Rút gọn các biểu thức sau :
a) \(A=\left(a-b\right)+\left(a+b-c\right)-\left(a-b-c\right)\)
b) \(B=\left(a-b\right)-\left(b-c\right)+\left(c-a\right)-\left(a-b-c\right)\)
c) \(C=\left(-a+b+c\right)-\left(a-b+c\right)-\left(-a+b-c\right)\)
Rút gọn biểu thức :
\(\frac{a^2\left(a+b\right)\left(a+c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{b^2\left(b+a\right)\left(b+c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{c^2\left(c+a\right)\left(c+b\right)}{\left(c-a\right)\left(c-b\right)}\)
Rút gọn biểu thức: \(A=\dfrac{2}{a-b}+\dfrac{2}{b-c}+\dfrac{2}{c-a}+\dfrac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)
Rút gọn biểu thức sau: \(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)+\left(c+a\right)\)
Rút gọn biểu thức sau: \(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
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 :)
Rút gọn biểu thức :
\(C=\left(a+b+c\right)\left(a+b-c\right)+\left(a+c+b\right)\left(a+c-b\right)+\left(a+b+c\right)\left(a+c-b\right)\)
\(C=\left(a+b+c\right)\left(a+b-c\right)+\left(a+b+c\right)\left(a+c-b\right)+\left(a+b+c\right)\left(a+c-b\right)\)
\(=\left(a+b+c\right)\left[\left(a+b-c\right)+\left(a+c-b\right)+\left(a+c-b\right)\right]\)
\(=\left(a+b+c\right)\left(3a-b+c\right)\)
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)
mk thấy có một số cái giống đó bn nhóm lại thành một được không???
mk cx chả bít!! 876869789780
Rút gọn biểu thức\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Rút gọn biểu thức sau :
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(\frac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
Rút gọn biểu thức sau
\(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{a^2\left(b-c\right)-b^2\left(c+a\right)-c^2\left(a-b\right)+2abc}\)
Rút gọn biểu thức :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}\)
\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{-\left(a-b\right)\left(a+b\right)+c\left(a-b\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}\)