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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: A= √x/( √x-5) – 10 √x/(x-25) – 5/ (√x +5 ) a,rút gọn b,tìm giá trị của a khi x=9 c,tìm x để a <1/3

Toán Lớp 9: A= √x/( √x-5) – 10 √x/(x-25) – 5/ (√x +5 )
a,rút gọn
b,tìm giá trị của a khi x=9
c,tìm x để a <1/3

Comments ( 2 )

  1. a) ĐK : x>=0 ; x ne 25
    A=(sqrtx)/(sqrtx-5)-(10sqrtx)/(x-25)-5/(sqrtx+5)
    =(sqrtx(sqrtx+5)-10sqrtx-5(sqrtx-5))/((sqrtx-5)(sqrtx+5))
    =(x+5sqrtx-10sqrtx-5sqrtx+25)/((sqrtx-5)(sqrtx+5))
    =(x-10sqrtx+25)/((sqrtx-5)(sqrtx+5))
    =((sqrtx-5)^2)/((sqrtx-5)(sqrtx+5))=(sqrtx-5)/(sqrtx+5)
    b)
    Tại x=9 \ \ text{(tmđk)} giá trị của A là :
    A=(sqrt9-5)/(sqrt9+5)=(3-5)/(3+5)=(-2)/8=-1/4
    Vậy với x=9 thì A=-1/4
    c)
    A<1/3 <=> (sqrtx-5)/(sqrtx+5)<1/3
    <=> (sqrtx-5)/(sqrtx+5)-1/3<0
    <=> (3(sqrtx-5)-(sqrtx+5))/(3(sqrtx+5))<0
    <=> (3sqrtx-15-sqrtx-5)/(3(sqrtx+5))<0
    <=> (2sqrtx-20)/(3(sqrtx+5))<0
    Mà : 3(sqrtx+5) > 0 ( vì x>=0 )
    to 2sqrtx-20<0
    <=> 2sqrtx<20
    <=> sqrtx<10
    <=> x<100
    Kết hợp với điều kiện : x>=0
    to 0<=x<100
    Vậy với 0<=x<100;x ne 25 thì A<1/3

  2. Giải đáp+Lời giải và giải thích chi tiết:
    A=sqrtx/(sqrtx-5)-(10sqrtx)/(x-25)-5/(sqrtx+5)(x>=0,x ne 25)
    A=(sqrtx(sqrtx+5))/((sqrtx-5)(sqrtx+5))-(10sqrtx)/((sqrtx-5)(sqrtx+5))-(5(sqrtx-5))/((sqrtx-5)(sqrtx+5))
    A=(x+5sqrtx-10sqrtx-5sqrtx+25)/((sqrtx-5)(sqrtx+5))
    A=(x-10sqrtx+25)/((sqrtx-5)(sqrtx+5))
    A=(sqrtx-5)^2/((sqrtx-5)(sqrtx+5))
    A=(sqrtx-5)/(sqrtx+5)
    b)x=9(tmđk)
    =>sqrtx=3
    =>A=(3-5)/(3+5)
    =>A=(-2)/8=-1/4
    c)A<1/3
    <=>(sqrtx-5)/(sqrtx+5)-1/3<0
    <=>(3sqrtx-15-sqrtx-5)/(3(sqrtx+5))<0
    <=>(2sqrtx-20)/(3(sqrtx+5))<0
    Mà 3(sqrtx+5)>=3.5=15>0
    <=>2sqrtx-20<0
    <=>2sqrtx<20
    <=>sqrtx<10
    <=>x<100
    Mà x>=0,x ne 25
    =>0<=x<100,x ne 25

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222-9+11+12:2*14+14 = ? ( )

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