cmr:
A)\(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
B)\(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
Những câu hỏi liên quan
Chứng minh rằng :
a) \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)=2a^3\)
b) \(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
c) \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(a,\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)\)\(=\left(a^3+b^3\right)+\left(a^3-b^3\right)=2a^3\Rightarrowđpcm\)
\(b,\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left(a^2-ab+b^2\right)\)\(=\left(a^3+b^3\right)\Rightarrowđpcm\)
\(c,\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2=\left(a^2c^2+2abcd+b^2d^2\right)+\left(a^2d^2-2abcd+b^2c^2\right)\)\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\Rightarrowđpcm\)
Đúng 1
Bình luận (0)
a) (a+b)(a2-ab+b2)+(a-b)(a2+ab+b2)
= a3+b3+a3-b3 = 2a3
b) a3+b3
= (a+b)(a2-ab+b2)
= (a+b)(a2- 2ab+b2)+ab
= (a+b)(a2-b2)+ab
Đúng 0
Bình luận (0)
a. Biến đổi vế trái:
(a+b)(a2−ab+b2)+(a−b)(a2+ab+b2)=a3+b3+a3−b3=2a3
=>VT bằng VP (đpcm)
b. Biến đổi vế phải:
(a+b)[(a−b)2+ab]=(a+b)[a2−2ab+b2+ab]
=(a+b)(a2−ab+b2)=a3+b3
=>VP bằng VT (đpcm)
c. Biến đổi vế phải:
(ac+bd)2+(ad−bc)2=a2c2+2abcd+b2d2+a2d2−2abcd+b2c2
=a2c2+b2d2+a2d2+b2c2=c(a2+b2)+d2(a2+b2)=(a2+b2)(c2+d2)
=>VP bằng VT (đpcm)
Đúng 0
Bình luận (0)
Bài tập về hằng đẳng thức
CMR:
a) \(\left(a+b\right)\left(a^2-b+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)=2a^3\)
b) \(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
c) \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
cmr:
A)\(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
B)\(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
A) Ta có :
Vế phải = ( a + b ) ( a2 - 2ab + b2 +ab )
= ( a + b ) ( a2 - ab + b2 )
= a3 + b3 = Vế trái ( điều phải chứng minh )
Chúc bạn học tốt ^^
Đúng 0
Bình luận (0)
Câu a) thôi nhé
Ta có (a+b) [(a-b)2+ab] = (a+b)(a2-ab-b2) = a3-a2b + ab2 + ba2 - ab2 +b3
Thu gọn lại ta được a3 + b3
Đúng 0
Bình luận (0)
B ,
Vế trái = a2c2 + a2d2 + b2c2 + b2c2 (1)
Vế phải = [(ac)2 + 2acbd + (bd)2] + [(ad)2 -2adbc + (bc)2]
= a2c2 + 2acbd + b2d2 + a2d2 -2adbc + b2c2
= a2c2 + b2d2 + a2d2 + b2c2 (2)
Từ (1) và (2) suy ra Vế trái = Vế phải ( đpcm )
Đúng 0
Bình luận (0)
Chứng minh rằng :
b) \(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
c) \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
b)
VP=(a+b)[(a-b)2+ab]
=(a+b)(a2-2ab+b2+ab)
=(a+b)(a2-ab+b2)
=a3+b3=VT
Vậy x3+y3=(a+b)[(a-b)2+ab]
c)
VP=(ac+bd)2+(ad-bc)2
=a2c2+2abcd+b2d2+a2d2-2abcd+b2c2
=a2c2+b2d2+a2d2+b2c2
=(a2c2+a2d2)+(b2d2+b2c2)
=a2.(c2+d2)+b2.(c2+d2)
=(a2+b2)(c2+d2)
Vậy (a2+b2)(c2+d2)=(ac+bd)2+(ad-bc)2
Đúng 0
Bình luận (0)
Afrac{a^2+bc}{b+ac}+frac{b^2+ca}{c+ab}+frac{c^2+ab}{a+bc}frac{3left(a^2+bcright)}{left(a+b+cright)b+3ac}+frac{3left(b^2+caright)}{left(a+b+cright)c+3ab}+frac{3left(c^2+abright)}{left(a+b+cright)a+3bc}gefrac{3left(a^2+bcright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(b^2+caright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(c^2+abright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}3
Đọc tiếp
\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
