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Đặt x = a+b , y = b+c , z = c+a
Thì biểu thức trên trở thành \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy-3xyz\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Từ đó thay a,b,c vào rồi rút gọn :)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{c-b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{a-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{b-a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{c-b+b-a+a-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
a,Ta đặt :
a-b-c=x ; b-c-a=y ; c-a-b=z
Ta có:
\(\text{x+y+z=a-b-c+b-c-a+c-a-b=-(a+b+c)}\)
\(\Rightarrow\left(x+y+z\right)^2=\left(a+b+c\right)^2\)
\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=\left(x+y+z\right)^2+x^2+y^2+z^2\)
\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=\left(x+y\right)^2+\left(y+z\right)^2+\left(x+z\right)^2\)\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=4\left(a^2+b^2+c^2\right)\)
\(C=\left(a+b+c\right)\left(a+b-c\right)+\left(a+b+c\right)\left(a+c-b\right)+\left(a+b+c\right)\left(a+c-b\right)\)
\(=\left(a+b+c\right)\left[\left(a+b-c\right)+\left(a+c-b\right)+\left(a+c-b\right)\right]\)
\(=\left(a+b+c\right)\left(3a-b+c\right)\)
C=(a+b+c)(a+b-c+a+c-b+a+c-b)
C=(a+b+c)(3a-b+c)
C=a(3a-b+c)+b(3a-b+c)+c(3a-b+c)
C=3a2-ab+ac+3ab-b2+bc+3ac-bc+c2
C=3a2-b2+c2+2ab+4ac
C=3a2-b2+c2+2a(b+2c)