早教吧作业答案频道 -->数学-->
附加题:设a−b=2,a−c=12,求整式(c−b)2+3(b−c)+94的值.
题目详情
a−b=2,a−c=
,求整式(c−b)2+3(b−c)+
的值.
1 1 2 2 (c−b)2+3(b−c)+
的值.)2+3(b−c)+
的值.)2+3(b−c)+
的值.2+3(b−c)+
的值.
9 9 4 4
1 |
2 |
9 |
4 |
1 |
2 |
9 |
4 |
9 |
4 |
9 |
4 |
9 |
4 |
9 |
4 |
▼优质解答
答案和解析
∵a-b=2,a-c=
,
∴b-c=
-2=-
,
(c-b)2+3(b-c)+
,
=(b-c+
)2,
=(-
+
)2,
=0.
1 1 12 2 2,
∴b-c=
-2=-
,
(c-b)2+3(b-c)+
,
=(b-c+
)2,
=(-
+
)2,
=0.
1 1 12 2 2-2=-
,
(c-b)2+3(b-c)+
,
=(b-c+
)2,
=(-
+
)2,
=0.
3 3 32 2 2,
(c-b)22+3(b-c)+
,
=(b-c+
)2,
=(-
+
)2,
=0.
9 9 94 4 4,
=(b-c+
)2,
=(-
+
)2,
=0.
3 3 32 2 2)22,
=(-
+
)2,
=0.
3 3 32 2 2+
)2,
=0.
3 3 32 2 2)22,
=0.
1 |
2 |
∴b-c=
1 |
2 |
3 |
2 |
(c-b)2+3(b-c)+
9 |
4 |
=(b-c+
3 |
2 |
=(-
3 |
2 |
3 |
2 |
=0.
1 |
2 |
∴b-c=
1 |
2 |
3 |
2 |
(c-b)2+3(b-c)+
9 |
4 |
=(b-c+
3 |
2 |
=(-
3 |
2 |
3 |
2 |
=0.
1 |
2 |
3 |
2 |
(c-b)2+3(b-c)+
9 |
4 |
=(b-c+
3 |
2 |
=(-
3 |
2 |
3 |
2 |
=0.
3 |
2 |
(c-b)22+3(b-c)+
9 |
4 |
=(b-c+
3 |
2 |
=(-
3 |
2 |
3 |
2 |
=0.
9 |
4 |
=(b-c+
3 |
2 |
=(-
3 |
2 |
3 |
2 |
=0.
3 |
2 |
=(-
3 |
2 |
3 |
2 |
=0.
3 |
2 |
3 |
2 |
=0.
3 |
2 |
=0.
看了 附加题:设a−b=2,a−c...的网友还看了以下:
观察:1+2=3=2^2-1,1+2+2^2=7=2^3-1,1+2+2^2+2^3=15=2^4- 2020-03-31 …
给出算式:3^2-2^2=(3+2)(3-2)=2+34^2-3^2=(4+3)(4-3)=3+4 2020-04-09 …
1.1-1又1/2-3/8+7/122.(-5.5)+(-3.2)-(-2.5)-4.83.1-2 2020-05-17 …
1.1-1又1/2-3/8+7/122.(-5.5)+(-3.2)-(-2.5)-4.83.1-2 2020-05-17 …
急用!1.2(x-2)-3(4x-1)=9(1-x)2.11x+64-2x=100-9x3.15- 2020-05-17 …
观察:1+2=3=2^2-1,1+2+2^2=7=2^3-1,1+2+2^2+2^3=15=2^4 2020-05-20 …
f(x)=1+x+cosx在[-3π/2,π/2]上是:A.单增函数B.单减函数c.在[-3π/2 2020-06-27 …
1/2+(1/3+2/3)+(1/4+2/4+3/4)+……+(1/60+2/60+……59/60 2020-07-17 …
1.(x^2-3)^2+2(x^2-3)(x-3)+(x-3)^22.(x+1)(x+3)(x+5 2020-07-18 …
1/2+1/3+2/3+1/4+2/4+3/4+1/5+2/5+3/5+4/5+...+1/60+ 2020-07-19 …