:(((((((help meeeeeeeee

Question

:(((((((help meeeeeeeee
help-meeeeeeeee

in progress 0
Philomena 5 years 2021-05-22T07:56:01+00:00 2 Answers 34 views 0

Answers ( )

  1. Acacia
    0
    2021-05-22T07:57:02+00:00
    Reply

    Đáp án
    help-meeeeeeeee

  2. thachthao
    0
    2021-05-22T07:57:45+00:00
    Reply

    Giải thích các bước giải:

    Bài 3:

    Từ hệ suy ra:

    $(x^2+y^2-2x-2y)+2(x+y-xy)=6+5\cdot 2$

    $\to x^2+y^2-2xy=16$

    $\to (x-y)^2=16$

    $\to |x-y|=4$

    Bài 4:

    Ta có $\dfrac{x}{x^2+x+1}=\dfrac14$

    $\to x^2+x+1=4x$

    $\to x^2-3x+1=0$

    Ta có:

    $x^5-4x^3-3x+9$

    $=(x^5-3x^4+x^3)+(3x^4-9x^3+3x^2)+(4x^3-12x^2+4x)+(9x^2-27x+9)+20x$

    $=x^3(x^2-3x+1)+3x^2(x^2-3x+1)+4x(x^2-3x+1)+9(x^2-3x+1)+20x$

    $=(x^3+3x^2+4x+9)(x^2-3x+1)+20$

    $=20x$ vì $x^2-3x+1=0$

    Ta có:

    $x^4+3x^2+11$

    $=(x^4-9x^2)+12x^2+11$

    $=(x^2-3x)(x^2+3x)+12x^2+11$

    $=-(x^2+3x)+12x^2+11$ vì $x^2-3x+1=0\to x^2-3x=-1$

    $=11x^2-3x+11$

    $=11(x^2+1)-3x$

    $=11\cdot 3x-3x$ vfi $x^2-3x+1=0\to x^2+1=3x$

    $=30x$

    Ta có:

    $P=\dfrac{x^5-4x^3-3x+9}{x^4+3x^2+11}=\dfrac{20x}{30x}=\dfrac23$

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )