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Bài 209 : đăng tách ra cho mn cùng làm nhé
a,sửa đề : \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)
b, \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2B=3^{64}-1\Rightarrow B=\frac{3^{64}-1}{2}\)
c, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\)
Ta có: a + b + c = 0
<=> a2 + b2 + c2 + 2(ab + bc + ac) = 0
<=> a2 + b2 + c2 = -2(ab + bc + ac)
<=> a4 + b4 + c4 + 2(a2b2 + b2c2 + a2c2 = 4[a2b2 + b2c2 + a2c2 + 2abc(a + b + c)] (vì a + b + c= 0)
<=> a4 + b4 + c4 + 2(a2b2 + b2c2 + a2c2) = 4(a2b2 + b2c2 + a2c2)
<=> a4 + b4 + c4 = 2(a2b2 + b2c2 + a2c2) (đpcm)
b) Từ a4 + b4 + c4 = 2(a2b2 + b2c2 + a2c2)
<=> (a4 + b4 + c4)/2 = a2b2 + b2c2 + a2c2 + 2abc(a + b + c) (vì a + b + c) = 0
<=> (a4 + b4 + c4)/2 = (ab + bc + ac)2
<=> a4 + b4 + c4 = 2(ab + bc + ac)2 (đpcm)
c) Từ a4 + b4 + c4 = 2(a2b2 + b2c2 + a2c2)
<=> 2(a4 + b4 + c4) = a4+ b4 + c4 + 2(a2b2 + b2c2 + a2c2)
<=> 2(a4 + b4 + c4) = (a2 + b2 + c2)2
<=> a4 + b4 + c4 = (a2 + b2 + c2)2/2 (đpcm)
Câu 1: \(a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=\left(a^4+b^4+c^4-2a^2b^2-2c^2a^2+2b^2c^2\right)-4b^2c^2=\left(a^2-b^2-c^2\right)^2-4b^2c^2=\left(a^2-b^2-c^2-2bc\right)\left(a^2-b^2-c^2+2bc\right)=\left[a^2-\left(b+c\right)^2\right]\left[a^2-\left(b-c\right)^2\right]=\left(a-b-c\right)\left(a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)\)Câu 2: \(a^3+a^2-ab^2-b^2=a^2\left(a+1\right)-b^2\left(a+1\right)=\left(a^2-b^2\right)\left(a+1\right)=\left(a+b\right)\left(a-b\right)\left(a+1\right)\)
Câu 3: \(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=a\left(b^3-c^3\right)-b\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]+c\left(a^3-b^3\right)=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(b-c\right)\left[b\left(c-a\right)+\left(c-a\right)\left(c+a\right)\right]=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
Câu 1.
a4 + b4 + c4 - 2a2b2 - 2b2c2 - 2a2c2
= [ ( a4 - 2a2b2 + b4 ) - 2a2c2 + 2b2c2 + c4 ] - 4b2c2
= [ ( a2 - b2 )2 - 2( a2 - b2 )c2 + ( c2 )2 ] - ( 2bc )2
= ( a2 - b2 - c2 ) - ( 2bc )2
= ( a2 - b2 - c2 - 2bc )( a2 - b2 - c2 + 2bc )
= [ a2 - ( b2 + 2bc + c2 ) ][ a2 - ( b2 - 2bc + c2 ) ]
= [ a2 - ( b + c )2 ][ a2 - ( b - c )2 ]
= ( a - b - c )( a + b + c )( a - b + c )( a + b - c )
Câu 2.
a3 + a2 - ab2 - b2
= a2( a + 1 ) - b2( a + 1 )
= ( a + 1 )( a2 - b2 )
= ( a + 1 )( a - b )( a + b )
ta có
a. \(ab+bc+ac\le a^2+b^2+c^2\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\) luôn đúng
mà ta lại có :
\(\hept{\begin{cases}a^2< a\left(b+c\right)\\b^2< b\left(a+c\right)\\c^2< c\left(a+b\right)\end{cases}\Rightarrow a^2+b^2+c^2\le2\left(ab+bc+ac\right)}\) vậy ta có điều phải chứng minh.
b. ta có :
\(\hept{\begin{cases}\left(a+b-c\right)\left(a+c-b\right)\le\left(\frac{a+b-c+a+c-b}{2}\right)^2=a^2\\\left(a+b-c\right)\left(b+c-a\right)\le b^2\\\left(a+c-b\right)\left(b+c-a\right)\le c^2\end{cases}}\)
nhân lại ta sẽ có : \(\left(a+b-c\right)\left(a+c-b\right)\left(b+c-a\right)\le abc\) vậy ta có dpcm
b. Xét vế trái \(\left(a^2+b^2\right)\left(x^2+y^2\right)\)
\(=a^2x^2-2axby+b^2y^2+a^2y^2+2aybx+b^2x^2\)
\(=\left(ax-by\right)^2+\left(ay+bx\right)^2\)(đpcm)
c. Xét vế trái \(a^3-b^3+ab\left(a-b\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+ab\left(a-b\right)\)
\(=\left(a-b\right)\left(a^2+2ab+b^2\right)\)
\(=\left(a-b\right)\left(a+b\right)^2\)(đpcm)