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\(A=\left[\left(a+b\right)+\left(c+d\right)\right]^2+\left[\left(a+b\right)-\left(c+d\right)\right]^2+\left[\left(a-b\right)+\left(c-d\right)\right]^2+\left[\left(a-b\right)-\left(c-d\right)\right]^2\)
Ta có
\(\left[\left(a+b\right)+\left(c+d\right)\right]^2=\left(a+b\right)^2+2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
\(\left[\left(a+b\right)-\left(c+d\right)\right]^2=\left(a+b\right)^2-2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
\(\left[\left(a-b\right)+\left(c-d\right)\right]^2=\left(a-b\right)^2+2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)
\(\left[\left(a-b\right)-\left(c-d\right)\right]^2=\left(a-b\right)^2-2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)
\(A=2\left(a+b\right)^2+2\left(a-b\right)^2+2\left(c+d\right)^2+2\left(c-d\right)^2\)
\(A=2\left(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2\right)\)
\(A=4\left(a^2+b^2+c^2+d^2\right)\)
Cách khác cho bài 1, 2 nha! Akai Haruma em tháy nó nhanh hơn!
1/Đặt \(a=x;b-c=y\)
biểu thức trở thành \(\left(x+y\right)^2+\left(x-y\right)^2-2y^2=2\left(x^2+y^2\right)-2y^2=2x^2=2a^2\)
2/ Đặt \(a-b-c=x;b-c-a=y;c-a-b=z\Rightarrow\left(a+b+c\right)^2=\left(-\left(a+b+c\right)\right)^2=\left(x+y+z\right)^2\)
Khi đó \(B=\left(x+y+z\right)^2+x^2+y^2+z^2\)
\(=2\left(x^2+y^2+z^2+xy+yz+zx\right)\)
\(=\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\)
\(=4\left(a^2+b^2+c^2\right)\)(thay x, y, z bởi các biến đã đặt rồi rút gọn thôi:))
Lời giải:
1.
\((a+b-c)^2+(a-b+c)^2-2(b-c)^2\)
\(=a^2+b^2+c^2+2ab-2ac-2bc+a^2+b^2+c^2-2ab+2ac-2bc-2(b^2-2bc+c^2)\)
\(=2(a^2+b^2+c^2)-4bc-2(b^2+c^2)+4bc\)
\(=2a^2\)
2.
\((a+b+c)^2+(a-b-c)^2+(b-c-a)^2+(c-a-b)^2\)
\(=(a+b+c)^2+a^2+(b+c)^2-2a(b+c)+b^2+(a+c)^2-2b(a+c)+c^2+(a+b)^2-2c(a+b)\)
\(=(a+b+c)^2+a^2+b^2+c^2+[(a+b)^2+(b+c)^2+(c+a)^2]-4(ab+bc+ac)\)
\(=a^2+b^2+c^2+2(ab+bc+ac)+a^2+b^2+c^2+(2a^2+2b^2+2c^2+2ab+2bc+2ac)-4(ab+bc+ac)\)
\(=4(a^2+b^2+c^2)\)
3.
\((a+b+c+d)^2+(a+b-c-d)^2+(a+c-b-d)^2+(a+d-b-c)^2\)
\(=(a+b)^2+(c+d)^2+2(a+b)(c+d)+(a+b)^2+(c+d)^2-2(a+b)(c+d)+(a-b)^2+(c-d)^2+2(a-b)(c-d)+(a-b)^2+(d-c)^2+2(a-b)(d-c)\)
\(=2(a+b)^2+2(c+d)^2+2(a-b)^2+2(c-d)^2\)
\(=2[(a+b)^2+(a-b)^2+(c+d)^2+(c-d)^2]\)
\(=2(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2)\)
\(=2(2a^2+2b^2+2c^2+2d^2)=4(a^2+b^2+c^2+d^2)\)
2(a+b)2+2(c+d)2+2(a−b)2+2(d−c)2=2(2a2+2b2+2d2+2c2=4(∑a2)⇔2(a+b)2+2(c+d)2+2(a−b)2+2(d−c)2=2(2a2+2b2+2d2+2c2=4(∑a2)
a)(a2+b2+c2)2- (a2-b2-c2)2 = ((a2)+(b2)+(c2) + 2ab + 2ac+2bc)2-((a2)+(b2)+(c2)-2ab-2ac+2c)2
=4ab +4ac
b)(a+b+c)2- (a-b-c)2-4ac = (a2+b2+c2+2ab+2ac+2bc) - (a2+b2+c2- 2ab - 2ac +2bc)
= (2ab + 2ac) - [(-2ab) - 2ac)=..........
c)(a+b+c)2-(a+b)2- (a+c)2- (b+c)2= (a2+b2+c2+2ab+2ac+2bc)-(a2+b2+2ab)-(a2+c2+2ac)-(b2+c2+2bc)
= a2 + b2 + c2
d)(a+b+c)2+(a-b+c)2+(a+b-c)2+(-a+b+c)2 = (a2+b2+c2+2ab+2ac+2bc) +(a2+b2+c2-2ab-2ac+2bc)+(a2+b2+c2+2ab-2ac-2bc)+(a2+b2+c2-2ab-2ac+2bc) = 4a2+4b2+4c2- 4ac +4bc
Mình không biết đúng hay sai đâu nha mình chỉ làm theo hiểu biết vì mình mới học lớp 7 thui!!!!!!!!!
