\(1.\left(x^2-x+1\right)\left(x^2-x+2\right)-12\)

Đặt : \(x^2-x+1=t\) , ta có :

\(t\left(t+1\right)-12=t^2+t-12=t^2-3t+4t-12=t\left(t-3\right)+4\left(t-3\right)=\left(t-3\right)\left(t+4\right)\)

Thay : \(x^2-x+1=t\) vào biểu thức trên , ta có :
\(\left(x^2-x+1-3\right)\left(x^2-x+1+4\right)=\left(x^2-x-2\right)\left(x^2-x+5\right)\)

\(2.\) Ta có : \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\Rightarrow x^3+y^3+z^3=3xyz\)

\(\left(x-15\right)\left(y+12\right)\left(z-3\right)=0\)

=>\(\left[{}\begin{matrix}x-15=0\\y+12=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\y=-12\\z=3\end{matrix}\right.\)

TH1: x=15

x+1=y+2=z+3

=>y+2=z+3=15+1=16

=>y=16-2=14;z=16-3=13

TH2: y=-12

x+1=y+2=z+3

=>x+1=z+3=-12+2=-10

=>x=-10-1=-11; z=-10-3=-13

TH3: z=3

x+1=y+2=z+3

=>x+1=y+2=3+3=6

=>x=6-1=5; y=6-2=4

a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=\left(x-y\right)\left(x-y+2\right)+37\)(1)

Thay x-y=7 vào biểu thức (1), ta được:

\(A=7\cdot\left(7+2\right)+37=7\cdot9+37=100\)

Vậy: Khi x-y=7 thì A=100

b) Ta có: \(x+y=2\)

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

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

\(\Leftrightarrow2xy+10=4\)

\(\Leftrightarrow2xy=-6\)

\(\Leftrightarrow xy=-3\)

Ta có: \(A=x^3+y^3\)

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

Thay x+y=2; \(x^2+y^2=10\) và xy=-3 vào biểu thức (2), ta được:

\(A=2\cdot\left(10+3\right)=2\cdot13=26\)

Vậy: Khi x+y=2 và \(x^2+y^2=10\) thì A=26

\(\Rightarrow A=x^2+2x+y^2-2y-2xy+37=x^2-2xy+y^2+2\left(x-y\right)+37=\left(x-y\right)^2+2\left(x-y\right)+37=7^2+2\cdot7+37=100\)

\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[x^2+y^2-\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}\right]=2\cdot\left[10+3\right]=2\cdot13=26\) \(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{x+z}{x}\right)=-\dfrac{z}{y}\cdot\dfrac{-x}{z}\cdot-\dfrac{y}{x}=-1\)

Bài 2: 

Ta có: \(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)

\(\Leftrightarrow\left(\dfrac{x-1}{65}-1\right)+\left(\dfrac{x-3}{63}-1\right)=\left(\dfrac{x-5}{61}-1\right)+\left(\dfrac{x-7}{59}-1\right)\)

\(\Leftrightarrow\left(x-66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)

=>x-66=0

hay x=66

qua de