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Ta có:
M = a³ + b³ + 3ab(a² + b²) + 6a²b²(a + b)
= (a+b)(a² - ab + b²) + 3ab[(a+b)² - 2ab] + 6a²b²(a +b )
= (a+b) [(a +b)² - 3ab] + 3ab[(a+b)² - 2ab] + 6a²b²(a +b )
_______thay a + b = 1 __________________:
M = 1.(1 - 3ab) + 3ab(1 - 2ab) + 6a²b²
M = 1 - 3ab + 3ab - 6a²b² + 6a² b² = 1
a) \(x^2+2x+1=\left(x+1\right)^2\)
b) \(9x^2+y^2+6xy=\left(3x+y\right)^2\)
c) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)
d) \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
e) \(\left(2x+3y\right)^3+2\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
f) mk chỉnh lại đề nha:
\(2xy^2+x^2y^4+1=\left(xy^2+1\right)^2\)
g) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
h) \(x^2-10xy+25y^2=\left(x-5y\right)^2\)
a) Vì \(x-y=1\)
\(\Rightarrow\left(x-y\right)^3=1\)
\(\Leftrightarrow x^3-y^3-3xy\left(x-y\right)=1\)
\(\Leftrightarrow x^3-y^3-3xy=1\)
b) \(B=2\left(x^3-y^3\right)-3\left(x+y\right)^2\)
\(=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(=4\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(=4x^2+4xy+4y^2-3x^2-6xy-3y^2\)
\(=x^2-2xy+y^2\)
\(=\left(x-y\right)^2\)
\(=4\)
1) \(\left(a+b\right)^3=\left(a+b\right)\left(a+b\right)^2=\left(a+b\right)\left(a^2+2ab+b^2\right)\)
\(=a^3+2a^2b+ab^2+a^2b+2ab^2+b^3\)
\(=a^3+3a^2b+3ab^2+b^3\)
2) \(\left(a-b\right)^3=\left(a-b\right)\left(a-b\right)^2=\left(a-b\right)\left(a^2-2ab+b^2\right)\)\(=a^3-2a^2b+ab^2-a^2b+2ab^2-b^3\)
\(=a^3-3a^2b+3ab^2-b^3\)
Có: M = a3 + b3 + 3ab(a2 + b2) + 6a2b2(a + b)
=> M = (a + b)(a2 - ab + b2) + 3ab((a + b)2 - 2ab) + 6a2b2(a + b)
=> M = (a + b)[(a + b)2 - 3ab] + 3ab[(a + b)2 - 2ab] + 6a2b2(a + b)
=> M = 1 - 3ab + 3ab(1 - 2ab) + 6a2b2 (vì a+b=1)
=> M = 1 - 3ab + 3ab - 6a2b2 + 6a2b2
=> M = 1
Vậy M = 1
M = \(a^3\)+ \(b^3\)+ 3ab ( \(a^2\)+ \(b^2\)) + \(6a^2\)\(b^2\)(a+b)
M = ( a + b ) ( \(a^2\)- ab + \(b^2\)) + 3ab [ \(a^2\)+ \(b^2\)+ 2ab( a + b )
M = \(a^2\)- ab + \(b^2\)+ 3ab ( \(a^2\)+ 2ab + \(b^2\))
Với a + b = 1
M= \(a^2\)- ab + \(b^2\)+ 3ab\(\left(a+b\right)^2\)
M = \(a^2\)- ab + \(b^2\)+ 3ab
M = \(a^2\)+ \(b^2\)+ 2ab
M = \(a^2\)+ 2ab + \(b^2\)
M = \(\left(a+b\right)^2\)
M = 1
Vậy M = 1
Câu a : \(-15a^2b^2\left(3ab^2-2a^2b+4b\right)=-45a^3b^4+30a^4b^3-60a^2b^3\)
Câu b : \(\left(a+2\right)\left(a-2\right)-\left(a+1\right)\left(a-1\right)=a^2-4-a^2+1=-3\)
Chúc bạn học tốt
ta có
M = a³ + b³ + 3ab(a² + b²) + 6a²b²(a + b)
= (a+b)(a² - ab + b²) + 3ab[(a+b)² - 2ab] + 6a²b²(a +b )
= (a+b) [(a +b)² - 3ab] + 3ab[(a+b)² - 2ab] + 6a²b²(a +b )
_______thay a + b = 1 __________________:
M = 1.(1 - 3ab) + 3ab(1 - 2ab) + 6a²b²
M = 1 - 3ab + 3ab - 6a²b² + 6a² b² = 1
giải cho ai vậy ông nội :) =_=?