toán đầu hè có thưởng!!!!!!!

W

welcome_yoyo

N

nguyenphuongthao28598

MẤY BÀI NÀY CŨNG THƯỜNG ĐÂY LÀ HOÁN VỊ VÒNG c-a=-(b-c)-(a-b) BẠN CỨ LÀM NHƯ VẬY LÀ OK
CÒN PHẦN d THÌ TẠO RA -3abc + abc sử dụng hẳng đẳng thức đệp chắc bạn biết nhóm được (a+b+c) tớ lười làm lên gợi ý bạn tự làm nha!
 
H

hellangel98

a/kết quả:
(b-a)[a^2b+ab^2+abc-c(a+b)^2+c^3]
b/
(a-b)(c-a)(b-c)(b+c+a)
c/
(a-b)(a-c)(b-c)(ab+ac+bc)
d/
(a+b-c)(b+c-a)(a+c-b)
 
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vansang02121998

$a(b+c)^2(b-c)+b(c+a)^2(c-a)+c(a+b)^2(a-b)$

$=a(b+c)^2(b-c)-b(c+a)^2(b-c+a-b)+c(a+b)^2(a-b)$

$=a(b+c)^2(b-c)-b(c+a)^2(b-c)-b(c+a)^2(a-b)+c(a+b)^2(a-b)$

$=(b-c)[a(b+c)^2-b(c+a)^2]-(a-b)[b(c+a)^2-c(a+b)^2]$

$=(b-c)[a(b^2+2bc+c^2)-b(a^2+2ac+c^2)]-(a-b)[b(a^2+2ac+c^2)-c(a^2+2ab+b^2)]$

$=(b-c)(ab^2+2abc+ac^2-a^2b-2abc-bc^2)-(a-b)(a^2b+2abc+bc^2-a^2c-2abc-b^2c)$

$=(b-c)(ab^2+ac^2-a^2b-bc^2)-(a-b)(a^2b+bc^2-a^2c-b^2c)$

$=(b-c)[c^2(a-b)-ab(a-b)]-(a-b)[a^2(b-c)-bc(b-c)]$

$=(b-c)(a-b)(c^2-ab)-(a-b)(b-c)(a^2-bc)$

$=(b-c)(a-b)(c^2-ab-a^2+bc)$

$=(b-c)(a-b)[(c^2-a^2)+(bc-ab)]$

$=(a-b)(b-c)[(c-a)(a+c)+b(c-a)]$

$=(a-b)(b-c)(c-a)(a+b+c)$
 
V

vansang02121998

$a(b-c)^3+b(c-a)^3+c(a-b)^3$

$=a(b-c)^2(b-c)+b(c-a)^2(c-a)+c(a-b)^2(a-b)$

$=a(b-c)^2(b-c)-b(c-a)^2(b-c+a-b)+c(a-b)^2(a-b)$

$=a(b-c)^2(b-c)-b(c-a)^2(b-c)-b(c-a)^2(a-b)+c(a-b)^2(a-b)$

$=(b-c)[a(b-c)^2-b(c-a)^2]-(a-b)[b(c-a)^2-c(a-b)^2]$

$=(b-c)[a(b^2-2bc+c^2)-b(a^2-2ac+c^2)]-(a-b)[b(a^2-2ac+c^2)-c(a^2-2ab+b^2)]$

$=(b-c)(ab^2-2abc+ac^2-a^2b+2abc-bc^2)-(a-b)(a^2b-2abc+bc^2-a^2c+2abc-b^2c)$

$=(b-c)(ab^2+ac^2-a^2b-bc^2)-(a-b)(a^2b+bc^2-a^2c-b^2c)$

$=(b-c)[c^2(a-b)-ab(a-b)]-(a-b)[a^2(b-c)-bc(b-c)]$

$=(b-c)(a-b)(c^2-ab)-(a-b)(b-c)(a^2-bc)$

$=(a-b)(b-c)(c^2-ab-a^2+bc)$

$=(a-b)(b-c)[(c^2-a^2)+(bc-ab)]$

$=(a-b)(b-c)[(c-a)(a+c)+b(c-a)]$

$=(a-b)(b-c)(c-a)(a+b+c)$
 
V

vansang02121998

$a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)$

$=a^2b^2(a-b)-b^2c^2(a-b+c-a)+c^2a^2(c-a)$

$=a^2b^2(a-b)-b^2c^2(a-b)-b^2c^2(c-a)+c^2a^2(c-a)$

$=b^2(a-b)(a^2-c^2)-c^2(c-a)(b^2-a^2)$

$=b^2(a-b)(a-c)(a+c)-c^2(c-a)(b-a)(b+a)$

