Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
a) \(\frac{x^2-16}{4x-x^2}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
c)
\(\frac{(x+y)^2-z^2}{x+y+z}=\frac{(x+y-z)(x+y+z)}{x+y+z}=x+y-z\)
d)
Biểu thức không rút gọn được
e)
\(\frac{a^3+b^3+c^3}{a^2+b^2+c^2-ab-bc-ac}=\frac{(a+b)^3-3ab(a+b)+c^3}{a^2+b^2+c^2-ab-bc-ac}=\frac{(a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\frac{(a+b+c)(a^2+b^2+c^2-ac-bc+2ab)-3ab(a+b+c)+3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc}{a^2+b^2+c^2-ab-bc-ac}=a+b+c+\frac{3abc}{a^2+b^2+c^2-ab-bc-ac}\)
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
Ta có
x+1=b2+c2−a22bc+1=b2+2bc+c2−a22bc=(b+c)2−a22bcx+1=b2+c2−a22bc+1=b2+2bc+c2−a22bc=(b+c)2−a22bc
Suy ra
y(x+1)=a2−(b−c)2(b+c)2−a2.(b+c)2−a22bc=a2−(b−c)22bcy(x+1)=a2−(b−c)2(b+c)2−a2.(b+c)2−a22bc=a2−(b−c)22bc
Do đó
P=x+y+xy=x+y(x+1)=b2+c2−a22bc+a2−(b−c)22bc=b2+c2−a2+a2−(b−c)22bc=1
Làm như bạn trên hướng dẫn ấy:
Ta có: \(x+1=\frac{b^2+c^2-a^2}{2bc}+1=\frac{\left(b+c\right)^2-a^2}{2bc}\)
\(y+1=\frac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}+1=\frac{4bc}{\left(b+c\right)^2-a^2}\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)=\frac{\left(b+c\right)^2-a^2}{2bc}.\frac{4bc}{\left(b+c\right)^2-a^2}=2\)
\(\Rightarrow P=\left(x+1\right)\left(y+x\right)-1=2-1=1\)