第8类RD Sharma解决方案–第7章分解–练习7.4

问题1.分解： qr – pr + qs – ps

解决方案：

Given: qr- pr + qs – ps

Grouping similar terms together we get = qr + qs – pr -ps

Taking the similar terms common we get = q(r + s) – p[r + s]

Therefore, as (r + s) is common = (r + s) (q – p) 

问题2。分解： p²q–pr²– pq +r²

解决方案：

Given: p²q – pr² – pq + r²

Grouping similar terms together we get = p²q – pq – pr² + r²

Taking the similar terms common we get = pq(p – 1)-r²(p – 1)

Therefore, as (p – 1) is common = (p – 1) (pq – r²)

问题3：分解： 1 + x + xy + x ² y

解决方案：

Given: 1 + x + xy + x²y

Taking the similar terms common we get = 1 (1 + x) + xy(1 + x)

Therefore, as (1 + x) is common = (1 + x) (1 + xy) 

问题4：分解： ax + ay – bx –由

解决方案：

Given: ax + ay – bx – by

Taking the similar terms common we get = (1 + x) (1 + xy) 

Therefore, as (x + y) is common = (x + y) (a – b)

问题5：分解： xa ² + xb ² – ya ² – yb ²

解决方案：

Given: xa² + xb² – ya² – yb²

Taking the similar terms common we get = x (a² + b²) – y (a² + b²)

Therefore, as (a² + b²) is common = (a² + b²) (x – y)

问题6：分解： x ² + xy + xz + yz

解决方案：

Given: x² + xy + xz + yz

Taking the similar terms common we get = x (x + y) + z(x + y)

Therefore, as (x + y) is common = (x + y) (x + z)

问题7：分解为： 2ax + bx + 2ay + by

解决方案：

Given: 2ax + bx + 2ay + by

Taking the similar terms common we get = x(2a + b) + y (2a + b)

Therefore, as (2a + b) is common = (2a + b) (x + y)

问题8.分解： ab-byy + y ²

解决方案：

Given: ab – by – ay + y²

Taking the similar terms common we get = b(a – y) – y(a – y)

Therefore, as (a – y) is common = (a – y) (b – y)

问题9：分解： axy + bcxy – az – bcz

解决方案：

Given: axy + bcxy – az – bcz

Taking the similar terms common we get = xy (a + bc) – z (a + bc)

Therefore, as (a + bc) is common = (a + bc) (xy – z)

问题10：分解： lm ² – mn ² – lm + n ²

解决方案：

Given: lm² – mn² – lm + n²

Taking the similar terms common we get = m (lm – n²) – 1 (lm – n²)

therefore, as (lm – n²) is common = (lm – n²) (m – 1)

问题11：分解： x ³ – y ² + x – x ² y ²

解决方案：

Given: x³ – y² + x – x²y²

Grouping similar terms together we get = x³ + x – x²y² – y²

Taking the similar terms common we get = x(x² + 1) – y²(x²+ 1)

Therefore, as (x² + 1) is common = (x² + 1) (x – y²)

问题12.分解： 6xy + 6 – 9y- 4x

解决方案：

Given: 6xy + 6 – 9y – 4x

Grouping similar terms together we get = 6xy – 4x – 9y + 6

Taking the similar terms common we get = 2x (3y – 2) – 3 (3y – 2)

Therefore, as (3y – 2) is common = (3y – 2) (2x – 3)

问题13：分解： x ² – 2ax – 2ab + bx

解决方案：

Given: x² – 2ax – 2ab + bx

Grouping similar term together we get = x² – 2ax + bx – 2ab

Taking the similar terms common we get = x (x – 2a) + b (x – 2a)

Therefore, as (x – 2a) is common = (x – 2a) (x + b)

问题14.分解： x ³ – 2x ² y + 3xy ² – ^{6y 3}

解决方案：

Given: x³ – 2x²y + 3xy² – 6y³

Taking the similar terms common we get = x² (x – 2y) + 3y² (x – 2y)

Therefore, as (x – 2y) is common = (x – 2y) (x² + 3y²)

问题15：分解： abx ² +(ay – b)x – y

解决方案：

Given: abx² + (ay – b) x – y

After solving the bracket we get = abx² + ayx – bx – y  

Taking the similar terms common we get = ax (bx + y) – 1 (bx + y)   

Therefore, as (bx + y) is common = (bx + y) (ax – 1)

问题16：分解： (ax + by) ² +(bx – ay) ²

解决方案：

Given: (ax + by)² + (bx – ay)²

After solving the bracket by using the formula ((a + b)² = a² + b² + 2ab) 

we get = a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy

Grouping similar terms together we get = a²x² + a²y²+ b²x² + b²y² 

Taking the similar terms common we get = x² (a² + b²) + y² (a² + b²)

Therefore, as (a² + b²) is common = (a² + b²) (x² + y²)

问题17：分解： 16(a – b) ³ – 24(a – b) ²

解决方案：

Given: 16 (a – b)³ – 24 (a – b)²  

Taking the similar terms common we get = 8 (a – b)² {2 (a – b) – 3}

Therefore, as (8(a – b)²) is common = 8 (a – b)² (2a – 2b – 3)

问题18.分解： ab(x ² +1)+ x(a ² + b ² )

解决方案：

Given: ab (x² + 1) + x(a² + b²)

After solving the bracket we get = abx² + ab + a²x + b²x

Grouping similar terms together we get = abx² + b²x + a²x + ab

Taking the similar terms common we get = bx (ax + b) + a (ax + b)

Therefore, as (ax + b) is common = (ax + b) (bx + a)

问题19.分解： a ² x ² +(ax ² +1)x + a

解决方案：

Given: a²x² + (ax² + 1) x + a

After solving the bracket we get = a²x² + ax³ + x + a

Grouping similar terms together we get = ax³ + a²x² + x + a

Taking the similar terms common we get = ax² (x + a) + 1 (x + a)

Therefore, as (x + a) is common = (x + a) (ax² + 1)

问题20：分解： a(a – 2b – c)+ 2bc

解决方案：

Given: a(a – 2b – c) + 2bc

After solving the bracket we get = a² – 2ab – ac + 2bc

Taking the similar terms common we get = a (a – 2b) – c (a – 2b) 

Therefore, as (a – 2b) is common = (a – 2b) (a – c)

问题21：分解： a(a + b – c)– bc

解决方案：

Given: a (a + b – c) – bc

After solving the bracket we get = a² + ab – ac – bc

Taking the similar terms common we get = a (a + b) – c (a + b)

Therefore, as (a + b) is common = (a + b) (a – c)

问题22：分解： x ² – 11xy – x + 11y

解决方案：

Given: x² – 11xy – x + 11y

Grouping similar terms together we get = x² – x – 11 xy + 11 y

Taking the similar terms common we get = x (x – 1) – 11y (x – 1)

Therefore, as (x – 1) is common = (x – 1) (x – 11y)

问题23：分解： ab – a – b + 1

解决方案：

Given: ab – a – b + 1

Taking the similar terms common we get = a (b – 1) – 1 (b – 1)

Therefore, as (b – 1) is common = (b – 1) (a – 1)

问题24.分解： x ² + y – xy – x

解决方案：

Given: x² + y – xy – x

Grouping similar terms together we get = x² – x – xy + y

Taking the similar terms common we get = x (x – 1) – y (x – 1)

Therefore, as (x – 1) is common = (x – 1) (x – y)