a) x³ - 64x = 0

x(x² - 64) = 0

x(x - 8)(x + 8) = 0

x = 0 hoặc x - 8 = 0 hoặc x + 8 = 0

*) x - 8 = 0

x = 8

*) x + 8 = 0

x = -8

Vậy x = -8; x = 0; x = 8

b) x³ - 4x² = -4x

x³ - 4x² + 4x = 0

x(x² - 4x + 4) = 0

x(x - 2)² = 0

x = 0 hoặc (x - 2)² = 0

*) (x - 2)² = 0

x - 2 = 0

x = 2

Vậy x = 0; x = 2

c) x² - 16 - (x - 4) = 0

(x - 4)(x + 4) - (x - 4) = 0

(x - 4)(x + 4 - 1) = 0

(x - 4)(x + 3) = 0

x - 4 = 0 hoặc x + 3 = 0

*) x - 4 = 0

x = 4

*) x + 3 = 0

x = -3

Vậy x = -3; x = 4

d) (2x + 1)² = (3 + x)²

(2x + 1)² - (3 + x)² = 0

(2x + 1 - 3 - x)(2x + 1 + 3 + x) = 0

(x - 2)(3x + 4) = 0

x - 2 = 0 hoặc 3x + 4 = 0

*) x - 2 = 0

x = 2

*) 3x + 4 = 0

3x = -4

x = -4/3

Vậy x = -4/3; x = 2

e) x³ - 6x² + 12x - 8 = 0

(x - 2)³ = 0

x - 2 = 0

x = 2

f) x³ - 7x - 6 = 0

x³ + 2x² - 2x² - 4x - 3x - 6 = 0

(x³ + 2x²) - (2x² + 4x) - (3x + 6) = 0

x²(x + 2) - 2x(x + 2) - 3(x + 2) = 0

(x + 2)(x² - 2x - 3) = 0

(x + 2)(x² + x - 3x - 3) = 0

(x + 2)[(x² + x) - (3x + 3)] = 0

(x + 2)[x(x + 1) - 3(x + 1)] = 0

(x + 2)(x + 1)(x - 3) = 0

x + 2 = 0 hoặc x + 1 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x + 1 = 0

x = -1

*) x - 3 = 0

x = 3

Vậy x = -1; x = -1; x = 3

a,x\(^3\)-64=0
   x\(^3\)     =64
  =>x=3
b,x\(^3\)-4x\(^2\)=-4x

   x\(^3\)-4x\(^2\)+4x=0
   x(x\(^2\)-4x+4)=0
   x(x-2)\(^2\)=)
TH1:x=0
TH2:x-2=0
     =>x=2

c,x\(^2\)-16-(x-4)=0

   (x+4)(x-4)-(x-4)=0
   (x-4)(x+4-1)=0

   (x-4)(x+3)=0
TH1:x-4=0
     =>x=4

TH2:x+3=0

    =>x=-3

d,(2x+1).2=3+x
    4x+2-3-x=0
    3x-1=0
     x=\(\dfrac{1}{3}\)

e,x\(^3\)-6x\(^2\)+12x-8=0
   (x-2)\(^3\)=0
=>x-2=0
=>x=2

f,x\(^3\)-7x+6=0
  x\(^3\)-x-6x+6=0
  x(x\(^2\)-1)-6(x-1)=0
  x(x+1)(x-1)-6(x-1)=0
  (x-1)(x\(^2\)+x-6)=0
TH1:x-1=0

  =>x=1
TH2:x\(^2\)+x-6=0

       x\(^2\)+3x-2x-6=0

       x(x+3)-2(x+3)=0

       (x+3)(x-2)=0

=>x+3=0                =>x-2=0

+>x=-3                   =>x=2

d,(2x+1)\(^2\)=(3+x)\(^2\)

4x\(^2\)+4x+1-9-6x-x\(^2\)=0
3x\(^2\)-2x-8=0
3x\(^2\)-6x+4x-8=0
3x(x-2)+4(x-2)=0

(3x+4)(x-2)=0

TH1:3x+4=0                     TH2:x-2=0
=>x=\(\dfrac{-4}{3}\)                              =>x=2

a: \(\Leftrightarrow\left(x-2010\right)\left(7x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2010\\x=-\dfrac{1}{7}\end{matrix}\right.\)

b: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

a)7x(x-2010)+(x-2010)=-

(x-2010)(7x+1)=0

x=2010 hoặc x=\(-\dfrac{1}{7}\)

Vậy \(x\in\left\{2010;-\dfrac{1}{7}\right\}\)

b)\(2x^2-2x=x^2-2x+1\)

\(2x^2-2x-x^2+2x-1=0\)

\(x^2-1=0\)

\(x^2=1\)

\(x=\pm1\)

Vậy \(x\in\left\{1;-1\right\}\)

???????????????????????

