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计算:$(a - b + c)^2 = $______.
答案
【易错点睛】
$a^{2}+b^{2}+c^{2}-2ab+2ac-2bc$
$a^{2}+b^{2}+c^{2}-2ab+2ac-2bc$
1.(教材变式)在括号内填上适当的整式:
(1)$a + b - c = a +$(______); (2)$a - b + c = a -$(______);
(3)$a - b - c = a -$(______); (4)$- a + b - c = b -$(______).
(1)$a + b - c = a +$(______); (2)$a - b + c = a -$(______);
(3)$a - b - c = a -$(______); (4)$- a + b - c = b -$(______).
答案
【基础题夯实】
1. (1)$b-c$ (2)$b-c$ (3)$b+c$ (4)$a+c$
1. (1)$b-c$ (2)$b-c$ (3)$b+c$ (4)$a+c$
2.(2025大连)运用平方差公式计算$(x - 2y - 1)(x - 2y + 1)$时,下列变形正确的是()
A.$[x - (2y - 1)]^2$
B.$[x + (2y - 1)][x - (2y - 1)]$
C.$[(x - 2y) - 1][(x - 2y) + 1]$
D.$[x + (2y + 1)]^2$
A.$[x - (2y - 1)]^2$
B.$[x + (2y - 1)][x - (2y - 1)]$
C.$[(x - 2y) - 1][(x - 2y) + 1]$
D.$[x + (2y + 1)]^2$
答案
2. C
3.运用完全平方公式计算$(x - y + 2z)^2$,下列变形不正确的是()
A.$[(x - y) + 2z]^2$
B.$[(x + 2z) - y]^2$
C.$[x - (y + 2z)]^2$
D.$[x + (2z - y)]^2$
A.$[(x - y) + 2z]^2$
B.$[(x + 2z) - y]^2$
C.$[x - (y + 2z)]^2$
D.$[x + (2z - y)]^2$
答案
3. C
4.(教材变式)运用乘法公式计算:
(1)$(a + b + 1)(a + b - 1)$; (2)$(x - y - z)(x + y + z)$.
(1)$(a + b + 1)(a + b - 1)$; (2)$(x - y - z)(x + y + z)$.
答案
4. 解:(1)原式$=(a+b)^{2}-1$
$=a^{2}+2ab+b^{2}-1$;
(2)原式$=[x-(y+z)][x+(y+z)]$
$=x^{2}-(y+z)^{2}$
$=x^{2}-y^{2}-2yz-z^{2}$.
$=a^{2}+2ab+b^{2}-1$;
(2)原式$=[x-(y+z)][x+(y+z)]$
$=x^{2}-(y+z)^{2}$
$=x^{2}-y^{2}-2yz-z^{2}$.
5.(教材变式)运用乘法公式计算:
(1)$(a - b - c)^2$; (2)$(a + b - 1)^2$.
(1)$(a - b - c)^2$; (2)$(a + b - 1)^2$.
答案
5. 解:(1)原式$=(a-b)^{2}-2(a-b)c+c^{2}$
$=a^{2}+b^{2}+c^{2}-2ab-$
$2ac+2bc$;
(2)原式$=(a+b)^{2}-2(a+b)+1$
$=a^{2}+b^{2}+1+2ab-2a-2b$.
$=a^{2}+b^{2}+c^{2}-2ab-$
$2ac+2bc$;
(2)原式$=(a+b)^{2}-2(a+b)+1$
$=a^{2}+b^{2}+1+2ab-2a-2b$.
6.运用乘方公式计算:
(1)$[(x + 1)(x - 1)]^2$; (2)$(a + \frac{1}{2})(a^2 - \frac{1}{4})(a - \frac{1}{2})$.
(1)$[(x + 1)(x - 1)]^2$; (2)$(a + \frac{1}{2})(a^2 - \frac{1}{4})(a - \frac{1}{2})$.
答案
6. 解:(1)原式$=(x^{2}-1)^{2}$
$=x^{4}-2x^{2}+1$;
(2)原式$=(a^{2}-\frac {1}{4})^{2}$
$=a^{4}-\frac {1}{2}a^{2}+\frac {1}{16}$.
$=x^{4}-2x^{2}+1$;
(2)原式$=(a^{2}-\frac {1}{4})^{2}$
$=a^{4}-\frac {1}{2}a^{2}+\frac {1}{16}$.
