\(\Leftrightarrow4x-8+7⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{3;1;9;-5\right\}\)

⇔4x−8+7⋮x−2⇔4x−8+7⋮x−2

hay 

Sửa đề bài 1 : k => x  P/s : đề sai r :)) 

\(A=\left(3-2x\right)3x^2-8+\left(2x+5\right)\left(3x-2\right)-20x\)

\(=9x^2-6x^3-8+6x^2-4x+15x-10-20x=15x^2-6x^3-18-9x\)

Vậy biểu thức phụ thuộc biến x 

\(B=\left(3-5x\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

\(=6x+33-10x^2-55x-6x^2-14x-9x-21=-72x+12-16x^2\)

Vậy biểu thức phụ thuộc biến x 

Bài 2 : 

a, \(2x\left(x-1\right)-x^2+6=0\Leftrightarrow2x^2-2x-x^2+6=0\)

\(\Leftrightarrow x^2-2x+6=0\)( vô nghiệm )

b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^2-9-x\left(x^2-4\right)=15\Leftrightarrow x^2-9-x^3+12=15\)

\(\Leftrightarrow-x^3+x^2-12=0\Leftrightarrow x=2\)

Sai gi 

`#3107.101107`

\(x(x+5)(x-5) - (x+2)(x^2-2x+4)=5\)

`<=> x(x^2 - 25) - (x^3 + 2^3) = 5`

`<=> x^3 - 25x - x^3 - 8 = 5`

`<=> -25x - 8 = 5`

`<=> -25x = 13`

`<=> x = -13/25`

Vậy, `x = -13/25`

_____

\((x+1)^3 - (x-1)^3 -6(x-1)^2 = -19\)

`<=> x^3 + 3x^2 + 3x + 1 - (x^3 - 3x^2 + 3x - 1) - 6(x^2 - 2x + 1) = -19`

`<=> x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 - 6x^2 + 12x - 6 = -19`

`<=> (x^3 - x^3) + (3x^2 + 3x^2 - 6x^2) + (3x - 3x + 12x) + (1 + 1 - 6) = -19`

`<=> 12x - 4 = -19`

`<=> 12x = -15`

`<=> x = -15/12 = -5/4`

Vậy, `x = -5/4.`

________

`@` Sử dụng các hđt:

`1)` `A^2 + B^2 = (A - B)(A + B)`

`2)` `A^3 + B^3 = (A + B)(A^2 - AB + B^2)`

`3)` `(A - B)^3 = A^3 - 3A^2B + 3AB^2 - B^3`

`4)` `(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3`

`5)` `(A - B)^2 = A^2 - 2AB + B^2.`

a: \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=5\)

=>\(x\left(x^2-25\right)-x^3-8=5\)

=>\(x^3-25x-x^3-8=5\)

=>-25x=13

=>\(x=-\dfrac{13}{25}\)

b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)

=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-19\)

=>\(6x^2+2-6x^2+12x-6=-19\)

=>12x-4=-19

=>12x=-15

=>x=-5/4

3x(x-1)-2(x+2)=4(1-x)-6

=> 3x²-3x-2x-4=4-4x-6

=> 3x²-3x-2x+4x=4+4-6

=> 3x²-x=2

=> x.(3x-1)=2

=>

đúng. mik thử rồi. cho đúng nha^^

a) |2x - 5| + |3x + 1| = 6

Ta có: (5 - 2x) + (-3x - 1) = 6

=> 4 - 5x = 6

=> 5x = 4 - 6 = -2

\(\Rightarrow x=\frac{-2}{5}\), thỏa mãn \(x< \frac{-1}{3}\)

Ta có: (5 - 2x) + (3x + 1) = 6

=> 6 + x = 6

=> x = 6 - 6 = 0, thỏa mãn \(\frac{-1}{3}\le x< \frac{5}{2}\)

Ta có: (2x - 5) + (3x + 1) = 6

=> 5x - 4 = 6

=> 5x = 6 + 4 = 10

=> x = 10 : 5 = 2, không thỏa mãn \(x\ge\frac{5}{2}\)

Vậy \(\left[\begin{array}{nghiempt}x=\frac{-2}{5}\\x=0\end{array}\right.\) thỏa mãn đề bài

mình k có time

2x-5+3x+1=6

suy ra: x(2+3)+(-5+1) =6

suy ra: x5 + (-4) =6

suy ra: x5 = 10

suy ra: x=2

x-4+x-6=2

suy ra: 2x + ( -4-6) =2

suy ra:2x+(-10)=2

suy ra:2x = 12

suy ra: x=6

a, 2,8

b, -4

\(a,0.25x\) - \(\frac{2}{3}x\) =\(1\frac{1}{6}\)

\(\Rightarrow x\cdot\left(0.25-\frac{2}{3}\right)=1\frac{1}{6}\)

\(\Rightarrow x\cdot\frac{-5}{12}=\frac{7}{6}\)

\(\Rightarrow x=\frac{7}{6}:\frac{-5}{12}\)

\(\Rightarrow x=\frac{-14}{5}=-2.8\)

BẠN VÀO CÂU HỎI TƯƠNG TỰ RỒI TÌM .

segerheth

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