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a) \(\frac{x^2-xy-x+y}{x^2+xy-x-y}\)=\(\frac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)=\(\frac{\left(x-1\right)\left(x-y\right)}{\left(x-1\right)\left(x+y\right)}\)=\(\frac{x-y}{x+y}\)
b) \(\frac{x^2-xy}{5y^2-5xy}\)=\(\frac{x\left(x-y\right)}{-5y\left(x-y\right)}\)=\(\frac{-x}{5y}\)
c) \(\frac{3x^2-12x+12}{x^4-8x}\)=\(\frac{3\left(x^2-4x+4\right)}{x\left(x^3-2^3\right)}\)=\(\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)=\(\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
a) \(\frac{x^2+5x+6}{x^2+7x+12}\)=\(\frac{x^2+2x+3x+6}{x^2+3x+4x+12}\)=\(\frac{x\left(x+2\right)+3\left(x+2\right)}{x\left(x+3\right)+4\left(x+3\right)}\)=\(\frac{\left(x+3\right)\left(x+2\right)}{\left(x+4\right)\left(x+3\right)}\)
b) \(\frac{7x^2+14x+7}{3x^2+3x}\)=\(\frac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)^2}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)\left(x+1\right)}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)}{3x}\)
a)\(\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}\)=\(\frac{x+1}{5x^2}\)
b)\(\frac{10y}{15\left(x+y\right)^2}\)
mk ko biết làm
xin lỗi bn nhae
xin lỗi vì đã ko giúp được bn
chcus bn học gioi!
nhae@@@
1) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2-12\)
\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)
\(=x^3.\left(x+2\right)+4x.\left(x+2\right)-5.\left(x+2\right)=\left(x+2\right)\left(x^3+4x-5\right)\)
\(=\left(x+2\right)\left(x^3-x^2+x^2-x+5x-5\right)=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)
2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)
Đặt \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)
\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)
Thay a , ta có :
\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)
\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
bn hãy xem cách tui làm phía duoi la làm dc, cứ xem có j chung thi rút nó ra
a) \(\frac{x^2-xy}{5y^2-5xy}\)=\(\frac{x\left(x-y\right)}{-5y\left(x-y\right)}\)=\(\frac{-x}{5y}\)
b) \(\frac{3x^2-12x+12}{x^4-8x}\)=\(\frac{3\left(x^2-4x+4\right)}{x\left(x^3-2^3\right)}\)=\(\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)=\(\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)