哇!急急急急急急急!!!!!!!!!!恆等式

Question

我有條數不明:
試證明(10i-8)2次+10=100i2次-160i+71是恆等式
左=(10i-8)2次+10=100i2次-160i+64+10
=100i2次-160i+74
右=100i2次-160i+71
因为左不等右方
为什么100i2次-160i+『64+10』
为什么64+10

和有一条數不懂
1:試證明(2g+4)2次+4=4g2次+16g+20是恆等式



*****
今天一定要

Answer 1

1.
你打錯了！
正確證明 :
LHS = (10i - 8) ^ 2 + 10
= 100i^ 2 - 160i + 64 +10
= 100i^2 - 160i + 74
RHS = 100i2次-160i+71
LHS = RHS
所以(10i-8)2次+10=100i2次-160i+71是恆等式

2.
(2g+4)2次+4=4g2次+16g+20
LHS = 4g^ 2 + 16g + 16 + 4
= 4g^ 2 + 16g + 20
RHS = 4g^ 2 + 16g + 20
LHS = RHS
所以(2g+4)2次+4=4g2次+16g+20是恆等式

Answer 2

An identity is an equation which left hand side (L.H.S.) is always equal to right hand side ( R.H.S.) for all values of the variables ( in the equation given by you, the variable is i).
To prove that an equation is not an identity, you can simply substitue a random value of i (e.g. i =0 ) into the L.H.S. and R.H.S. of the equation respectively. If L.H.S. is not equal to R.H.S., then you can conclude that the equation is not an identity. In the equation given by you,

substituting i =0,

L.H.S. = 74 but R.H.S. = 71
L.H.S. is not equal to R.H.S.

Therefore the equation is not an identity.

Second question

L.H.S. = (4g^2 + 16g +16) +4 = 4g^2 +16g+ 20 = R.H.S.

Therefore it is an identity.

Answer 3

證明(10i - 8)² + 10 = 100i² - 160i + 71
L.H.S = 10i² - 64
R.H.S = 100i² - 160i + 71
L.H.S ≠ R.H.S
所以(10i - 8)² + 10 = 100i² - 160i + 71不是恆等式
證明(2g + 4)² + 4 = 4g² + 16g + 20
L.H.S =
(2g + 4)² + 4
= (2g)² + 2(2g)(4) + 4² + 4
= 4g² + 16g + 16 + 4
= 4g² + 16g + 20
R.H.S = 4g² + 16g + 20
所以L.H.S = R.H.S
所以(2g + 4)² + 4 = 4g² + 16g + 20是恆等式