9类RD Sharma解决方案–第四章代数身份-练习4.2

问题1：以展开形式编写以下内容：

(i)(a + 2b + c) ²

(ii)(2a-3b-c) ²

(iii)(−3x + y + z) ²

(iv)(m + 2n−5p) ²

(v)(2 + x−2y) ²

(vi)(a ² + b ² + c ² ) ²

(vii)(ab + bc + ca) ²

(viii)(x / y + y / z + z / x) ²

(ix)(a / bc + b / ac + c / ab) ²

[x)(x + 2y + 4z) ²

(xi)(2x−y + z) ²

(xii)(−2x + 3y + 2z) ²

解决方案：

By following the identity,

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2xz

(i) (a + 2b + c)²

= a² + (2b)² + c² + 2a(2b) + 2ac + 2(2b)c

= a² + 4b² + c² + 4ab + 2ac + 4bc

(ii) (2a − 3b − c)²

= [(2a) + (−3b) + (−c)]²

= (2a)² + (−3b)² + (−c)² + 2(2a)(−3b) + 2(−3b)(−c) + 2(2a)(−c)

= 4a² + 9b² + c² − 12ab + 6bc − 4ca

(iii) (−3x+y+z)²

= [(−3x)² + y² + z² + 2(−3x)y + 2yz + 2(−3x)z]

= 9x² + y² + z² − 6xy + 2yz − 6xz

(iv) (m+2n−5p)²

= m² + (2n)² + (−5p)² + 2m × 2n + (2×2n×−5p) + 2m × (−5p)

= m² + 4n² + 25p² + 4mn − 20np − 10pm

(v) (2+x−2y)²

= 2² + x² + (−2y) 2 + 2(2)(x) + 2(x)(−2y) + 2(2)(−2y)

= 4 + x² + 4y² + 4 x − 4xy − 8y

(vi) (a² +b² +c²)²

= (a²)² + (b²)² + (c²)² + 2a² b² + 2b²c² + 2a²c²

= a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2c²a²

(vii) (ab+bc+ca)²

= (ab)² + (bc)² + (ca)² + 2(ab)(bc) + 2(bc)(ca) + 2(ab)(ca)

= a²b² + b²c² + c²a² + 2(ac)b² + 2(ab)(c)² + 2(bc)(a)²

(viii) (x/y+y/z+z/x)²

$=(\frac{x}{y})^2+(\frac{y}{z})^2+(\frac{z}{x})^2+2\frac{x}{y}×\frac{y}{z}+2\frac{y}{z}×\frac{z}{x}+2\frac{z}{x}×\frac{x}{y}\\=(\frac{x^2}{y^2})+(\frac{y^2}{z^2})+(\frac{z^2}{x^2})+2\frac{x}{z}+2\frac{y}{x}+2\frac{z}{y}$

(ix) (a/bc + b/ac + c/ab)²

$=(\frac{a}{bc})^2+(\frac{b}{ca})^2+(\frac{c}{ab})^2+2(\frac{a}{bc})(\frac{b}{ca})+2(\frac{b}{ca})(\frac{c}{ab})+2(\frac{a}{bc})(\frac{c}{ab})\\=(\frac{a^2}{b^2c^2})+(\frac{b^2}{c^2a^2})+(\frac{c^2}{a^2b^2})+\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}$

(x) (x+2y+4z)²

= x² + (2y)² + (4z)² + (2x)(2y) + 2(2y)(4z) + 2x(4z)

= x² + 4y² + 16z² + 4xy + 16yz + 8xz

(xi) (2x−y+z)²

= (2x)² + (−y)² + (z)² + 2(2x)(−y) + 2(−y)(z) + 2(2x)(z)

= 4x² + y² + z² − 4xy − 2yz + 4xz

(xii) (−2x+3y+2z)²

= (−2x)² + (3y)² + ( 2z)² + 2(−2x)(3y) + 2(3y)(2z) + 2(−2x)(2z)

= 4x² + 9y² + 4z² −12xy + 12yz −8xz

问题2：简化

(i)(a + b + c) ² +(a − b + c) ²

(ii)(a + b + c) ² −(a − b + c) ²

(iii)(a + b + c) ² +(a – b + c) ² +(a + b − c) ²

(iv)(2x + p-c) ^2- (2x-p + c) ²

(v)(x ² + y ² -z ² ) ^2- (x ² -y ² + z ² ) ²

解决方案：

(i) (a + b + c)² + (a − b + c)²

= (a² + b² + c² + 2ab+2bc+2ca) + (a² + (−b)² + c² −2ab−2bc+2ca)

= 2a² + 2 b² + 2c² + 4ca

(ii) (a + b + c)² − (a − b + c)²

= (a² + b² + c² + 2ab+2bc+2ca) − (a² + (−b)² + c² −2ab−2bc+2ca)

= a2 + b2 + c2 + 2ab + 2bc + 2ca − a2 − b2 − c2 + 2ab + 2bc − 2ca

= 4ab + 4bc

(iii) (a + b + c)² + (a – b + c)² + (a + b − c)

= a² + b² + c² + 2ab + 2bc + 2ca + (a² + b² + (c)² − 2ab − 2cb + 2ca) + (a² + b² + c² + 2ab − 2bc – 2ca)

= 3a² + 3b² + 3c² + 2ab − 2bc + 2ca

(iv) (2x + p − c)² − (2x − p + c)²

= [4x² + p² + c² + 4xp − 2pc − 4xc] − [4x² + p² + c² − 4xp− 2pc + 4xc]

= 4x² + p² + c² + 4xp − 2pc − 4cx − 4x² − p² − c² + 4xp + 2pc− 4cx

= 8xp − 8xc

= 8(xp − xc)

(v) (x² + y² − z²)² − (x² − y² + z²)²

= (x² + y² + (−z)²)² − (x² − y² + z²)²

= [x⁴ + y⁴ + z⁴ + 2x2y² – 2y2z² – 2x2z² − [x⁴ + y⁴ + z⁴ − 2x2y² − 2y2z² + 2x2z²]

= 4x²y² – 4z²x²

问题3：如果a + b + c = 0且a ² + b ² + c ² = 16，则求ab + bc + ca的值。

解决方案：

Given,

a + b + c = 0 and a² + b² + c² = 16

Choose a + b + c = 0

Squaring both sides,

(a + b + c)² = 0

a² + b² + c² + 2(ab + bc + ca) = 0

16 + 2(ab + bc + c) = 0

2(ab + bc + ca) = -16

ab + bc + ca = -16/2 = -8

or

ab + bc + ca = -8