Toán Lớp 9: p=2cănx -9 phần x-5cănx+6-cănx+3 phầncănx-2-2cănx+1 phần 3-cănx

Question

Toán Lớp 9:  p=2cănx -9 phần x-5cănx+6-cănx+3 phầncănx-2-2cănx+1 phần 3-cănx

haiyemy · Answer

ĐK:{(x ge 0),(x ne 4),(x ne 9):}   P=(2 sqrt{x}-9)/(x-5 sqrt{x}+6)-( sqrt{x}+3)/( sqrt{x}-2)-(2 sqrt{x}+1)/(3- sqrt{x)}   =(2 sqrt{x}-9)/(( sqrt{x}-2)( sqrt{x}-3))-( sqrt{x}+3)/( sqrt{x}-2)+(2 sqrt{x}+1)/( sqrt{x}-3)   =(2 sqrt{x}-9-( sqrt{x}+3)( sqrt{x}-3)+(2 sqrt{x}+1)( sqrt{x}-2))/(( sqrt{x}-2)( sqrt{x}-3))   =(2 sqrt{x}-9-x+9+2x-3 sqrt{x}-2)/(( sqrt{x}-2)( sqrt{x}-3))   =(- sqrt{x}+x-2)/(( sqrt{x}-2)( sqrt{x}-3))   =-(x- sqrt{x}+2)/(( sqrt{x}-2)( sqrt{x}-3))   =(( sqrt{x}+1)( sqrt{x}-2))/(( sqrt{x}-2)( sqrt{x}-3))   =( sqrt{x}+1)/( sqrt{x}-3)    a)t&igrave;m x để p=2     ->( sqrt{x}+1)/( sqrt{x}-3)=2     &harr; sqrt{x}+1=2( sqrt{x}-3)     &harr; sqrt{x}+1=2 sqrt{x}-6     &harr;- sqrt{x}=-7     &harr; sqrt{x}=7     &harr;x=49     Vậy x=49 th&igrave; P=2     b)t&igrave;m p để x=0     ->P=( sqrt{0}+1)/( sqrt{0}-3)     &harr;P=(0+1)/(0-3)     &harr;P=-1/3     Vậy P=-1/3 th&igrave; x=0     c)t&igrave;m x thuộc z để p nhận gi&aacute; trị nguy&ecirc;n     ( sqrt{x}+1)/( sqrt{x}-3)     =( sqrt{x}-3+4)/( sqrt{x}-3)     =1+4/( sqrt{x}-3)     -> sqrt{x}-3&isin;Ư(4)     &harr; ( left[  begin{array}{l} sqrt{x}-3=-4   sqrt{x}-3=-2   sqrt{x}-3=-1   sqrt{x}-3=1   sqrt{x}-3=2   sqrt{x}-3=4 end{array}  right. )      &harr; ( left[  begin{array}{l} sqrt{x}=-1(KTM)   sqrt{x}=1   sqrt{x}=2   sqrt{x}=4(KTM)   sqrt{x}=5   sqrt{x}=7 end{array}  right. )      Vậy x={1;2;5;7} th&igrave; P nguy&ecirc;n