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Bài 2:
a: \(A=1999\cdot2001\)
\(=\left(2000-1\right)\left(2000+1\right)\)
\(=2000^2-1< 2000^2=B\)
Do đó: B lớn hơn
b: \(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1< 2^{16}=D\)
Do đó: D lớn hơn
Bài 1 :
\(=\left(x^3-x\right)-\left(6x+6\right)\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=\left(x^2-x\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x^2-x-6\right)\left(x+1\right)\)
a, ta có (a2+ b2)2 = (a2)2 + 2a2b2+ (b2)2( hđt)
= (a2)2 + 2a2b2+(b2)2 - 4a2b2 + 4a2b2
=(a2)2 - 2a2b2 + (b2)2 + 4a2b2
= (a2-b2)2 + 22a2b2 = (a2 -b2) + (2ab)2
vậy .........
b, ta co :(ax +b)2 +(a-bx)2 + c2x2 + c2 = [(ax)2 + 2axb +b2 ] + [ a2 -2abx + (bx)2 ] + (cx)2 + c2 = (ax)2+ (bx)2 + (cx)2 +a2 + b2 +c2 +( 2axb- 2axb)
= x2.(a2+b2+c2) + (a2+b2+c2)
= (x2+1) . (a2+b2+c2)
vay .................
a.) \\(\\left(a+b+c\\right)^3-a^3-b^3-c^3\\)
\\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc-a^3-b^3-c^3\\)\\(=3\\left(3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\right)\\)
\\(=3\\left(abc+a^2b+a^2c+ac^2+b^2c+ab^2+abc+bc^2\\right)\\)
\\(=3\\left[ab\\left(a+c\\right)+ac\\left(a+c\\right)+b^2\\left(a+c\\right)+bc\\left(a+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(ab+ac+bc+b^2\\right)\\)
\\(=3\\left(a+c\\right)\\left[a\\left(b+c\\right)+b\\left(b+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(a+b\\right)\\left(b+c\\right)\\)
b) 4a2b2-(a2 +b2-c2)2
=（2ab+a2+b2-c2)(2ab-a2-b2+c2）
=[(a+b)2-c2][c2-(a-b)2]
=(a+b+c)(a+b-c)(c+a-b)(c-a+b)
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc-a^3-b^3-c^3\)
\(=3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)
\(=3\left(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\right)\)
\(=3\left(ab\left(a+b\right)+b^2c+abc+bc^2+c^2a+ca^2+abc\right)\)
\(=3\left(ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\right)\)
\(=3\left(a+b\right)\left(ab+bc+c^2+ac\right)\)
\(=3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
Bài 1:
\(a,\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^2\left(x^2-1\right)\)
\(b,\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3+x^2\right)^3\)
\(=x^6+27-27-27x^2-9x^4-x^6\)
\(=-9x^2\left(3-x^2\right)\)
Bài 5:
\(A=x^2-2x+1\)
\(=\left(x^2-2x+1\right)-2\)
\(=\left(x-1\right)^2-2\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2-2\ge-2\)
Vậy Min A = -2
Để A = -2 thì \(x-1=0\Rightarrow x=1\)
b, \(B=4x^2+4x+5\)
\(=\left(4x^2+4x+1\right)+4\)
\(=\left(2x+1\right)^2+4\)
Với mọi giá trị của x ta có:
\(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+4\ge4\)
Vậy Min B = 4
Để B = 4 thì \(2x+1=0\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
c, \(C=2x-x^2-4\)
\(=-\left(x^2-2x+1\right)-3\)
\(=-\left(x-1\right)^2-3\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-3\le-3\)Vậy Max C = -3
để C = -3 thì \(x-1=0\Rightarrow x=1\)
\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
Ta có:
\(\left(a-1\right)^2\ge0\Leftrightarrow a^2-2a+1\ge0\Leftrightarrow a^2+1\ge2a\) (1)
\(\left(b-1\right)^2\ge0\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow b^2+1\ge2b\) (2)
\(\left(c-1\right)^2\ge0\Leftrightarrow c^2-2c+1\ge0\Leftrightarrow c^2+1\ge2c\) (3)
Từ (1), (2) và (3) suy ra;
\(a^2+1+b^2+1+c^2+1\ge2a+2b+2c\)
<=> \(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)
<=> \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3\) \(\xrightarrow[]{}\) đpcm
Dấu "=" xảy ra khi a=b=c=1
Ta có: \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
\(\Leftrightarrow a^2-2a+1+b^2-2b+1+c^2-2c+1\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge+2a+2b+2c-3\)
\(\Leftrightarrow a^2+b^2+c^2\ge2\left(a+b+c\right)-3\) (đpcm)
Vậy \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3\)
a) \(x^2-6x+3\)
\(=x^2-2.x.3+9-6\)
\(=\left(x-3\right)^2-\left(\sqrt{6}\right)^2\)
\(=\left(x-3-\sqrt{6}\right)\left(x-3+\sqrt{6}\right)\)
b) \(9x^2+6x-8\)
\(=\left(3x\right)^2+2.3x+1-9\)
\(=\left(3x+1\right)^2-3^2\)
\(=\left(3x+1-3\right)\left(3x+1+3\right)\)
\(=\left(3x-2\right)\left(3x+4\right)\)
d) \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
e) \(x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-3x-x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+8\right)\left(x-3\right)\left(x-1\right)\)
c, Ta có:
\(VP=\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=\left(a+b\right)^3+c^3+3\left(a+b\right).c.\left(a+b+c\right)\)
\(=a^3+b^3+3a^2b+3ab^2+c^3+3\left(a+b\right).c.\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3ab.\left(a+b\right)+3\left(a+b\right).c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right).\left[ab+c.\left(a+b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right).\left(ab+ac+cb+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right).\left(ab+ac\right)+\left(cb+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right).a.\left(b+c\right)+c.\left(b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=VT\)
\(\rightarrow\) đpcm
Chúc bạn học tốt!!!
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)
\( (a + b)(a^2 - ab + b^2) + (a - b)(a^2 + ab + b^2) = 2a^3 = a^3 + b^3 + a^3 - b^3 = 2a^3\)
b)
\(a^3 + b^3 = (a + b)(a^2 - ab + b^2) = (a + b)(a^2 - (2ab - ab) + b^2) = (a + b)(a^2 - 2ab + b^2 + ab) = (a + b)[(a - b)^2 + ab] \)