Chứng minh các hằng đẳng thức sau :
a, left(a^2-b^2right)+left(2abright)^2left(a^2+b^2right)^2
b, left(a^2+b^2right).left(c^2+d^2right)left(ac+bdright)^2+left(ad-bcright)^2
c, left(ax+bright)^2+left(a-bxright)^2+c^2x^2left(a^2+b^2+c^2right).left(x^2+1right)
d, dfrac{1}{2}.left(a+b+cright).left[left(a-bright)^2+left(b-cright)^2+left(c-aright)^2right]a^3+b^3+c^3-3abc
e, 1000^2+1003^2+1005^2+1006^21001^2+1002^2+1004^2+1007^2
Đọc tiếp
Chứng minh các hằng đẳng thức sau :
a, \(\left(a^2-b^2\right)+\left(2ab\right)^2=\left(a^2+b^2\right)^2\)
b, \(\left(a^2+b^2\right).\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
c, \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2=\left(a^2+b^2+c^2\right).\left(x^2+1\right)\)
d, \(\dfrac{1}{2}.\left(a+b+c\right).\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=a^3+b^3+c^3-3abc\)
e, \(1000^2+1003^2+1005^2+1006^2=1001^2+1002^2+1004^2+1007^2\)
a: \(\left(a^2-b^2\right)^2+\left(2ab\right)^2\)
\(=a^4-2a^2b^2+b^4+4a^2b^2\)
\(=a^4+2a^2b^2+b^4=\left(a^2+b^2\right)^2\)
b: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
c: \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2\)
\(=a^2x^2+b^2+a^2+b^2x^2+c^2x^2\)
\(=a^2\left(x^2+1\right)+b^2\left(x^2+1\right)+c^2x^2\)
\(=\left(x^2+1\right)\left(a^2+b^2\right)+c^2x^2\)
Đúng 0
Bình luận (0)
Cho a + b + c + d = 0 và ab + bc + ca = 1
Tính \(P=\dfrac{\left(ab-cd\right)\left(bc-ad\right)\left(ac-bd\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\)
left(a+b+cright)^3-a^3-b^3-c^3left[left(a+bright)+cright]^3-a^3-b^3-c^3left(a+bright)^3+3left(a+bright)^2c+3left(a+bright)c^2+c^3-a^3-b^3-c^3a^3+3a^2b+3ab^2+b^3+3cleft(a^2+2ab+b^2right)+3ac^2+3bc^2-a^3-b^33a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^23left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abcright)3left[left(a^2b+ab^2right)+left(a^2c+abcright)+left(ac^2+bc^2right)+left(b^2c+abcright)right]3left[ableft(a+bright)+acleft(a+bright)+c^2left(a+bright)+bcleft(a+bright)right]3left(a+bright)left(ab+ac+c^2+bcrigh...
Đọc tiếp
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc\right)\)
\(=3\left[\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(ac^2+bc^2\right)+\left(b^2c+abc\right)\right]\)
\(=3\left[ab\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+b\right)\)
Châu ơi!đăng làm j z
65. Phân tích đa thức thành nhân tửa) ableft(a+bright)-bcleft(b+cright)+acleft(a-cright)b) aleft(b^2+c^2right)+bleft(c^2+a^2right)+cleft(a^2+b^2right)+2abcc) left(a+bright)left(a^2-b^2right)+left(b+cright)left(b^2+c^2right)+left(c+aright)left(c^2+a^2right)d) a^3left(b-cright)+b^3left(c-aright)+c^3left(a-bright)e) a^3left(c-b^2right)+b^3left(a-c^2right)+c^3left(b-a^2right)+abcleft(abc-1right)
Đọc tiếp
65. Phân tích đa thức thành nhân tử
a) \(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)
b) \(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+2abc\)
c) \(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2+c^2\right)+\left(c+a\right)\left(c^2+a^2\right)\)
d) \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
e) \(a^3\left(c-b^2\right)+b^3\left(a-c^2\right)+c^3\left(b-a^2\right)+abc\left(abc-1\right)\)