Tìm x a, (2x + 3)(x – 4) + (x+5)(x-2) = (3x-5)(x-4) b, (8x-3)(3x+2) – (4x+7)(x+4) = (4x+1)(5x-1) c, 4(x – 1)(x+5) – (x+2)(x+5) = 3(x-1)(x+2)

Question

Tìm x
a, (2x + 3)(x – 4) + (x+5)(x-2) = (3x-5)(x-4)
b, (8x-3)(3x+2) – (4x+7)(x+4) = (4x+1)(5x-1)
c, 4(x – 1)(x+5) – (x+2)(x+5) = 3(x-1)(x+2)

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Amity 9 months 2020-11-02T09:21:49+00:00 2 Answers 107 views 0

Answers ( )

    0
    2020-11-02T09:23:05+00:00

    `a)(2x+3)(x−4)+(x+5)(x−2)=(3x−5)(x−4)`

    `⟺2x^2−5x−12+x^2+3x−10=3x^2−17x+20`

    `⟺15x−42=0`

    `⟺x=42/15`

    `b)(8x-3)(3x+2) – (4x+7)(x+4) = (4x+1)(5x-1)`

    `<=> 24x^2+16x−9x−6−(4x^2+16x+7x+28)=20x^2−4x+5x−1`

    `<=>20x^2−16x−34=20x^2−x−1`

    `<=>-15x=33`

    `<=> x=-33/15`

    `c)4(x−1)(x−5)−(x+2)(x+5)=3(x−1)(x+2)`

    `<=>4(x^2+4x−5)−(x^2+2x+5x+10)=3(x^2+x−^2)`

    `<=>4x^2+16x−x^2−7x−3x^2−3x=−6+30`

    `<=>6x=24`

    `<=>x=4`

    0
    2020-11-02T09:23:48+00:00

    Đáp án:

     

    Giải thích các bước giải:

    a) (2x + 3)(x – 4) + (x+5)(x-2) = (3x-5)(x-4)

    ⇒ x(2x+3) – 4(2x+3) + x(x+5) – 2(x+5) = x(3x-5) – 4(3x-5)

    ⇒ 2$x^{2}$ + 3x – 8x – 12 + $x^{2}$ + 5x – 2x – 10 = 3$x^{2}$ – 5x – 12x + 20

    ⇒ (2$x^{2}$ + $x^{2}$ ) + ( 3x – 8x + 5x – 2x ) + ( -12 – 10 ) = 3$x^{2}$ – 5x – 12x + 20

    ⇒ 3$x^{2}$ – 2x – 22 = 3$x^{2}$ – 5x – 12x + 20

    ⇒ ( 3$x^{2}$ – 3$x^{2}$ ) + ( -2x + 12x + 5x ) = 22 + 20 < quy tắc chuyển vế >

    ⇒ 15x = 42

    ⇒ x = 42 : 15

    ⇒ x = $\frac{42}{15}$

    Vậy x = $\frac{42}{15}$

    b, (8x-3)(3x+2) – (4x+7)(x+4) = (4x+1)(5x-1)

    ⇒ 3x(8x-3)+2(8x-3) – x(4x-7)+4(4x-7) = 4x(5x-1)+(5x-1)

    ⇒ 24$x^{2}$ – 9x + 16x – 6 – 4$x^{2}$ + 7x + 16x – 28 = 20$x^{2}$ – 4x + 5x – 1

    ⇒ (24$x^{2}$ – 4$x^{2}$) + (-9x + 16x + 7x + 16x ) + ( -6-28 ) = 20$x^{2}$ + ( -4x + 5x ) -1

    ⇒ 20$x^{2}$ + 30x – 34 = 20$x^{2}$ + x – 1

    ⇒ (20$x^{2}$ – 20$x^{2}$) + (30x – x) = 34 – 1 < qui tắc chuyển vế >

    ⇒ 29x = 33

    ⇒ x = $\frac{33}{29}$

    Vậy x = $\frac{33}{29}$

    c, 4(x – 1)(x+5) – (x+2)(x+5) = 3(x-1)(x+2)

    ⇒ 4[x(x+5) – (x+5)] – [x(x+5) + 2(x+5)] = 3[x(x+2) – (x+2)]

    ⇒ 4($x^{2}$ – x – 5) – ( $x^{2}$ + 5x + 2x + 10 ) = 3($x^{2}$ + 2x – x – 2)

    ⇒ 4$x^{2}$ -4x – 20 – $x^{2}$ – 7x – 10 = 3$x^{2}$ + 3x – 6

    ⇒ (4$x^{2}$ – $x^{2}$) + ( -4x-7x ) + ( -20-10 ) = 3$x^{2}$ + 3x – 6

    ⇒ 3$x^{2}$ – 11x -30 = 3$x^{2}$ + 3x – 6

    ⇒ (3$x^{2}$ – 3$x^{2}$) + (- 3x – 11x ) = -6 + 30 < qui tắc chuyển vế >

    ⇒ -14x = 24

    ⇒ x = $\frac{12}{7}$

    Vậy x= $\frac{12}{7}$

    #ham xin ctlhn ạ

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