a)ta có \(120⋮x,240⋮x,300⋮x,x\ge10\)

=> \(x\inƯC\left(120;240;300\right)\)

120=2³.3.5

240=2⁴.3.5

300=2².5².3

ƯCLN(120,240,300)=2².3.5=60

b) ta có\(x⋮16;x⋮15;x⋮11,x< 3000\)

=>\(x\in BCNN\left(16;15;11\right)\)

16=2⁴

15=3.5

11=11

BCNN(16;15;11)=2⁴.5.3.11=2640

các bạn nhớ tích đúng cho mình nhé

a)\(120⋮x;240⋮x;300⋮x;x\ge10\)

=> x\(\inƯC\left(120;240;300\right)\)

120=2³.3.5

240=2⁴.3.5

300=2².5².3

Ưc(120;240;300)=2².5.3=60

b)\(x⋮16;x⋮15;x⋮11;x< 3000\)

=>x\(\in BC\left\{16;15;11\right\}\)

16=2⁴

15=3.5

11=11

BC(16;15;11)=2⁴.3.5.11=2640

a) (x+1)+(x+2)+....+(x+100)=5750

<=> (x+x+x+....+x)+(1+2+....+100)=5750

<=> 100x+5050=5750

<=> 100x=700

<=> x=7

b) A=7-Ix-1I

Ta có Ix-1I =<0 với mọi x thuộc Z
=> 7-Ix-1I =<7 với mọi x thuộc Z hay A =< 7

Dấu "=" <=> Ix-1I=0

<=> x-1=0

<=> x=1

Vậy MaxA=7 đạt được khi x=1

(𝑥+1)+(𝑥+2)+⋯+(𝑥+100)=5750

=) x.100 + ( 100 + 99 + .... + 2 + 1 ) = 5750

=) x.100 + 5050 = 5750

=) x.100 = 5750 - 5050 = 200

=) x = 200/100 = 2 

Vậy x = 2

Nếu mình sai thì các bạn sẽ cùng góp ý với mình nhoa !

b)Ta có:\(\left|x-1\right|\ge0\)

\(\Rightarrow-\left|x-1\right|\le0\)

\(\Rightarrow7-\left|x-1\right|\le7\)

\(\Rightarrow A\le7\)

Dấu bằng xảy ra khi:\(\left|x-1\right|=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(\left(x+4\right)+\left(x+6\right)+\left(x+8\right)+...+\left(x+26\right)=210\)

Số các chữ số x: \(\left(26-4\right):2+1=12\left(số\right)\)

\(\Rightarrow12x+\left(4+6+8+...+26\right)=210\)

\(\Rightarrow12x+\dfrac{\left(26+4\right)\left(\dfrac{26-4}{2}+1\right)}{2}=210\)

\(\Leftrightarrow12x+180=210\Leftrightarrow12x=30\Leftrightarrow x=\dfrac{5}{2}\)

(x + 4) + (x + 6) + (x + 8) + ... + (x + 26) = 210
<=> x[(26 - 4) : 2 + 1] + (26 + 4)[(26 - 4) : 2 + 1]  : 2 = 210
<=> 12x + 180 = 210
<=> 12x  = 210 - 180
<=> 12x = 30
<=> x = 2,5

Ta có: \(\left(x+4\right)+\left(x+6\right)+...+\left(x+26\right)=210\)

\(\Leftrightarrow12x=30\)

hay \(x=\dfrac{5}{2}\)

a) 2x + 15 = 45

2x = 45 - 15

2x = 30

x = 30 : 2

x = 15 (nhận)

Vậy x = 15

b) 120 - 2.(x + 3) = 22.52

120 - 2.(x + 3) = 1144

2.(x + 3) = 120 - 1144

2.(x + 3) = - 1024

x + 3 = -1024 : 2

x + 3 = -512

x = - 512 - 3

x = -515 (loại)

Vậy không tìm được x thỏa mãn x là số tự nhiên

c) 11 ⋮ (x - 2)

⇒ x - 2 ∈ Ư(11) = {-11; -1; 1; 11}

⇒ x ∈ {-9; 1; 3; 13}

Do x là số tự nhiên

⇒ x ∈ {1; 3; 13}

d) Do 12 ⋮ x và 18 ⋮ x nên x ∈ ƯC(12; 18)

12 = 2².3

18 = 2.3²

ƯCLN(12; 18) = 2.3 = 6

⇒ x ∈ ƯC(12; 18) = {1; ; 3; 6}

a)x:[(1800+600):30] = 560:(315 -35)

x: [2400:30] = 560:280

x:80=560:280

x:80=2

x=2x80

=160

*sai thì srr nha*

ax:[(1800+600):30]=560:(315-35)

x:[2400:30]=560:280

x:80=2

x=2.80

x=160

b[(250-25):15]:x=(450-60):130

[225:15]:x=390:130

15:x=3

  x=15:3

x=5