Cho a+b = 1 . Tính M = a^3 + b^3 - 3ab ( a^2 + b^2 ) + 6a^2 b^2 ( a+b)
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Ta có :
M = 2( a3 + b3 ) - 3( a2 + b2 )
= 2( a + b ) ( a2 - ab + b2 ) - 3( a2 + b2 )
= 2( a2 - ab + b2 ) - 3 ( a2 + b2 )
= 2a2 - 2ab + 2b2 - 3a2 - 3b2
= -a2 - 2ab - b2
= - ( a + b )2
= -1
Ta có: \(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\cdot\left(a+b\right)\)
\(\Leftrightarrow M=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+b^2\right)+6a^2b^2\)
\(\Leftrightarrow M=a^2-ab+b^2+3ab\left(a^2+2ab+b^2\right)\)
\(\Leftrightarrow M=a^2-ab+b^2+3ab\cdot\left(a+b\right)^2\)
\(\Leftrightarrow M=a^2-ab+3ab+b^2\)
\(\Leftrightarrow M=\left(a+b\right)^2=1^2=1\)
Vậy: Khi a+b=1 thì M=1
M=(a+b)^3-3ab(a+b)+3ab[(a+b)^2-2ab]+6a^2b^2
=1-3ab+3ab(1-2ab)+6a^2b^2
=1
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a^2+b^2\right)+6a^2b^2\)
\(=1-3ab+3ab\cdot\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)
\(=1-3ab-6a^2b^2+6a^2b^2=1-3ab\)
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\\ M=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a^2+b^2\right)+6a^2b^2\\ M=1-3ab+3ab\left(a^2+b^2+2ab\right)=1-3ab+3ab\left(a+b\right)^2\\ M=1-3ab+3ab=1\)
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab\left[1-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)
=1
ta có : M=2.(a^3 +b^3) -3.(a^2 + b^2)
<=>M=2.(a+b)(a^2 -ab +b^2) - 3(a^2 +3b^2)
<=>M=2(a^2 -ab +b^2) -3(a^2 +b^2) vì a+b=1(gt)
<=>M=-(a^2 +b^2 +2ab)
<=>M=-(a+b)^2
<=>M=-1 (vì a+b=1)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+b^2+2ab\right)\)
\(=a^2-ab+b^2+3ab\left(a+b\right)^2=a^2-ab+b^2+3ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2=1\)
M=1 ( chtt ) có đó vô mà tham khảo