$=(a-b)(a-c)(ab^2+b^2c)-(a-c)(a-b)(ac^2+bc^2)$

$=(a-b)(a-c)(ab^2+b^2c-ac^2-bc^2)$

$=(a-b)(a-c)[(ab^2-ac^2)+bc(b-c)]$

$=(a-b)(a-c)[(ab+ac)(b-c)+bc(b-c)]$

$=(a-b)(a-c)(b-c)(ab+ac+bc)$
 
W

welcome_yoyo

cảm ơn bạn nhiều nha vansang02121998^^, mong những lần sau bạn có thể giúp mình nhiều hơn nữa:khi (67):
 
C

changruabecon

Còn phần d sao không làm nốt****************************?????? để tớ làm vậy.
 
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changruabecon

welcome_yoyo said:
Phân tích thanh nhân tử
[TEX]a)a(b+c)^2(b-c)+b(c+a)^2(c-a)+c(a+b)^2(a-b)[/TEX]
[TEX]b)a(b-c)^3+b(c-a)^3+c(a-b)^3[/TEX]
[TEX]c)a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)[/TEX]
[TEX]d)a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)-2abc-a^3-b^3-c^3[/TEX]
Bấm để xem đầy đủ nội dung ...

d,$a(b^2+c^2)$ + $b(c^2+a^2)$ + $ c(a^2+b^2)$ - $2abc$ - $a^3$ - $b^3$ - $c^3$
=$a(b^2-2bc+c^2-a^2)$ + $b(a^2+2ac +c^2-b^2)$ + $c(a^2-2ab+b^2-c^2)$
=$a[(b-c)^2-a^2]$ + $b[(a+c)^2-b^2]$ + $c[(a-b)^2-c^2]$
=$-a(c-b+a)(c-b-a)$ + $b(a+c+b)(a+c-b)$ + $c(a-b+c)(a-b-c)$
=$(a+c-b)(ab+bc+b^2+ac-bc-c^2-ac+ab+a^2)$
=$(a+c-b)[(a^2-2ab+b^2)-c^2]$
=$(a+c-b)[(a+b)^2-c^2]$
=$(a+c-b)[(a+b+c)(a+b-c)]$
 
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welcome_yoyo

Nữa nè

Rút gọn các biểu thức:
[TEX]a) 2x(2x-1)^2-3x(x+3)(x-3)-4x(x+1)^2[/TEX]
[TEX]b)(a-b+c)^2-2(3x+1)(3x+5)+(3x+5)^2[/TEX]
[TEX]c)(3x+1)(3^2+1)(3^4+1)(3^8+1)(3^16 +1)(3^32+1)[/TEX]
[TEX]d)(3c+1)^2-2(3x+1)(3x+5)+(3x+5)^2[/TEX]
[TEX]e)((a+b-c)^2+(a-b+c)^2-2(b-c)^2[/TEX]
[TEX]g)(a+b+c)^2+(a-b-c)^2+(b-c-a)^2+(c-a-b)^2[/TEX]
[TEX]h)(a+b+c+d)^2+(a+b-c-d)^2+(a+c-b-d)^2+(a+d-b-d)^2[/TEX]
 
C

changruabecon

welcome_yoyo said:
Rút gọn các biểu thức:
[TEX]a) 2x(2x-1)^2-3x(x+3)(x-3)-4x(x+1)^2[/TEX]
[TEX]b)(a-b+c)^2-2(3x+1)(3x+5)+(3x+5)^2[/TEX]
[TEX]c)(3x+1)(3^2+1)(3^4+1)(3^8+1)(3^16 +1)(3^32+1)[/TEX]
[TEX]d)(3c+1)^2-2(3x+1)(3x+5)+(3x+5)^2[/TEX]
[TEX]e)((a+b-c)^2+(a-b+c)^2-2(b-c)^2[/TEX]
[TEX]g)(a+b+c)^2+(a-b-c)^2+(b-c-a)^2+(c-a-b)^2[/TEX]
[TEX]h)(a+b+c+d)^2+(a+b-c-d)^2+(a+c-b-d)^2+(a+d-b-d)^2[/TEX]
Bấm để xem đầy đủ nội dung ...




