Lời giải:

a.

$12x+19x-123=280$

$31x=280+123$

$31x=403$

$x=403:31=13$

b.

$5-(17-3)=x-(2-5)$

$-9=x-(-3)$

$x=(-9)+(-3)=-12$

a.12x+19x-123=280

12x+19x =280+123

12x+19x =403

31x =403

x=403:31

x=13

b. 5-(17-3)=x-(2-15)

5-14 =x-(-13)

-9 =x+13

=> x=(-9)-13=-22

Tìm x:

a. 12x + 19x - 123 = 280

(12 + 19).x - 123 = 280

31.x - 123 = 280

31.x = 280 + 123

31.x = 403

x = 403 : 31

x = 13

b) 5 - (17 - 3) = x - (2 - 15)

5 - 14 = x - (-13)

(- 9) = x + 13

=> x = (-9) - 13 = -22

Ko bt làm

1 3 x − 1 + 1 2 x + 4 = 1 9 x − 2 + 1 5 − 4 x Đ K :   x ≠ 1 3 , x ≠ − 2 , x ≠ 2 9 , x ≠ 5 4

Ta có pt:  5 x + 3 ( 3 x − 1 ) ( 2 x + 4 ) = 5 x + 3 ( 9 x − 2 ) ( 5 − 4 x )

< = > x = − 3 5 ( 3 x − 1 ) ( 2 x + 4 ) = ( 9 x − 2 ) ( 5 − 4 x ) < = > x = − 3 5 6 x 2 + 12 x − 2 x − 4 = − 36 x 2 + 45 x + 8 x − 10 < = > x = − 3 5 ( T M ) x = 6 7 ( T M ) x = 1 6 ( T M )

Vậy phương trình đã có có 3 nghiệm phân biệt như trên.

\(6x^2-x-2=6x^2+3x-4x-2=3x\left(2x+1\right)-2\left(2x+1\right)=\left(2x+1\right)\left(3x-2\right)\)

\(12x^2+19x+4=12x^2+3x+16x+4=3x\left(4x+1\right)+4\left(4x+1\right)=\left(3x+4\right)\left(4x+1\right)\)

\(-6x^2+11x+4=\) / câu này thấy hơi kì :> bạn xem lại đề đc hog?

a, \(5y^2-5x^2+6x+6y=5\left(y-x\right)\left(x+y\right)+6\left(x+y\right)\)

\(=\left(x+y\right)\left(5y-5x+6\right)\)

b, \(12x^2+19x+7=12x^2+12x+7x+7\)

\(=12x\left(x+1\right)+7\left(x+1\right)=\left(12x+7\right)\left(x+1\right)\)

5y² - 5x² + 6x + 6y 

= 5(y² - x²) + 6(x + y) 

= 5(y - x)(x + y) + 6(x + y) 

= (x + y)(5y - 5x + 6) 

b) 12x² + 19x + 7 

= 12x² + 12x + 7x + 7 

= 12x(x + 1) + 7(x + 1) 

= (x + 1)(12x + 7) 

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)

\(=\frac{\left(x-\frac{2}{5}\right)\left(x+3\right)}{\left(x+\frac{1}{3}\right)\left(x+3\right)}\)

\(=\frac{x-\frac{2}{5}}{x+\frac{1}{3}}\)

=\(\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)

=\(\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)

=\(\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)

=\(\frac{2x^2-6x+5x-15}{3x^2-9x-x+3}\)

=\(\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)

=\(\frac{2x+5}{3x-1}\)

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)

\(=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)

\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)

\(=\frac{\left(2x^2-x-15\right)\left(x-3\right)}{\left(3x^2-10x+3\right)\left(x-3\right)}\)

\(=\frac{2x^2-x-15}{3x^2-10x+3}\)

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