1.
a) \((a + b + c)^2 + (a - b - c)^2 +( b - c - a) ^2 + (c - a - b)^2 \)
\(= (a + b + c)^2 + (a + b - c)^2 + (a - b - c)^2 + (a - b + c)^2 \)
\(= (a + b)^2 + 2c(a + b) + c^2 + (a + b)^2 - 2c(a + b) + c^2 + (a - b)^2 - 2c(a - b) + c^2 + (a - b)^2 + 2c(a - b) +c^2 \)
\(= 2(a + b)^2 + 2c^2 + 2(a - b)^2 + 2c^2 \)
\(= 2[(a + b)^2 + (a - b)^2] + 4c^2 \)
\(=2(2a^2 + 2b^2) + 4c^2 \)
\(= 4(a^2 + b^2 + c^2)\)
b) Đặt: \(x=a+b; y=c+d; z=a-b; t=c-d \)
Ta được:
\((x+y)^2+(x-y)^2+(z+t)^2+(z-t)^2 \)
\(= (x^2+2xy+y^2)+(x^2-2xy+y^2)+(z^2+2zt+t^2)+(z^2-2zt+t^2) \)
\(= 2x^2+2y^2+2z^2+2t^2 \)
\(= 2(x^2+y^2+z^2+t^2) \)
\(=2.\left[(a+b)^2+(c+d)^2+(a-b)^2+(c-d)^2 \right]\)
\(= 2(a^2+2ab+b^2+c^2+2cd+d^2+a^2-2ab+b^2+c^2-2cd+d^2) \)
\(= 2(2a^2+2b^2+2c^2+2d^2) \)
\(= 4(a^2+b^2+c^2+d^2)\)
#)Giải :
b) Ta có :
\(\left[\left(a+b\right)+\left(c+d\right)\right]^2=\left(a+b\right)^2+2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
Áp dụng hằng đẳng thức tương tự với ba đa thức còn lại, ta được :
\(2\left(a+b\right)^2+2\left(a-b\right)^2+2\left(c+d\right)^2+2\left(c-d\right)^2\)
\(=2\left(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2\right)\)
\(=4\left(a^2+b^2+c^2+d^2\right)\)
\(\Rightarrowđpcm\)
a , áp dụng a2 - b2 = ( a +b) ( a - b ) ta được
( a2 + b 2 - c2 + a 2 - b 2 + c2 ) ( a2 + b 2 - c2 - a2 + b2 - c2 )
= 2 a2 ( 2b2- 2c2) = 4a2b2- 4a2c2
b , ( a + b + c )2 + ( a + b -c ) 2 - 2 ( a +b )2
= ( a + b )2 + 2c ( a + b ) + c 2 + ( a +b )2 - 2c ( a +b ) + c2 - 2 ( a + b )2 = 2c2
c, ((a + b ) +c )2 + ( ( a - b ) +c )2 + ( ( a +b) -c )2 + ( c - ( a +b ))
= ( a + b )2 +2c ( a + b ) + c2 ( a - b ) 2 + 2c ( a-b ) + c 2 + ( a +b) 2 - 2c ( a + b ) + c 2 + c 2 - 2c ( a - b ) + ( a -b )2
= 2 ( a + b )2 + 2 ( a -b )2 + 4c 2
= 2 ( a2 + 2ab + b2 ) + 2 ( a2 - 2ab + b2 ) + 4c2
= 4 ( a2 + b2 + c2 )
Câu 4 :
Ta có : a+b+c=0
=> a+b=-c
Lại có : a3+b3=(a+b)3-3ab(a+b)
=> a3+b3+c3=(a+b)3-3ab(a+b)+c3
=-c3-3ab. (-c)+c3
=3abc
Vậy a3+b3+c3=3abc với a+b+c=0
a: \(=a^2+2a\left(b-c\right)+\left(b-c\right)^2+a^2-2a\left(b-c\right)+\left(b-c\right)^2-2\left(b-c\right)^2\)
\(=2a^2+2\left(b-c\right)^2-2\left(b-c\right)^2=2a^2\)
b: \(=a^2+2a\left(b+c\right)+\left(b+c\right)^2+a^2-2a\left(b+c\right)+\left(b+c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
\(=2a^2+2\left(b+c\right)^2+\left(a-b+c\right)^2+\left(a+b-c\right)^2\)
\(=2a^2+2\left(b+c\right)^2+a^2-2a\left(b-c\right)+\left(b-c\right)^2+a^2+2a\left(b-c\right)+\left(b-c\right)^2\)
\(=2a^2+2\left(b+c\right)^2+2a^2+2\left(b-c\right)^2\)
\(=4a^2+2\left(b^2+2bc+c^2+b^2-2bc+c^2\right)\)
\(=4a^2+4b^2+4c^2\)