Phần d hình như bạn viết nhầm thì phải****************************?????????
 
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changruabecon

$(3x+1)^2$ - $2(3x+1)^2$ - $2(3x+1)(3x+5)$ + $3x+5$
= $(3x+1)(3x+1-2)$ - $(3x+5)(6x+2+3x+5)$
= $(3x+1)(3x-1)$ - $(3x+5)(9x+7)$
= $9x^2$ - $1$ - $27x^2$ - $21x$ - $45x$ -$35$
= $-18x^2$ -$66x$ - $56$
 
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changruabecon

welcome_yoyo said:
Rút gọn các biểu thức:
[TEX]a) 2x(2x-1)^2-3x(x+3)(x-3)-4x(x+1)^2[/TEX]
[TEX]b)(a-b+c)^2-2(3x+1)(3x+5)+(3x+5)^2[/TEX]
[TEX]c)(3x+1)(3^2+1)(3^4+1)(3^8+1)(3^16 +1)(3^32+1)[/TEX]
[TEX]d)(3c+1)^2-2(3x+1)(3x+5)+(3x+5)^2[/TEX]
[TEX]e)((a+b-c)^2+(a-b+c)^2-2(b-c)^2[/TEX]
[TEX]g)(a+b+c)^2+(a-b-c)^2+(b-c-a)^2+(c-a-b)^2[/TEX]
[TEX]h)(a+b+c+d)^2+(a+b-c-d)^2+(a+c-b-d)^2+(a+d-b-d)^2[/TEX]
Bấm để xem đầy đủ nội dung ...


a, $2x(2x-1)^2$ - $3x(x+3)(x-3)$ - $4x(x+1)^2$
= $2x(4x^2-4x+1)$ - $3x(x^2-9)$ - $4x(x^2+2x+1)$
= $8x^3$ - $8x^2$ + $2x$ - $3x^3$ + $27$ - $4x^3$ - $8x^2$ - $4x$
= $x^3$ - $16x^2$ + $25x$


 
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V

veklhy

cac anh cac chi oi giai giup em bai nay vs
1: tren AB cua hinh vuong ABCD lay E tuy y. phan giac goc CDE giao vs BC la K . CM:AE+ KC=DE
 
D

dragon_promise

Mình Đã gặp dàng này rồi. Nhìn khó nhưng thực chất cũng bình thường

Mình Đã gặp dàng này rồi. Nhìn khó nhưng thực chất cũng bình thường.
Như bạn nguyenphuongthao28598 nói :Mình thấy bạn đấy nới đúng đó. Làm theo cách HOÁN VỊ VÒNG là phương pháp dễ nhất để giải bài này. Sau đây là bài giải của mình
a, a(b+c)2(b−c)+b(c+a)2(c−a)+c(a+b)2(a−b)

=a(b+c)2(b−c)−b(c+a)2(b−c+a−b)+c(a+b)2(a - b)

=a(b+c)2(b−c)−b(c+a)2(b−c)−b(c+a)2(a−b)+c(a+b) 2(a−b)

=(b−c)[a(b+c)2−b(c+a)2]−(a−b)[b(c+a)2−c(a+b)2]

=(b−c)[a(b2+2bc+c2)−b(a2+2ac+c2)]−(a−b)[b(a2+2ac+c2)−c(a2+2ab+b2)]

=(b−c)(ab2+2abc+ac2−a2b−2abc−bc2)−(a−b)(a2b+2abc+bc2−a2c−2abc−b2c)

=(b−c)(ab2+ac2−a2b−bc2)−(a−b)(a2b+bc2−a2c−b2c)

=(b−c)[c2(a−b)−ab(a−b)]−(a−b)[a2(b−c)−bc(b−c)]

=(b−c)(a−b)(c2−ab)−(a−b)(b−c)(a2−bc)

=(b−c)(a−b)(c2−ab−a2+bc)

=(b−c)(a−b)[(c2−a2)+(bc−ab)]

=(a−b)(b−c)[(c−a)(a+c)+b(c−a)]

=(a−b)(b−c)(c−a)(a+b+c)

b, a(b-c)³+b(c-a)³+c(a-b)³

=a(b−c)2(b−c)+b(c−a)2(c−a)+c(a−b)2(a−b)

=a(b−c)2(b−c)−b(c−a)2(b−c+a−b)+c(a−b)2(a−b)

