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a) a2 + b2 = a2 + b2 + 2ab - 2ab = (a + b)2 - 2ab = s2 - 2p
b) a3 + b3 = (a + b)(a2 - ab + b2) = (a + b)(a2 + 2ab + b2 - 3ab) = (a + b).[(a + b)2 - 3ab] = s.(s2 - 3p) = s3 - 3ps
c) a4 + b4 = a4 + b4 + 4a2b2 - 4a2b2 = (a2 + b2)2 - 4(ab)2 = (s2 - 2p)2 - 4p2
= (s2 - 2p - 2p)(s2 - 2p + 2p) = s2.(s2 - 4p) = s4 - 4ps2
a, \(a^2+b^2=\left(a+b\right)^2-2ab\)
Thay a+b=s; ab vào đa thức trên ta được:
\(\left(a+b\right)^2-2ab=s^2-2p\)
b, \(a^3+b^3=\left(a+b\right)^3+3a^2b-3ab^2\)
\(=\left(a+b\right)^3-3ab.\left(a+b\right)\)
Thay a+b=s; ab=p Ta được:
\(\left(a+b\right)^3-3ab.\left(a+b\right)=s^3-3sp\)
c, \(a^4+b^4=\left(x^2+y^2\right)^2-2x^2y^2\)
\(=\left(s^2-2p\right)^2-2p^2=s^4-4s^2p+2p^2\)
CHÚC HỌC TỐT!!
A = a2 + b2
= a2 + 2ab + b2 - 2ab
= ( a + b )2 - 2ab
= S2 - 2P
B = a3 + b3
= a3 + 3a2b + 3ab2 + b3 - 3a2b - 3ab2
= ( a + b )3 - 3ab( a + b )
= S3 - 3PS
= S( S2 - 3P )
C = a4 + b4
= ( a2 )2 + 2a2b2 + ( b2 )2 - 2a2b2
= ( a2 + b2 )2 - 2( a.b )2
=( a2 + 2ab + b2 - 2ab )2 - 2P2
= [ ( a + b )2 - 2ab ]2 - 2P2
= [ S2 - 2P ]2 - 2P2
= S4 - 4PS2 + 4P2 - 2P2
= S4 - 4PS + 2P2
Bài 1: \(A=2x^2-8x=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4\right)-8=2\left(x-2\right)^2-8\ge-8\)
Vậy MinA= -8 \(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
\(B=3x^2-3x=3\left(x^2-x\right)=3\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{3}{4}\)
\(=3\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\ge-\dfrac{3}{4}\)
Vậy \(Min_B=-\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
\(C=x^2+y^2-2x+4y+7=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+2\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+2\ge2\)
Vậy \(Min_C=2\Leftrightarrow x=1;y=-2\)
\(D=x^2+4y^2+x+4y+2=\left(x^2+x+\dfrac{1}{4}\right)+\left(4y^2+4y+1\right)+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\left(2y+1\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy \(Min_D=\dfrac{3}{4}\Leftrightarrow x=y=-\dfrac{1}{2}\)
Bài 2: \(A=x-x^2=-\left(x^2-x\right)=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Vậy \(Max_A=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
\(B=3x-2x^2=-2\left(x^2-\dfrac{3}{2}x\right)\)
\(=-2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{9}{8}\)
\(=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\)
Vậy \(Max_B=\dfrac{9}{8}\Leftrightarrow x=\dfrac{3}{4}\)
\(C=2x-2x^2-3=-2\left(x^2-x+\dfrac{3}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{5}{4}\right)=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{2}\le-\dfrac{5}{2}\)
Vậy \(Max_C=-\dfrac{5}{2}\Leftrightarrow x=\dfrac{1}{2}\)
a)
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)-3abc+c^3\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[a^2+b^2+c^2-ab-bc-ca\right]\)
\(=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
b/
\(a+b+c=0\Rightarrow c=-\left(a+b\right)\Rightarrow c^2=\left(a+b\right)^2\)
\(\Leftrightarrow c^2=a^2+b^2+2ab\)\(\Leftrightarrow a^2+b^2+ab=c^2-ab\)
\(2x^4=\left(a^2+b^2+ab\right)^2+\left(c^2-ab\right)^2\)
\(=a^4+b^4+a^2b^2+2a^2b^2+2a^3b+2ab^3+c^4-2abc^2+a^2b^2\)
\(=a^4+b^4+c^4+\left(4a^2b^2+2a^3b+2ab^3-2abc^2\right)\)
\(=a^4+b^4+c^4+2ab\left(2ab+a^2+b^2-c^2\right)\)
\(=a^4+b^4+c^4+0\)
\(=a^4+b^4+c^4\)
Với a,b,c ko âm
a^2 = b^2 + c^2 (1)
=> a^2 = (b+c)^2 - 2bc
=> a^2 <= (b+c)^2
=> a <= b+c (2)
Nhân (1) với (2), vế theo vế ta có:
a^3 = b^3 + c^3 + bc(b+c)
=> a^3 >= b^3 + c^3
Ta có:
a2 + b2=(a+b)2-2ab=S2-2P
a3+b3=(a+b)3-3ab(a+b)=S3-3.S.P
a4+b4=(a2+b2)2-2a2b2=(S2-2P)2-2S2
a. \(A=a^2+b^2=\left(a+b\right)^2-2ab=s^2-2p\)
b. \(B=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=s\left(s^2-2p-p\right)=s\left(s^2-3p\right)\)
c. \(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=\left(s^2-2p\right)^2-2p^2=s^4-4s^2p+2p^2\)