=\left(x-2y\right)^2\)

Khi x=1 và y=1/2 thì \(P=\left(1-2\cdot\dfrac{1}{2}\right)^2=\left(1-1\right)^2=0\)

\(P=\left(x+y\right)\left\{\left[\left(x+y\right)^2-2xy\right]\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\right\}\\ \)

Thây số vào

VÌ \(x+y=7;xy=10\)

\(\Rightarrow x,y=5\)và \(2\)

\(\Rightarrow P=\left(5+2\right)\left(5^2+2^2\right)\left(5^3+2^3\right)\)

\(\Rightarrow P=7.29.133\)

    \(P=26999\)

a) \(x-y=-3\)

\(\Rightarrow\left(x-y\right)^2=9\)

\(\Leftrightarrow x^2-2xy+y^2=9\)

\(\Leftrightarrow x^2+y^2-20=9\)

\(\Leftrightarrow x^2+y^2=29\)

b) \(\left(x-y\right)^3=-27\)

\(\Leftrightarrow x^3-y^3-3xy\left(x+y\right)=-17\)

\(\Leftrightarrow x^3-y^3-30\left(x+y\right)=-17\)

WHAT ??? 

        \(x+y=2\)

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

\(\Leftrightarrow\)\(x^2+y^2+2xy=4\)

\(\Leftrightarrow\)\(2xy=-6\)  do x² + y² = 10

\(\Leftrightarrow\)\(xy=-3\)

\(T=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=2^3-3.\left(-3\right).2=26\)

Vì \(\left(x+y\right)=2\Rightarrow\left(x+y\right)^2=4\Leftrightarrow x^2+y^2+2xy=4\Leftrightarrow2xy=-6\Leftrightarrow xy=-3\)

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

\(\Rightarrow T=2.\left(10-xy\right)\)

\(\Rightarrow T=20-2xy=20+6=26\)