Cho a-b=1.Cmr a^3-b^3=1-3ab
CMR :1,a2+b2=<a+b>2-2ab
2,a3+b3=<a+b>3-3ab.<a+b>
3,a3-b3=<a-b>3+3ab.<a+b>
Cho :a+b=1
Tính :A=a3+b3+3ab
2
Ta có:
VP=(a+b)3−3ab(a+b)VP=(a+b)3-3ab(a+b)
=a3+b3+3ab(a+b)−3ab(a+b)=a3+b3+3ab(a+b)-3ab(a+b)
=a3+b3=VT(dpcm)
1, \(VT=a^2+b^2=a^2+b^2+2ab-2ab=\left(a+b\right)^2-2ab=VP\left(đpcm\right)\)
Cho a+b=1. CMR a3+b3=1-3ab
Ta có: a+b=1(1)
=> (a+b)3=1
<=> \(a^3+3a^2b+3ab^2+b^3=1\)
<=> \(a^3+b^3+3ab\left(a+b\right)=1\)(2)
Từ (1)(2)=> \(a^3+b^3+3ab=1\)
<=> \(a^3+b^3=1-3ab\)(đpcm)
Cho \(\left(a+\sqrt{a^2+2}\right)\left(b-1+\sqrt{b^2-2b+3}\right)=2\)
CMR \(a^3+b^3+3ab=1\)
Cho \(\left(a+\sqrt{a^2+2}\right)\left(b-1+\sqrt{b^2-2b+3}\right)=2\)
CMR \(a^3+b^3+3ab=1\)
cmr (a+b)^3=a^3-a^2b+3ab^2-b^3 ; (a-b)^3= a^ - 3a^2b+3ab^2-b^3 ;
cho a, b dương thõa mãn : \(a^3+b^3=3ab-1\)
CMR:\(a^{2018}+b^{2018}=2\)
cho a-b=2.cmr a^3-b^3+2(4+3ab)
Cho a,b>0 thoả mãn a3+b3=3ab-1
CMR: a2018+b2019=2
Thanks nha! I love you!
\(a^3+b^3=3ab-1\)
\(\Rightarrow a^3+b^3+1-3ab=0\)
\(\Rightarrow\left(a+b\right)^3+1-3ab\left(a+b\right)-3ab=0\)
\(\Rightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1\right)-3ab\left(a+b\right)=0\)
\(\Rightarrow\left(a+b+1\right)\left(a^2-ab+b^2-a-b+1\right)=0\)
Mà \(a,b>0\Rightarrow a+b+1>0\)
\(\Rightarrow a^2-ab+b^2-a-b+1=0\)
\(\Rightarrow2a^2-2ab+2b^2-2a-2b+2=0\)
\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\)
\(\Rightarrow a=b=1\Rightarrow a^{2018}+b^{2019}=1+1=2\)
1. Tìm y, biết:
a) ( y + 3 )3 - ( y - 1 )3
2. Cho
m + n = 7. Tính: A= ( m + n )3 + 2m2 + 4mn + 2n2
3. Cho
a + b = 1. CMR: a3 + b3 = 1 - 3ab
1) \(\left(y+3\right)^3-\left(y-1\right)^3\)
=(y+3-y+1)\(\left[\left(y+3\right)^2+\left(y+3\right)\left(y-1\right)+\left(y-1\right)^2\right]\)
=4.(\(y^2+6y+9\)+\(y^2-y+3y-3\)+\(y^2-2y+1\))
=4(\(3y^2+6y+7\))
=\(12y^2+24y+28\)
3.
\(a^3+b^3=\left(a+b\right).\left(a^2-ab+b^2\right)\)
\(=1.\left(a^2+b^2-ab\right)\) (1)
Lại có : \(a^2+b^2=\left(a+b\right)^2-2ab=1-2ab\) thay vào (1) có :
\(a^3+b^3=1.\left(1-2ab-ab\right)\)
\(=1-3ab\left(đpcm\right)\)
2.
\(A=\left(m+n\right)^3+2m^2+4mn+2n^2\)
\(=\left(m+n\right)^3+\left(\sqrt{2}m\right)^2+2.\sqrt{2}m.\sqrt{2}n+\left(\sqrt{2}n\right)^2\)
\(=\left(m+n\right)^3+\left(\sqrt{2}m+\sqrt{2}n\right)^2\)
\(=7^3+\left(\sqrt{2}.7\right)^2=343+98=441\)
( Do \(m+n=7\) )