Question

[33. Nguyễn Minh Ngọc]

Cho x + y = 5. Tính giá trị biểu thức (sử dụng HĐT)

 a) P = 3x² - 2x + 3y² - 2y + 6xy - 100

 b) Q = x³ + y³ - 2x² - 2y² + 3xy(x + y) - 4xy + 3(x + y) + 10

Answer 1

[l҉o҉n҉g҉ d҉z҉]

P = 3x² - 2x + 3y² - 2y + 6xy - 100

= 3( x² + 2xy + y² ) - 2( x + y ) - 100

= 3( x + y )² - 2( x + y ) - 100

Với x + y = 5

=> P = 3.5² - 2.5 - 100 = 75 - 10 - 100 = -35

Q = x³ + y³ - 2x² - 2y² + 3xy( x + y ) - 4xy + 3( x + y ) + 10

= x³ + y³ - 2x² - 2y² + 3x²y + 3xy² - 4xy + 3( x + y ) + 10

= ( x³ + 3x²y + 3xy² + y³ ) - ( 2x² + 4xy + 2y² ) + 3( x + y )

= ( x + y )³ - 2( x² + 2xy + y² ) + 3( x + y ) + 10

= ( x + y )³ - 2( x + y )² + 3( x + y ) + 10

Với x + y = 5

=> Q = 5³ - 2.5² + 3.5 + 10 = 100

Answer 2

[Khánh Ngọc]

a. \(P=3x^2-2x+3y^2-2y+6xy-100\)

\(\Leftrightarrow P=\left(3x^2+6xy+3y^2\right)-\left(2x+2y\right)-100\)

\(\Leftrightarrow P=3\left(x+y\right)^2-2\left(x+y\right)-100\)

\(\Leftrightarrow P=3.5^2-2.5-100\)

\(\Leftrightarrow P=-35\)

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

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

\(\Leftrightarrow Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)

\(\Leftrightarrow Q=5^3-2.5^2+3.5+10\)

\(\Leftrightarrow Q=100\)