Toán Lớp 9: $sqrt{x+3}$ `+` $sqrt{4x+12}$`=x-1` tìm `x` -

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Toán Lớp 9: $\sqrt{x+3}$ `+` $\sqrt{4x+12}$`=x-1` tìm `x`

Huyền Trâm Tháng Sáu 8, 2022 Toán học 2 Comments 0 views

Toán Lớp 9: $\sqrt{x+3}$ + $\sqrt{4x+12}$=x-1
tìm x

Comments ( 2 )

chauhoangmy98
8 Tháng Sáu, 2022 at 11:03 sáng

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~ Bạn tham khảo ~

\sqrt(x+3) + \sqrt(4x+12) = x-1

ĐKXĐ : x>=1

<=> \sqrt(x+3) + 2\sqrt(x+3) = x-1

<=> 3\sqrt(x+3)=x-1

<=> 9(x+3)=(x-1)^2

<=> 9x+27=x^2-2x+1

<=> x^2-11x-26=0

<=> x^2-13x+2x-26=0

<=> x(x-13) + 2(x-13)=0

<=> (x-13)(x+2)=0

<=> [(x-13=0),(x+2=0):}

<=> [(x=13\quad(TMĐK)),(x=-2\quad(KTMĐK)):}

Vậy S={13}
khanhlanchau
8 Tháng Sáu, 2022 at 11:03 sáng

Reply

\sqrt{x+3}+\sqrt{4x+12}=x-1(x>=1)

<=>\sqrt{x+3}+\sqrt{4(x+3)}=x-1

<=>\sqrt{x+3}+2\sqrt{x+3}=x-1

<=>3\sqrt{x+3}=x-1

<=>9(x+3)=(x-1)^2

<=>x^2-2x+1-9x-27=0

<=>x^2-11x-26=0

<=>x^2+2x-13x-26=0

<=>x(x+2)-13(x-3)=0

<=>(x+2)(x-13)=0

<=>[(x=-2),(x=13):} ™

S={-2;13}

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