代数表达式

练习9.5

第1部分

问题1:使用合适的身份来获得以下产品。

(我)' (x + 3) (x + 3) '

答:用' (a + b)^2 = a^2+ 2ab + b^2 '得到如下方程:
' = x²+ 6x + 9 '

(2)' (2y + 5) (2y + 5) '

答:4y^2 + 20y + 25

(3)(2a - 7) (2a - 7) '

答:用' (a - b)^2 = a^2 -2ab + b^2 '得到如下方程:
' = 4a^2 - 28a + 49 '

(iv)“(3 1/2)(3 1/2)”

答:“9 ^ 2-3a + 1/4 '

(v)(1.1米- 0.4)(1.1米+ 0.4)

答:使用' (a - b)(a + b) = a^2 - b^2 '
' = 1.21m^2 - 0.16 '

(vi)' (a²+ b²)(- a²+ b²)'

答:' = (a²- a²)'
' = (b²+ a²)(b²- a²)'
' = a^4 - b^4 '

(七)' (6x - 7) (6x + 7) '

答:36x^2 - 49

(八)' (- a + c) (- a + c) '

答:' = (c - a) ^ 2 = c ^ 2-2ac + ^ 2 '

(第九)”(x / 2 + (3 y) / (4)) (x / 2 + (3 y) /(4))”

答:' (x ^ 2) / (4) + (9 y ^ 2) / (16) + (3 xy) /(4)”

(x)(7a - 9b) (7a - 9b)

答:' = 49a²- 126ab + 81b²'

问题2:用恒等式' (x + a) (x + b) = x^2+ (a + b) x + ab '求下列乘积。

(我)' (x + 3) (x + 7) '

答:' x^2 + (3+7)x + 21 '
' = x²+ 10x + 21 '

(2)' (4x + 5) (4x + 1) '

答:' = 16x^2 + (5 + 1)4x + 5 '
' = 16x^2 + 24x + 5 '

(3)' (4x - 5) (4x - 1) '

答:' = 16x^2 + (-5-1)4x + 5 '
' = 16x^2 - 20x + 5 '

(iv)' (4x + 5) (4x - 1) '

答:' = 16x^2 + (5-1)4x - 5 '
' = 16x^2 +16x - 5 '

(v)(2x + 5y) (2x + 3y)

答:' = 4x²+ (5y + 3y)4x + 15y²'
' = 4x²+ 32xy + 15y²'

(vi)(2a²+ 9)(2a²+ 5)'

答:' = 4a^4 + (9+5)2a^2 + 45 '
' = 4a^4 + 28a^2 + 45 '

(七)' (xyz - 4) (xyz - 2) '

答:' = x^2y^2z^2 + (-4 -2)xyz - 8 '
' = x^2y^2z^2 - 6xyz - 8 '

问题3:用恒等式求出下列正方形。

(我)' (b - 7)²'

答:' = b^2 - 14b + 49 '

(2)(xy + 3z)^2 '

答:' = x²y²+ 6xyz + 9z²'

(3)”(6 x ^ 2 - 5 y) ^ 2》

答:' = 36x^4 - 60x^2y + 25y^2 '

(iv)“(2/3m + 3/2 n) ^ 2”

答:“4/9m ^ 2 + 9/4n ^ 2 + 2 mn的

(v)' (0.4p - 0.5q)^2 '

答:' = 0.16p²- 0.4pq + 0.25q²'

(vi)(2xy + 5y)

答:' = 4x²y²+ 20xy²+ 25y²'

问题4:简化。

(我)' (a ^ 2 b ^ 2) ^ 2》

答:”= ^ 4-4a ^ 2 b b ^ ^ 2 + 4的

(2)' (2x + 5)²- (2x - 5)²'

答:' = 4x²+ 20x +25 - (4x²- 20x +25) '
' = 4x²+ 20x + 25 - 4x²+ 20x - 25= 40x '

(3)' (7m - 8n)²+ (7m + 8n)²'

答:' = 49m^2 - 112mn + 64n^2 + 49m^2 + 112mn + 64n^2 '
' = 98m²+ 128n²'

(iv)' (4m + 5n)²+ (5m + 4n)²'

答:' = 16m²+ 40mn + 25n²+ 25m²+ 40mn + 16n²'
' = 41m²+ 80mn + 41n²'

(v)' (2.5p - 1.5q)²- (1.5p - 2.5q)²'

答:' = 6.25便士^ 2 - 7.5 q pq + 2.25 ^ 2 - 2.25 p ^ 2 + 7.5 q pq - 6.25 ^ 2》
' = 4p^2 - 4q^2 '

(vi)' (ab + bc)^2 - 2ab^2c '

答:' = a^2b^2 + 2ab^2c + b^2c^2 - 2ab^2c '
' = a^2b²+ b^2c²'

(七)' (m^2 - n^2m)²+ 2m^3n^2 '

答:' = m^4 - 2m^3n^2 + m^2n^4 + 2m^3n^2 '
' = m^4 + m^2n^4 '