Bài khó:
Cho x + y – 2 = 0
Tính:
a) A = $x^{3}$ + $x^{2}y$ – 2 $x^{2}$ – xy – y ² + 3y + x – 1
b) B = x ³ – 2x ² – xy ² + 2xy + 2y + 2x – 2
Bài khó:
Cho x + y – 2 = 0
Tính:
a) A = $x^{3}$ + $x^{2}y$ – 2 $x^{2}$ – xy – y ² + 3y + x – 1
b) B = x ³ – 2x ² – xy ² + 2xy + 2y + 2x – 2
Đáp án + giải thích bước giải :
Bài 1
Ta có : `x + y -2 = 0`
`-> x + y = 2`
`A = x^3 + x^2y – 2x^2 – xy – y^2 + 3y + x – 1`
`-> A = x^2 (x + y) – 2x^2 – y (x + y) + 3y + x – 1`
`-> A = x^2 . 2 – 2x^2 – y . 2 + 3y + x – 1`
`-> A = x^2 (2 – 2) + y + x – 1`
`-> A = 1`
`B = x^3 – 2x^2 – xy^2 + 2xy + 2y + 2x – 2`
`-> B = x^2 (x – 2) – xy (y -2) + 2 (x + y) – 2`
`-> B = -x^2y + x^2y + 2`
`-> B = 2`
Đáp án:
$\begin{array}{l}
Ta có:x + y – 2 = 0\\
\Rightarrow x + y = 2\\
a)A = {x^3} + {x^2}y – 2{x^2} – xy – {y^2} + 3y + x – 1\\
A = {x^2}\left( {x + y} \right) – 2{x^2} – y(x + y) + 3y + x – 1\\
A = {x^2}.2 – 2{x^2} – y.2 + 3y + x – 1\\
A = {x^2}(2 – 2) + y + x – 1\\
A = 0 + 2 – 1\\
A = 1\\
b)B = {x^3} – 2{x^2} – x{y^2} + 2xy + 2y + 2x – 2\\
B = {x^2}(x – 2) – xy(y – 2) + 2(x + y) – 2\\
B = {x^2}( – y) – xy( – x) + 4 – 2\\
B = – {x^2}y + {x^2}y + 2\\
B = 2
\end{array}$