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Bài làm :
a) \(\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2=a^2+2ab+b^2\)
b) \(\left(a-b\right)^2=a^2-2ab+b^2\)
c) \(\left(a+b\right)\left(a-b\right)=a^2-b^2\)
d) \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)
e) \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)
f) \(\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\)
g) \(\left(a-b\right)\left(a^2+ab+b^2\right)=a^3-b^3\)
1) (a+b).(a+b)=(a+b)2=a2+2ab+b2
2) (a-b)2=a2-2ab+b2
3) (a+b).(a-b)=a2-b2
4) (a+b)3=a3+3a2b+3ab2+b3
5) (a-b)3=a3-3a2b+3ab2-b3
6) (a+b).(a2-ab+b2)=a3+b3
7) (a-b).(a2+ab+b2)=a3-b3
mấy cái ày là hằng đẳng thức đáng nhớ mà
lấy a+a b+b
lấy b^2-a
lấy a.b b.a
a^3 +b
b^3-a
hai câu cuối thì mình k biết
1) (a+b)3=(a+b)(a+b)(a+b)=(a2+ab+ab+b2)(a+b)=(a2+2ab+b2)(a+b)(a3+2a2b+ab2)+(a2b+2ab2+b3)=a3+2a2b+ab2+a2b+2ab2+b3
=a3+3a2b+3ab2+b3
1) \(\left(a+b\right).\left(a+b\right)=a.\left(a+b\right)+b.\left(a+b\right)=a^2+ab+b^2+ab\)
2) \(\left(a-b\right)^2=\left(a-b\right).\left(a-b\right)=a.\left(a-b\right)-b.\left(a-b\right)=a^2-ab-ab+b^2\)
\(=a^2+\left(-ab\right)+\left(-ab\right)+b^2\)
3) \(\left(a+b\right).\left(a-b\right)=a.\left(a-b\right)+b.\left(a-b\right)=a^2-ab+ab-b^2=a^2-b^2\)
\(=a^2+-\left(b^2\right)\)
4) \(\left(a+b\right)^3=\left(a+b\right).\left(a+b\right).\left(a+b\right)=a.\left(a+b\right).\left(a+b\right)+b.\left(a+b\right).\left(a+b\right)\)
\(=\left[a.\left(a+b\right)\right].\left(a+b\right)+\left[b.\left(a+b\right)\right].\left(a+b\right)=\left(a^2+ab\right).\left(a+b\right)+\left(ab+b^2\right).\left(a+b\right)\)
\(=a^2.\left(a+b\right)+ab.\left(a+b\right)+ab.\left(a+b\right)+b^2.\left(a+b\right)\)
\(=a^3+a^2b+a^2b+ab^2+a^2b+ab^2+b^2a+b^3\)
5) \(\left(a-b\right)^3=\left(a-b\right).\left(a-b\right).\left(a-b\right)=a.\left(a-b\right).\left(a-b\right)-b.\left(a-b\right).\left(a-b\right)\)
\(=\left(a^2-ab\right).\left(a-b\right)-\left(ba-b^2\right).\left(a-b\right)\)
\(=a^2.\left(a-b\right)-ab.\left(a-b\right)-ba.\left(a-b\right)+b^2.\left(a-b\right)\)
\(=a^3-a^2b-a^2b+ab^2-ba^2+b^2a-ba^2+b^2a-b^3\)
6) \(\left(a+b\right).\left(a^2-ab+b^2\right)=a.\left(a^2-ab+b^2\right)+b.\left(a^2-ab+b^2\right)\)
\(=a^3-a^2b+ab^2+ba^2-ab^2+b^3\)
\(=a^3+b^3\)
7) \(\left(a-b\right).\left(a^2+ab+b^2\right)=a.\left(a^2+ab+b^2\right)-b.\left(a^2+ab+b^2\right)\)
\(=a^3+a^2b+ab^2-ba^2-ab^2-b^3\)
\(=a^3-b^3\)
1 a^2+2ab+b^2
2 a^2-2ab+b^2
3 a^2-b^2
4 a^3+3a^2b+3ab^2+b^3
5 a^3-3a^2b+3ab^2-b^3
6 a^3+b^3
7 a^3-b^3
1) (a+b).(a+b)=a^2+ab+ba+b^2
=a^2+2ab+b^2
2)(a-b)^2=(a-b).(a-b)=a^2-ab-ab+b^2=a^2-2ab+b^2
