8类NCERT解决方案-第9章代数表达式和恒等式-练习9.4

问题1.将二项式相乘。

解决方案：

当我们将两个二项式相乘时，必须进行四个乘法。尽管我们需要注意前两个项中的每一个都乘以第二个项中的每一个，但是这些乘法可以以任何顺序进行。

例如： (2x + 3)(3x – 1) ，如果我们必须将这两个二项式相乘。

Step 1: Multiply the first term of each binomial together. 
            (2x)(3x) = 6x² 

Step 2: Multiply the outer terms together. 
            (2x)(–1) = –2x 

Step 3: Multiply the inner terms together. 
            (3)(3x) = 9x 

Step 4: Multiply the last term of each expression together. 
            (3)(–1) = –3

(i)(2x + 5)×(4x – 3)

解决方案：

= 2x × (4x – 3) + 5 × (4x – 3) # Here, we used distributive property of multiplication. 
= (2x × 4x) – (3 × 2x) + (5 × 4x) – (5 × 3) # Expanding the terms. 
= 8x² – 6x + 20x – 15 # Adding or subtracting the like terms. 
= 8x² + 14x – 15

(ii)(y – 8)×(3y – 4)

解决方案：

= y × (3y – 4) – 8 × (3y – 4) # Here, we used distributive property of multiplication. 
= (y × 3y) – (y × 4) – (8 × 3y) + (-8 × -4) # Expanding the terms. 
= 3y² – 4y – 24y + 32 # Adding or subtracting the like terms. 
= 3y² – 28y + 32

(iii)(2.5l – 0.5m)×(2.5l + 0.5m)

解决方案：

= (2.5l × 2.5l) + (2.5l × 0.5m) – (0.5m × 2.5l) – (0.5m × 0.5m) 
# Here, we used distributive property of multiplication. 
= 6.25l² + 1.25ml – 1.25ml – 0.25m² 
# Expanding the terms. 
= 6.25l² + 0 – 0.25m² # Subtracting the like terms. 
= 6.25l² – 0.25m²

(iv)(a + 3b)×(x + 5)

解决方案：

= a × (x + 5) + 36 × (x + 5) 
# Here, we used distributive property of multiplication. 
= (a × x) + (a × 5) + (36 × x) + (36 × 5) 
# Expanding the terms. 
= ax + 5a + 3bx + 15b # Adding the like terms.

(v)(2pq + 3q ² )×(3pq – 2q ² )

解决方案：

= 2pq × (3pq – 2q²) + 3q² (3pq – 2q²) 
# Here, we used distributive property of multiplication. 
= (2pq × 3pq) – (2pq × 2q²) + (3q² × 3pq) – (3q² × 2q²) 
# Expanding the terms. 
= 6p²q² – 4pq³ + 9pq³ – 6q⁴ # Subtracting the like terms. 
= 6p²q² + 5pq³ – 6q⁴

(vi)(3/4 a ² + 3b ² )x 4(a ² – 2 / 3b ² )

解决方案：

= 3/4a² x (4a² – 8/3b² ) + 3b² x (4a² – 8/3b² ) # Here , we used distributive property of multiplication. 
= 3a⁴ – 2a²b² + 12a²b² – 8b⁴ # Subtracting the like terms. 
= 3a⁴ + 10a²b² – 8b⁴

问题2：找到产品。

(i)(5 – 2x)(3 + x)

解决方案：

= 5(3 + x) – 2x(3 + x) 
# Here, we used distributive property of multiplication. 
= (5 × 3) + (5 × x) – (2x × 3) – (2x × x) 
# Expanding the terms. 
= 15 + 5x – 6x – 2x² # Subtracting the like terms. 
= 15 – x – 2x²

(ii)(x + 7y)(7x – y)

解决方案：

= x(7x – y) + 7y(7x – y) 
# Here, we used distributive property of multiplication. 
= (x × 7x) – (x × y) + (7y × 7x) – (7y × y) 
# Expanding the terms. 
= 7x² – xy + 49xy – 7y² # Subtracting the like terms. 
= 7x² + 48xy – 7y²

