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.
\((a+b+c+d)^2+(a+b-c-d)^2+(a+c-b-d)^2+(a+d-b-c)^2\)
\(=(a+b)^2+(c+d)^2+2(a+b)(c+d)+(a+b)^2+(c+d)^2-2(a+b)(c+d)+(a-b)^2+(c-d)^2+2(a-b)(c-d)+(a-b)^2+(d-c)^2+2(a-b)(d-c)\)
\(=2(a+b)^2+2(c+d)^2+2(a-b)^2+2(c-d)^2\)
\(=2[(a+b)^2+(a-b)^2+(c+d)^2+(c-d)^2]\)
\(=2(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2)\)
\(=2(2a^2+2b^2+2c^2+2d^2)=4(a^2+b^2+c^2+d^2)\)
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.
\((a+b+c+d)^2+(a+b-c-d)^2+(a+c-b-d)^2+(a+d-b-c)^2\)
\(=(a+b)^2+(c+d)^2+2(a+b)(c+d)+(a+b)^2+(c+d)^2-2(a+b)(c+d)+(a-b)^2+(c-d)^2+2(a-b)(c-d)+(a-b)^2+(d-c)^2+2(a-b)(d-c)\)
\(=2(a+b)^2+2(c+d)^2+2(a-b)^2+2(c-d)^2\)
\(=2[(a+b)^2+(a-b)^2+(c+d)^2+(c-d)^2]\)
\(=2(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2)\)
\(=2(2a^2+2b^2+2c^2+2d^2)=4(a^2+b^2+c^2+d^2)\)
