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

Toán Lớp 9: Cho A = 2 √x +1/ √x-3 + 2 √x-9/ x-5 √x +6 – √x+3/ √x-2 a, tìm đkxđ, rút gọn b, tìm A biết x=6 +2 √5 c, tìm x để A =3

Toán Lớp 9: Cho A = 2 √x +1/ √x-3 + 2 √x-9/ x-5 √x +6 – √x+3/ √x-2
a, tìm đkxđ, rút gọn
b, tìm A biết x=6 +2 √5
c, tìm x để A =3

Comments ( 2 )

  1. \qquad A=(2\sqrt{x}+1)/(\sqrt{x}-3)+(2\sqrt{x}-9)/(x-5\sqrt{x}+6)-(\sqrt{x}+3)/(\sqrt{x}-2)
    a) \text{ĐKXĐ:} {(x>=0),(\sqrt{x}-3\ne0),(\sqrt{x}-2\ne0),(x-5\sqrt{x}+6\ne0):}<=>{(x>=0),(\sqrt{x}\ne3),(\sqrt{x}\ne2):}<=>{(x>=0),(x\ne9),(x\ne4):}
    Với x>=0,x\ne9,x\ne4 thì
    A=((2\sqrt{x}+1)(\sqrt{x}-2)+2\sqrt{x}-9-(\sqrt{x}-3)(\sqrt{x}+3))/((\sqrt{x}-3)(\sqrt{x}-2))
    A=(2x-4\sqrt{x}+\sqrt{x}-2+2\sqrt{x}-9-x+9)/((\sqrt{x}-3)(\sqrt{x}-2))
    A=(x-\sqrt{x}-2)/((\sqrt{x}-3)(\sqrt{x}-2))
    A=(x-2\sqrt{x}+\sqrt{x}-2)/((\sqrt{x}-3)(\sqrt{x}-2))
    A=((\sqrt{x}-2)(\sqrt{x}+1))/((\sqrt{x}-3)(\sqrt{x}-2))
    A=(\sqrt{x}+1)/(\sqrt{x}-3)
    Vậy A=(\sqrt{x}+1)/(\sqrt{x}-3) với x>=0;x\ne9;x\ne4
    b) x=6+2\sqrt{5}->\sqrt{x}=\sqrt{6+2\sqrt{5}}=\sqrt{(\sqrt{5}+1)^2}=\sqrt{5}+1
    Thay \sqrt{x}=\sqrt{5}+1 vào A ta có:
    A=(\sqrt{5}+1+1)/(\sqrt{5}+1-3)=(\sqrt{5}+2)/(\sqrt{5}-2)=(\sqrt{5}+2)^2/(5-4)=9+4\sqrt{5}
    c) A=(\sqrt{x}+1)/(\sqrt{x}-3)=3
    <=> \sqrt{x}+1=3(\sqrt{x}-3)
    <=> \sqrt{x}+1=3\sqrt{x}-9
    <=> 2\sqrt{x}=10
    <=>\sqrt{x}=5
    <=> x=25(\text{tm})
    Vậy x=25 thỏa mãn A=3

  2. Giải đáp:
    A=(2sqrtx+1)/(sqrtx-3)+(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)
    a) Điều kiện:{(x>=0),(sqrtx-2 ne 0),(sqrtx-3 ne 0),(x-5sqrtx+6 ne 0):}
    <=>{(x>=0),(x ne 4),(x ne 9):}
    Phân tích:
    x-5sqrtx+6=x-2sqrtx-3sqrtx+6
    =sqrtx(sqrtx-2)-3(sqrtx-2)
    =(sqrtx-2)(sqrtx-3)
    A=((2sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))+(2sqrtx-9)/((sqrtx-2)(sqrtx-3))-((sqrtx+3)(sqrtx-3))/((sqrtx-2)(sqrtx-3))
    A=(2x-3sqrtx-2+2sqrtx-9-x+9)/((sqrtx-2)(sqrtx-3))
    A=(x-sqrtx-2)/((sqrtx-2)(sqrtx-3))
    A=((sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))
    A=(sqrtx+1)/(sqrtx-3)
    b)x=6+2sqrt5
    x=5+2sqrt5+1
    x=(sqrt5+1)^2
    =>sqrtx=sqrt5+1
    =>A=(sqrt5+1+1)/(sqrt5+1-3)
    A=(sqrt5+2)/(sqrt5-2)
    A=((sqrt5+2)^2)/(5-4)
    A=(sqrt5+2)^2=9+4sqrt5
    c)A=3
    <=>(sqrtx+1)/(sqrtx-3)=3
    <=>sqrtx+1=3sqrtx-9
    <=>2sqrtx=10
    <=>sqrtx=5
    <=>x=25(tmđk)

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

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