cho a+b=1. Tính M=\(2\left(a^3+b^3\right)-3\left(a^2+b^2\right)\)
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ND
6
29 tháng 4 2018
\(M=2\left(a+b\right)\left(a^2-ab+b^2\right)-3\left(a^2+b^2\right)\)
\(=2\left(a^2-ab+b^2\right)-3\left(a^2+b^2\right)=2a^2-2ab+2b^2-3a^2-3b^2\)
\(=-a^2-2ab-b^2=-\left(a^2+2ab+b^2\right)=-\left(a+b\right)^2=-1^2=-1\)
KT
29 tháng 4 2018
\(M=2\left(a^3+b^3\right)-3\left(a^2+b^2\right)\)
\(=2a^3+2b^3-3a^2-3b^2\)
\(=a^2\left(2a-3\right)+b^2\left(2b-3\right)\)
\(=a^2\left[2a-3\left(a+b\right)\right]+b^2\left[2b-3\left(a+b\right)\right]\) (do a+b=1 )
\(=a^2\left(2a-3a-3b\right)+b^2\left(2b-3a-3b\right)\)
\(=a^2\left(-a-3b\right)+b^2\left(-b-3a\right)\)
\(=-a^3-3a^2b-b^3-3ab^2\)
\(=-\left(a^3+3a^2b+3ab^2+b^3\right)\)
\(=-\left(a+b\right)^3\)
\(=-3\)
19 tháng 5 2022
Ta có \(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2021}\left(1\right)\)
\(\Rightarrow\dfrac{3}{2}A=\dfrac{3}{4}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2013}\left(2\right)\)
Lấy (2) - (1) ta được:
\(\dfrac{3}{2}A-A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}\)
\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{1}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}+\dfrac{1}{2}\)
Vậy \(B-A=\dfrac{3^{2013}}{2^{2014}}-\dfrac{3^{2013}}{2^{2012}}+\dfrac{5}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 7 2021
\(A=2\left(a+b\right)^3-6ab\left(a+b\right)-3\left(a+b\right)^2+6ab\)
\(=2-6ab-3+6ab=-1\)
30 tháng 1 2024
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab\left[1-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)
=1
AT
24 tháng 8 2018
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
9 tháng 12 2019
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
9 tháng 12 2019
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
\(M=2\left(a^3+b^3\right)-3\left(a^2+b^2\right)\)
\(=2\left(a+b\right)\left(a^2+ab+b^2\right)-3a^2-3b^2\)
\(=2a^2+2ab+2b^2-3a^2-3b^2\)
\(=-a^2+2ab-b^2\)
\(=-\left(a^2-2ab+b^2\right)\)
\(=-\left(a-b\right)^2\)
\(=-\left(1-b-b\right)^2=-\left(1-2b\right)^2\)