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Lời giải :
a) \(VP=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
\(=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=a^3+b^3=VT\)( đpcm )
b) \(VT=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2=VP\)( đpcm )
a)CM \(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
VT = \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
VP = \(\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left(a^2-ab+b^2\right)\)
Ta thấy VP = VT
=> \(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
b) CM \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
VT = \(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
VP = \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=ac^2+2acbd+bd^2+ad^2-2abcd+bc^2=ac^2+ad^2+bd^2+bc^2\)Ta thấy VP = VT
=> \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
a) \(a^2+b^2+c^2=ab+bc+ca\)
<=> \(2a^2+2b^2+2c^2=2ab+2bc+2ca\)
=> \(a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc-2ca=0\)
<=> (a2 - 2ab + b2) + (a2 - 2ac + c2) + (b2 -2bc + c2) = 0
<=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\) (1)
Mà \(\left(a-b\right)^2\ge0\); \(\left(a-c\right)^2\ge0\); \(\left(b-c\right)^2\ge0\) (2)
Từ (1); (2) =>
+ \(\left(a-b\right)^2=0\Leftrightarrow a=b\)
+ \(\left(a-c\right)^2=0\Leftrightarrow a=c\)
+ \(\left(b-c\right)^2=0\Leftrightarrow b=c\)
=> a = b = c => đpcm
2 Câu dưới tương tự bài a bn nhé
a) VT = (a+b)(\(a^2-ab+b^2\)) + \(\left(a-b\right)\left(a^2+ab+b^2\right)=a^3+b^3\)\(+a^3-b^3\) = \(2a^3=VP\) (đpcm)
b, VP =\(\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=\left(a+b\right)\left[a^2-2ab+b^2+ab\right]=\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3=VT\left(đpcm\right)\)
c, Ta có : \(VT=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)(1)
\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2acbd+b^2d^2+a^2d^2-2adbc+b^2c^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2\) (2)
Từ (1) và (2), ta có \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\left(đpcm\right)\)
a) Biến đổi VT ta có :
(a2-b2)2 + (2ab)2
= a4 -2a2+b4+4a2b2
= a4+2a2b2 +b4
= (a2b2)2 = VP (đpcm)
b) Biến đổi vế trái ta có :
(ax+b)2 + (a-bx)2+cx2+c2
= a2x2+2axb+b2 +a2 - 2axb+b2x2 +c2x2+ c2
= (a2+b2+c2) + x2(a2+b2+c2)
= (a2+b2+c2) (x2+1) = VP (đpcm)
a: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2+2bacd+a^2d^2+b^2c^2-2bacd\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
b: \(\Leftrightarrow2a^2+2b^2+2c^2=2ba+2ac+2bc\)
=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
=>(a-b)^2+(b-c)^2+(a-c)^2=0
=>a=b=c
2/ Ta có \(\left(a+b+c+d\right)^2\ge\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right)\)
\(\Leftrightarrow a^2+b^2+c^2+d^2+2\left(ab+ac+ad+bc+bd+cd\right)\ge\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2+d^2\right)+6\left(ab+ac+ad+bc+bd+cd\right)\ge8\left(ab+ac+ad+bc+bd+cd\right)\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(a^2-2ad+d^2\right)+\left(b^2-2bc+c^2\right)+\left(b^2-2bd+d^2\right)+\left(c^2-2cd+d^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(c-d\right)^2\ge0\)(luôn đúng)
Vậy bđt ban đầu được chứng minh.
a, Ta có: \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)\)
= \(a^3+b^3+a^3-b^3=a^3+a^3=2a^3\)
\(\xrightarrow[]{}\) đpcm
b, Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left(\left(a-b\right)^2+ab\right)\)
\(\xrightarrow[]{}\) đpcm
c, Ta có: \(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(\xrightarrow[]{}\) đpcm
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