=a(b−c)2(b−c)−b(c−a)2(b−c)−b(c−a)2(a−b)+c(a−b)2(a−b)

=(b−c)[a(b−c)2−b(c−a)2]−(a−b)[b(c−a)2−c(a−b)2]

=(b−c)[a(b2−2bc+c2)−b(a2−2ac+c2)]−(a−b)[b(a2−2ac+c2)−c(a2−2ab+b2)]

=(b−c)(ab2−2abc+ac2−a2b+2abc−bc2)−(a−b)(a2b−2abc+bc2−a2c+2abc−b2c)

=(b−c)(ab2+ac2−a2b−bc2)−(a−b)(a2b+bc2−a2c−b2c)

=(b−c)[c2(a−b)−ab(a−b)]−(a−b)[a2(b−c)−bc(b−c)]

=(b−c)(a−b)(c2−ab)−(a−b)(b−c)(a2−bc)

=(a−b)(b−c)(c2−ab−a2+bc)

=(a−b)(b−c)[(c2−a2)+(bc−ab)]

=(a−b)(b−c)[(c−a)(a+c)+b(c−a)]

=(a−b)(b−c)(c−a)(a+b+c)

c,a2b2(a−b)+b2c2(b−c)+c2a2(c−a)

=a2b2(a−b)−b2c2(a−b+c−a)+c2a2(c−a)

=a2b2(a−b)−b2c2(a−b)−b2c2(c−a)+c2a2(c−a)

=b2(a−b)(a2−c2)−c2(c−a)(b2−a2)

=b2(a−b)(a−c)(a+c)−c2(c−a)(b−a)(b+a)

=(a−b)(a−c)(ab2+b2c)−(a−c)(a−b)(ac2+bc2)

=(a−b)(a−c)(ab2+b2c−ac2−bc2)

=(a−b)(a−c)[(ab2−ac2)+bc(b−c)]

=(a−b)(a−c)[(ab+ac)(b−c)+bc(b−c)]

=(a−b)(a−c)(b−c)(ab+ac+bc)

Và câu cuối cùng là câu d

d,a(b2+c2) + b(c2+a2) + c(a2+b2) - 2abc - a3 - b3 - c3

=a(b2−2bc+c2−a2) + b(a2+2ac+c2−b2) + c(a2−2ab+b2−c2)

=a[(b−c)2−a2] + b[(a+c)2−b2] + c[(a−b)2−c2]

=−a(c−b+a)(c−b−a) + b(a+c+b)(a+c−b) + c(a−b+c)(a−b−c)

=(a+c−b)(ab+bc+b2+ac−bc−c2−ac+ab+a2)

=(a+c−b)[(a2−2ab+b2)−c2]

=(a+c−b)[(a+b)2−c2]

=(a+c−b)[(a+b+c)(a+b−c)]
Nhớ Thanks mình nha!!:M012::M012::M012::M012::M012::M012::M012::M012::M012:
 
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K

kiev

toan 8

a, a(b+c)2(b−c)+b(c+a)2(c−a)+c(a+b)2(a−b)

=a(b+c)2(b−c)−b(c+a)2(b−c+a−b)+c(a+b)2(a - b)

=a(b+c)2(b−c)−b(c+a)2(b−c)−b(c+a)2(a−b)+ c(a+b) 2(a−b)

=(b−c)[a(b+c)2−b(c+a)2]−(a−b)[b(c+a)2−c(a+b)2]

=(b−c)[a(b2+2bc+c2)−b(a2+2ac+c2)]−(a−b)[b(a2+2ac+c2)−c(a2+2ab+b2)]

=(b−c)(ab2+2abc+ac2−a2b−2abc−bc2)−(a−b )(a2b+2abc+bc2−a2c−2abc−b2c)

=(b−c)(ab2+ac2−a2b−bc2)−(a−b)(a2b+bc2− a2c−b2c)

=(b−c)[c2(a−b)−ab(a−b)]−(a−b)[a2(b−c)−bc(b−c)]

=(b−c)(a−b)(c2−ab)−(a−b)(b−c)(a2−bc)

=(b−c)(a−b)(c2−ab−a2+bc)

=(b−c)(a−b)[(c2−a2)+(bc−ab)]

=(a−b)(b−c)[(c−a)(a+c)+b(c−a)]

