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

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

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

a: \(\Leftrightarrow x\cdot\left(\dfrac{37}{99}+\dfrac{62}{99}\right)=10\)

=>x=10

b: \(\Leftrightarrow x:\dfrac{4}{9}=\dfrac{4}{33}:\dfrac{5}{3}=\dfrac{4}{33}\cdot\dfrac{3}{5}=\dfrac{4}{55}\)

\(\Leftrightarrow x=\dfrac{16}{495}\)

1

37.(43-51)-43.(37-51)

=37.43-37.51-43.37-43.51

=(37.43-43.57)-(37.51-43.51)

=0-(-306)

=306

2

a.(x-1).(2x+2)=0

=>x-1 hoặc 2x+2=0

=>x=1 hoặc 2x=-2

=>x=1 hoặc x=-1

Vậy x=+1

b.(6x-12).(x-3)=0

=>6x-12 hoặc x-3=0

=>6x=12 hoặc x=3

=>x=2 hoặc x=3

Vậy x=2 hoặc x=3

a x-37=0

x=37

b x-42=0 hoặc x-21=0

x=42,21

\(a.\left(x-37\right).25=0\\\Leftrightarrow x-27=0\\ \Leftrightarrow x=27\)

\(b.\left(x-42\right)\left(x-21\right)=0\\ \Leftrightarrow\orbr{\begin{cases}x-42=0\\x-21=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=42\\x=21\end{cases}}}\)

a, x=37

b, \(\orbr{\begin{cases}x=42\\x=21\end{cases}}\)