Các bạn giúp mik giải hai bài này vs nhé

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Các bạn giúp mik giải hai bài này vs nhé
cac-ban-giup-mik-giai-hai-bai-nay-vs-nhe

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Adela 4 years 2020-11-13T21:30:30+00:00 2 Answers 85 views 0

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    0
    2020-11-13T21:31:56+00:00

    Giải thích các bước giải:

    `1a,(2x-3)^2+(4x-5)(x-7)`

    `=4x^2-12x+9+4x^2-33x+35` 

    `=8x^2-45x+44`

    `b,(x-2)(x+1)(x+2)-(x-1)^3`

    `=x^3+x^2-4x-4-x^3+3x^2-3x+1`

    `=4x^2-7x-3`

    `c,(2x+5)(5-2x)+(3x-7)(x+2)`

    `=-4x^2+25+3x^2-x-14`

    `=-x^2-x+11`

    `d,(2x+4)^2-(3x-5)^2`

    `=4x^2+16x+10-9x^2+30x-25`

    `=-5x^2+46x-15`

    `2a,(2x-3)(x+1)-2(x-1)^2=3`

    `<=>2x^2+2x-3x-3-2x^2+4x-2=3`

    `<=>3x-5=3`

    `<=>x=8/3`

    `b,(4x+5)(2x-1)-(x+2)^2=(x-3)(x+3)`

    `<=>8x^2-4x+10x-5-x^2-4x-4=x^2-9`

    `<=>7x^2+2x=x^2`

    `<=>6x^2+2x=0`

    `<=>2x(3x+1)=0`

    `<=>x=0` hoặc `x=-1/3`

    `c,(x-2)^2+(x+3)^2=(x+1)^2+15`

    `<=>2x^2+2x+13=x^2+2x+1+15`

    `<=>x^2-16+13=0`

    `<=>x^2=3`

    `<=>x=` `+-`$\sqrt[]{3}$ 

    Vậy …………………

    0
    2020-11-13T21:32:21+00:00

    `1a,(2x-3)^2+(4x-5)(x-7)`

    `=4x^2-12x+9+4x^2-33x+35` 

    `=8x^2-45x+44`

    `b,(x-2)(x+1)(x+2)-(x-1)^3`

    `=x^3+x^2-4x-4-x^3+3x^2-3x+1`

    `=4x^2-7x-3`

    `c,(2x+5)(5-2x)+(3x-7)(x+2)`

    `=-4x^2+25+3x^2-x-14`

    `=-x^2-x+11`

    `d,(2x+4)^2-(3x-5)^2`

    `=4x^2+16x+10-9x^2+30x-25`

    `=-5x^2+46x-15`

    `2a,(2x-3)(x+1)-2(x-1)^2=3`

    `<=>2x^2+2x-3x-3-2x^2+4x-2=3`

    `<=>3x-5=3`

    `<=>3x=8`

    `<=>x=8/3`

    Vậy `x=8/3`

    `b,(4x+5)(2x-1)-(x+2)^2=(x-3)(x+3)`

    `<=>8x^2-4x+10x-5-x^2-4x-4=x^2-9`

    `<=>7x^2+2x=x^2`

    `<=>6x^2+2x=0`

    `<=>2x(3x+1)=0`

    `<=>x=0` hoặc `x=-1/3`

    Vậy `S={0;-1/3}`

    `c,(x-2)^2+(x+3)^2=(x+1)^2+15`

    `<=>2x^2+2x+13=x^2+2x+1+15`

    `<=>x^2-16+13=0`

    `<=>x^2=3`

    `<=>x=` `+-`$\sqrt[]{3}$ 

    Vậy `x=+-` $\sqrt[]{3}$ 

     

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