8x³+30x²+150x+125

=(2x)³+3. (2x)².5+ 3.2x.5²+5³

=(2x+5)³

8x³ + 30x² + 150x + 125 = (2x)³ + 3. (2x)².5 + 3. 2x. 5² + 5³ = (2x + 5)³

a) Ta có: 70 = 2.5.7; 84 =  2 2 . 3 . 7 => ƯCLN(70,84) = 2.7 = 14

=> ƯC(70,84) = Ư(14) = {1;2;7;14}

Mà x ∈ ƯC(70, 84) và x > 8.Vậy x = 14

b) Ta có: 64 =  2 6 ; 48 =  2 4 . 3 ; 88 =  2 3 . 11 => ƯCLN(64,48,88) =  2 3 = 8

=> ƯC(64,48,88) = Ư(8) = {1;2;4;8}

Mà x ∈ ƯC(64,48,88) và x > 4 . Vậy x = 8

c) Vì 126 ⋮ x; 210 ⋮ x nên x ∈ ƯC(126,210)

Ta có: 126 =  2 . 3 2 . 7 ; 210 = 2.3.5.7 => ƯCLN(126,210) = 2.3.7 = 42

=> ƯC(126,210) = Ư(42) = {1;2;3;6;7;14;21;42}

Mặt khác: 15 < x < 30. Vậy x = 21

d) Vì 150 ⋮ x; 84 ⋮ x; 30 ⋮ x nên x ∈ ƯC(150,84,30)

Ta có: 150 =  2 . 3 . 5 2 ; 84 =  2 2 . 3 . 7 ; 30 = 2.3.5 => ƯCLN(150,84,30) = 2.3 = 6

=> ƯC(150,84,30) = Ư(6) = {1;2;3;6}

Mặt khác: 2 < x < 6. Vậy x = 3

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2x\(^2\)+8x+6= 2x\(^2\)+2x+6x+6= 2x(x+1)+6(x+1)=(2x+6)(x+1)

3x\(^2\)-8x+5= 3x\(^2\)-3x-5x+5=3x(x-1)-5(x-1)= (3x-5)(x+1)

2x\(^2\)-30x+28= 2x\(^2\)-2x-28x+28= 2x(x-1)-28(x-1)= 2(x-19)(x-1)

4x\(^2\)+8x+3= 4x\(^2\)+2x+6x+3= 2x(2x+1)+3(2x+1)= (2x+3)(2x+1)

1.

12 - 10x = 25 - 30x

<=> 12 - 25 = -30x + 10x

<=> - 20x = -13

<=> x = 13/20

2.

3(2x+3) - 2(4x - 5) = 10x+21

<=> 6x + 9 - (8x - 10) = 10x + 21

<=> 6x + 9 - 8x + 10 = 10x + 21

<=> -2x + 19 = 10x + 21

<=> -12x = 2

<=> x = -1/6

1: =>-10x+30x=25-12

=>20x=13

=>x=13/20

2: =>6x+9-8x+10=10x+21

=>10x+21=-2x+19

=>12x=-2

=>x=-1/6

1: \(\Leftrightarrow x^3-9x^2+27x-27=-35\)

\(\Leftrightarrow\left(x-3\right)^3=-35\)

\(\Leftrightarrow x-3=\sqrt[3]{-35}\)

hay \(x=\sqrt[3]{-35}+3\)

2: \(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)

=>6x=6

hay x=1

4: \(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)

=>12x-4=-10

=>12x=-6

hay x=-1/2

\(2x^4-8x^3-7x^3+28x^2+7x^2-28x-2x+8\\ =2x^3\left(x-4\right)-7x^2\left(x-4\right)+7x\left(x-4\right)-2\left(x-4\right)\\ =\left(x-4\right)\left(2x^3-7x^2+7x-2\right)\\ =\left(x-4\right)\left(2x^3-4x^2-3x^2+6x+x-2\right)\\ =\left(x-4\right)\left[2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)\right]\\ =\left(x-4\right)\left(x-2\right)\left(2x^2-2x-x+1\right)\\ =\left(x-4\right)\left(x-2\right)\left(2x-1\right)\left(x-1\right)\)