3)(a+b).(a-b)=a^2-ab+ba-b^2=a^2-b^2
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K na
a) \(\left(a+b\right).\left(a-b\right)=a.\left(a-b\right)+b.\left(a-b\right)=a^2-ab+ba-b^2\)\(=a^2-b^2\)
b) \(\left(a+b\right)^3=\left(a+b\right).\left(a+b\right).\left(a+b\right)=a.\left(a+b\right).\left(a+b\right)+b.\left(a+b\right).\left(a+b\right)\)
\(=\left(a^2+ab\right).\left(a+b\right)+\left(ba+b^2\right).\left(a+b\right)\)\(=a^2.\left(a+b\right)+ab.\left(a+b\right)+ba.\left(a+b\right)+b^2.\left(a+b\right)\)
\(=a^3+a^2b+a^2b+ab^2+ba^2+b^2a+b^2a+b^3\)\(=a^3+3a^2b+3ab^2+b^3\)
c) \(\left(a+b\right).\left(a^2-ab+b^2\right)=a.\left(a^2-ab+b^2\right)+b.\left(a^2-ab+b^2\right)\)
\(=a^3-a^2b+ab^2+ba^2-ab^2+b^3\)\(=a^3+b^3\)
d) \(\left(a-b\right).\left(a^2+ab+b^2\right)=a.\left(a^2+ab+b^2\right)-b.\left(a^2+ab+b^2\right)\)
\(=a^3+a^2b+ab^2-ba^2-ab^2-b^3\)\(=a^3-b^3\)
e) \(\left(a-b\right)^3=\left(a-b\right).\left(a-b\right).\left(a-b\right)=a.\left(a-b\right).\left(a-b\right)-b.\left(a-b\right).\left(a-b\right)\)
\(=\left(a^2-ab\right).\left(a-b\right)-\left(ba-b^2\right).\left(a-b\right)\)\(=a^2.\left(a-b\right)-ab.\left(a-b\right)-ba.\left(a-b\right)+b^2.\left(a-b\right)\)
\(=a^3-a^2b-a^2b+ab^2-ba^2+b^2a+b^2a-b^3\)
\(=a^3-3a^2b+3ab^2-b^3\)
1) (a+b).(a+b)=(a+b)2=a2+2ab+b2
2) (a-b)2=a2-2ab+b2
3) (a+b).(a-b)=a2-b2
4) (a+b)3=a3+3a2b+3ab2+b3
5) (a-b)3=a3-3a2b+3ab2-b3
6) (a+b).(a2-ab+b2)=a3+b3
7) (a-b).(a2+ab+b2)=a3-b3
a) Cho \(3x^2-4x=0\)
\(\Rightarrow3.x.x-4x=0\)
\(\Rightarrow x.\left(3x-4\right)\) = 0
\(\left[{}\begin{matrix}x=0\\3x-4=0\end{matrix}\right.\)
Có \(3x - 4 =0\)
\(\Rightarrow3x=4\)
\(\Rightarrow x=\dfrac{4}{3}\)
Vậy x= 0 hoặc x =\(\dfrac{4}{3}\)là nghiệm của đa thức \(3x^2-4x\)
b) Cho \(x+3x^2=0\)
\(\Rightarrow x+3.x.x=0\)
\(\Rightarrow x.\left(3x+1\right)=0\)
Suy ra x =0
hoặc \(3x+1=0\)
\(\Rightarrow\)3x=-1
x=\(\dfrac{-1}{3}\)
Vậy ...
Bài 3: Tìm nghiệm các đa thức sau:
a. 3x2 - 4x
Gọi P(x) là đa thức 3x2 - 4x.
Cho P(x) = 0
=> 3x2 - 4x = 0
=> x (3x - 4)= 0
Suy ra:
TH1: x = 0
TH2: 3x - 4 = 0
_____3x___= 0 + 4
_____3x___= 4
______x___= \(\dfrac{4}{3}\)
Vậy x = \(\dfrac{4}{3}\) là nghiệm của đa thức 3x2 - 4x.
b. x + 3x2
Gọi Q(x) là đa thức x+3x2
Cho Q(x) = 0
=> x+3x2 = 0
=> x ( 3x) = 0
Suy ra:
TH1: x = 0
TH2: 3x = 0
=> x = 0.
Vậy x = 0 là nghiệm của đa thức x + 3x2 .
Chúc bn hx tốt!
Cái này lên lớp 8 mới hok nhưng bạn chịu khó hiểu nha :
\(\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)
Ta thấy dấu - vs dấu + triệt tiêu nha còn :
\(=a^3+b^3\)
Thế là xong
Ủng hộ mik nha
Thnaks
k còn cách khác s