(iii)(a ² + b)(a + b ² )

解决方案：

= a² (a + b²) + b(a + b²) 
# Here, we used distributive property of multiplication. 
= (a² × a) + (a² × b²) + (b × a) + (b × b²) 
# Expanding the terms. 
= a³ + a²b² + ab + b³

(iv)(p ² – q ² )(2p + q)

解决方案：

=p²(2p + q) – q²(2p + q) 
# Here, we used distributive property of multiplication. 
= (p² × 2p) + (p² × q) – (q² × 2p) – (q² × q) 
# Expanding the terms. 
= 2p³ + p²q – 2pq² – q³

问题3.简化。

(i)(x ² – 5)(x + 5)+ 25

解决方案：

= x²(x + 5) + 5(x + 5) + 25 
# Here, we used distributive property of multiplication. 
= x³ + 5x² – 5x – 25 + 25 
# Expanding the terms. 
= x³ + 5x² – 5x + 0 
# Subtracting the like terms. 
= x³ + 5x² – 5x

(ii)(a ² + 5)(b ³ + 3)+ 5

解决方案：

= a²(b³ + 3) + 5(b³ + 3) + 5 # Here, we used distributive property of multiplication. 
= a²b³ + 3a² + 5b³ + 15 + 5 # Expanding the terms. 
= a²b³ + 3a² + 5b³ + 20 # Adding the like terms.

(iii)(t + s ² )(t ² – s)

解决方案：

= t(t² – s) + s²(t² – s) # Here, we used distributive property of multiplication. 
= t³ – st + s²t² – s³ # Expanding the terms. 
= t³ + s²t² – st – s³

(iv)(a + b)(c – d)+(a – b)(c + d)+ 2(ac + bd)

解决方案：

= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2ac + 2bd # Here, we used distributive property of multiplication. 
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd # Expanding the terms. 
= ac + ac + 2ac + bc – bc – ad + ad – bd – bd + 2bd # Adding or subtracting the like terms. 
= 4ac + 0 + 0 + 0 
= 4ac

(v)(x + y)(2x + y)+(x + 2y)(x – y)

解决方案：

= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y) # Here, we used distributive property of multiplication. 
= 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y² # Expanding the terms. 
= 2x² + x² + xy + 2xy – xy + 2xy + y² – 2y² # Adding or subtracting the like terms. 
= 3x² + 4xy – y²

(vi)(x + y)(x ² – xy + y ² )

解决方案：

= x(x² – xy + y²) + y(x² – xy + y²) # Here, we used distributive property of multiplication. 
= x³ – x²y + x²y + xy² – xy² + y³ # Expanding the terms. 
= x³ – 0 + 0 + y³ # Adding or subtracting the like terms. 
= x³ + y³

(vii)(1.5x – 4y)(1.5x + 4y + 3)– 4.5x。+ 12y

解决方案：

= 1.5x (1.5x + 4y + 3) – 4y(1.5x + 4y + 3) – 4.5x + 12y # Here, we used distributive property of multiplication. 
= 2.25x² + 6xy + 4.5x – 6xy – 16y² – 12y – 4.5x + 12y # Expanding the terms. 
= 2.25x² + 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y² # Adding or subtracting the like terms. 
= 2.25x² + 0 + 0 + 0 – 16y² 
= 2.25x² – 16y²

(viii)(a + b + c)(a + b – c)

解决方案：

= a(a + b – c) + b(a + b – c) + c(a + b – c) # Here , we used distributive property of multiplication. 
= a² + ab – ac + ab + b² – bc + ac + bc – c² # Expanding the terms. 
= a² + ab + ab – bc + bc – ac + ac + b² – c² # Adding or subtracting the like terms. 
= a² + 2ab + b² – c² + 0 + 0 
= a² + 2ab + b² – c²