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Ta có:a2+b2=13(1)
ab=6=>2ab=12(2)
Lấy (1)-(2), vế theo vế ta đc:
a2+b2-2ab=13-12
=>a2-2ab+b2=1
=>(a-b)2=1=>a-b=1=>a= b+1
Vậy |a+b|=|2b+1|
mk ko chắc nhé bn
Ta có: \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(=\left(a^2+b^2\right)+\left(2.6\right)\)
\(=13+12\)
\(=25\)
\(\Rightarrow\left(a+b\right)^2=25\)
\(\Rightarrow a+b=\pm5\)
\(\Rightarrow\left|a+b\right|=5\)
Vậy \(\left|a+b\right|=5\)
1) \(\left(A+B\right)^2=\left(A+B\right)\left(A+B\right)=A\left(A+B\right)+B\left(A+B\right)\)
\(=A^2+AB+AB+B^2=A^2+2AB+B^2\)
2) \(\left(A-B\right)^2=\left(A-B\right)\left(A-B\right)=A\left(A-B\right)-B\left(A-B\right)\)
\(=A^2-AB-AB+B^2=A^2-2AB+B^2\)
3) \(A^2-B^2=A^2-AB-B^2+AB\)
\(=A\left(A-B\right)+B\left(A-B\right)=\left(A-B\right)\left(A+B\right)\)
p/s: mấy cái kia tương tự
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) \(\left(a+b\right).\left(a+b\right)=a.\left(a+b\right)+b.\left(a+b\right)=a^2+ab+b^2+ab\)
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.a+a.b+a.b+b.b=a2+2ab+b2
2,(a-b).(a-b)=a.a-a.b-a.b+b.b=a2-2ab+b2
3,(a+b).(a-b)=a.a-a.b+a.b-b.b=a2-b2
4,(a+b)3=(a+b).(a+b)2=(a+b).(a2+2ab+b2)=a3+2a2b+ab2+a2b+2ab2+b3
=a3+3a2b+3ab2+b3
5,(a-b)3=(a-b).(a-b)2=(a-b).(a2-2ab+b2)=a3-2a2b+ab2-a2b+2ab2-b3
=a3-3a2b+3ab2-b3
6,(a+b).(a2-ab+b2)=a3-a2b+ab2+a2b-ab2+b3=a3+b3
7,(a-b).(a2+ab+b2)=a3+a2b+ab2-a2b-ab2-b3=a3-b3
Các bước khai triển cũng đơn giản mà,có gì ko hiểu thì hỏi lại nhé