Tìm x 1) (x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x+4)(x-4)+3x² 2) (2x+3)²+(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)² 3) (-x+5)(x-2)+(x-7)(x+7)=(3x+1)²-(3x-2)(3x+

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Tìm x
1) (x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x+4)(x-4)+3x²
2) (2x+3)²+(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)²
3) (-x+5)(x-2)+(x-7)(x+7)=(3x+1)²-(3x-2)(3x+2)
4) (5x-1)(x+1)-2(x-3)²=(x+2)(3x-1)-(x+4)²+(x²-x)

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Alexandra 8 phút 2021-09-15T22:51:16+00:00 2 Answers 0 views 0

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    0
    2021-09-15T22:52:37+00:00

    Đáp án:

    $1) (x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x+4)(x-4)+3x²$

    $⇔ x²+5x-5x-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-4x+4x-16)+3x²$

    $⇔ x²-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+4x-4x+16+3x²$

    $⇔ x²-25-x²-6x-9+3x²-12x+12 – x²-2x-1+x²-4x+4x-16-3x²=0$

    $⇔ (x²-x²+3x²-x²+x²-3x²)+(-6x-12x-2x-4x+4x)+(-25-9+12-1-16)=0$

    $⇔ -20x-37=0$

    $⇔ -20x=39$

    $⇔ x=-39/20$

    $2) (2x+3)²+(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)²$

    $⇔ 4x²+12x+9+x²+x-x-1=5(x²+4x+4)-(x²+x-5x-5)+x²+8x+16$

    $⇔ 4x²+12x+9+x²+x-x-1=5x²+20x+20-x²-x+5x+5+x²+8x+16$

    $⇔ 4x²+12x+9+x²+x-x-1-5x²-20x-20+x²+x-5x-5-x²-8x-16=0$

    $⇔ (4x²+x²-5x²+x²-x²)+(12x+x-x-20x+x-5x-8x)+(9-1-20-5-16)=0$

    $⇔ -20x-33=0$

    $⇔ -20x=33$

    $⇔ x=-1,65$

    $3) (-x+5)(x-2)+(x-7)(x+7)=(3x+1)²-(3x-2)(3x+2)$

    $⇔ -x²+2x+5x-10+x²+7x-7x-49=9x²+6x+1-(9x²+6x-6x-4)$

    $⇔ -x²+2x+5x-10+x²+7x-7x-49 -9x²-6x-1+9x²+6x-6x-4=0$

    $⇔ (-x²+x²-9x²+9x²)+(2x+5x+7x-7x-6x+6x-6x)+(-10-49-1-4)=0$

    $⇔x-64=0$

    $⇔ x=64$

    $4) (5x-1)(x+1)-2(x-3)²=(x+2)(3x-1)-(x+4)²+(x²-x)$

    $⇔ 5x²+5x-x-1-2(x²-6x+9)=3x²-x+6x-2-(x²+8x+16)+x²-x$

    $⇔ 5x²+5x-x-1-2x²+12x-18=3x²-x+6x-2-x²-8x-16+x²-x$

    $⇔ 5x²+5x-x-1-2x²+12x-18 – 3x²+x-6x+2+x²+8x+16-x²+x=0$

    $⇔ (5x²-2x²-3x²+x²-x²)+(5x-x+12x+x-6x+8x+x)+(-1-18+2+16)=0$

    $⇔ 20x-1=0$

    $⇔ 20x=1$

    $⇔ x=1/20$

    BẠN THAM KHẢO NHA!!!

     

    0
    2021-09-15T22:53:13+00:00

    Đáp án:

     

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35:5x4+1-9:3 = ? ( )