=(a−b)(b−c)(c−a)(a+b+c)

b, a(b-c)³+b(c-a)³+c(a-b)³

=a(b−c)2(b−c)+b(c−a)2(c−a)+c(a−b)2(a−b )

=a(b−c)2(b−c)−b(c−a)2(b−c+a−b)+c(a−b )2(a−b)

=a(b−c)2(b−c)−b(c−a)2(b−c)−b(c−a)2(a −b)+c(a−b)2(a−b)

=(b−c)[a(b−c)2−b(c−a)2]−(a−b)[b(c−a)2−c(a−b)2]

=(b−c)[a(b2−2bc+c2)−b(a2−2ac+c2)]−(a−b)[b(a2−2ac+c2)−c(a2−2ab+b2)]

=(b−c)(ab2−2abc+ac2−a2b+2abc−bc2)−(a−b )(a2b−2abc+bc2−a2c+2abc−b2c)

=(b−c)(ab2+ac2−a2b−bc2)−(a−b)(a2b+bc2− a2c−b2c)

=(b−c)[c2(a−b)−ab(a−b)]−(a−b)[a2(b−c)−bc(b−c)]

=(b−c)(a−b)(c2−ab)−(a−b)(b−c)(a2−bc)

=(a−b)(b−c)(c2−ab−a2+bc)

=(a−b)(b−c)[(c2−a2)+(bc−ab)]

=(a−b)(b−c)[(c−a)(a+c)+b(c−a)]

=(a−b)(b−c)(c−a)(a+b+c)

c,a2b2(a−b)+b2c2(b−c)+c2a2(c−a)

=a2b2(a−b)−b2c2(a−b+c−a)+c2a2(c−a)

=a2b2(a−b)−b2c2(a−b)−b2c2(c−a)+c2a2(c− a)

=b2(a−b)(a2−c2)−c2(c−a)(b2−a2)

=b2(a−b)(a−c)(a+c)−c2(c−a)(b−a)(b+a)

=(a−b)(a−c)(ab2+b2c)−(a−c)(a−b)(ac2+bc2)

=(a−b)(a−c)(ab2+b2c−ac2−bc2)

=(a−b)(a−c)[(ab2−ac2)+bc(b−c)]

=(a−b)(a−c)[(ab+ac)(b−c)+bc(b−c)]

=(a−b)(a−c)(b−c)(ab+ac+bc)

d,a(b2+c2) + b(c2+a2) + c(a2+b2) - 2abc - a3 - b3 - c3

=a(b2−2bc+c2−a2) + b(a2+2ac+c2−b2) + c(a2−ab+b2−c2)

=a[(b−c)2−a2] + b[(a+c)2−b2] + c[(a−b)2−c2]

=−a(c−b+a)(c−b−a) + b(a+c+b)(a+c−b) + c(a−b+c)(a−b−c)

=(a+c−b)(ab+bc+b2+ac−bc−c2−ac+ab+a2)

=(a+c−b)[(a2−2ab+b2)−c2]

=(a+c−b)[(a+b)2−c2]

=(a+c−b)[(a+b+c)(a+b−c)]_


Nho cam on nha
 
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V

vansang02121998

$(a+b+c)^2+(a-b-c)^2+(b-c-a)^2+(c-a-b)^2$

$=a^2+b^2+c^2+2(ab+ac+bc)+a^2+b^2+c^2+2(bc-ac-ab)+a^2+b^2+c^2+2(ac-ab-bc)+a^2+b^2+c^2+2(ab-ac-bc)$

$=4(a^2+b^2+c^2)+2(ab+ac+bc-ab-ac+bc-ab+ac-bc+ab-ac-bc)$

$=4(a^2+b^2+c^2)$




$(a+b+c+d)^2+(a+b-c-d)^2+(a+c-b-d)^2+(a+d-b-c)^2$

$=[(a+b)+(c+d)]^2+[(a+b)-(c+d)]^2+[(a-b)+(c-d)]^2+[(a-b)-(c-d)]^2$

$=(a+b)^2+2(a+b)(c+d)+(c+d)^2+(a+b)^2-2(a+b)(c+d)+(c+d)^2+(a-b)^2+2(a-b)(c-d)+(c-d)^2+(a-b)^2-2(a-b)(c-d)+(c-d)^2$

$=2(a+b)^2+2(c+d)^2+2(a-b)^2+2(c-d)^2$

$=4(a^2+b^2+c^2+d^2)$
 
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