2,4 x 1,24 + 7,6 x 4,2 + 2,4 x 7,26 + 7,6 x 4,3

=  2,4 x (1,24 + 7,26) + 7,6 x ( 4,2 + 4,3)

= 2,4 x 8,5 + 7,6 x 8,5

= 8,5 x ( 2,4 + 7,6)

= 8,5 x 10

= 85

1-3+5-7+...+21-23
=(1-3)+(5-7)+...+(21-23)
Tổng số cặp là:
[(23-1):2+1]:2=6
Nháp: [(23-1):2+1]:2=(22:2+1):2=12:2=6
=1-3=-2=-2*6=-12
Đây nha bạn.
Chúc bạn học tốt!

\(1-3+5-7+...+21-23\)

\(=\left(1+5+9+13+17+21\right)-\left(3+7+11+15+19+23\right)\)

\(=\left[\left(21-1\right):2+1\right].\left(21+1\right):2-\left[\left(23-3\right):2+1\right].\left(23+3\right):2\)

\(=11.11-11.13\)

\(=121-143\)

\(=-22\)

giúp mik vs ạ

câu 1 phải ko bạn

Bài 1: 

a) Ta có: \(A=\left(x+3\right)^2+2\left(x+3\right)\left(x-3\right)-3x\left(x-3\right)\)

\(=x^2+6x+9+2\left(x^2-9\right)-3x^2+9x\)

\(=-2x^2+15x+9+2x^2-18\)

\(=15x-9\)

b) Ta có: \(\dfrac{2x}{3x-3y}-\dfrac{x^2}{x-y}\)

\(=\dfrac{2x}{3\left(x-y\right)}-\dfrac{3x^2}{3\left(x-y\right)}\)

\(=\dfrac{2x-3x^2}{3\left(x-y\right)}\)

c) Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=0\)

\(\Leftrightarrow x^2-2x+1+4x-x^2=0\)

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

Ta có: \(\frac{x^2+y^2}{x^2-2xy+y^2}-\frac{2}{xy}\)

\(=\frac{x^2+y^2}{\left(x-y\right)^2}-\frac{2}{xy}\)

\(=\frac{xy\left(x^2+y^2\right)-2\left(x-y\right)^2}{xy\left(x-y\right)^2}=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}\)

TA có: \(\left(\frac{x^2+y^2}{x^2-2xy+y^2}-\frac{2}{xy}\right):\left(\frac{1}{x}-\frac{1}{y}\right)^2\)

\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}:\left(\frac{y-x}{xy}\right)^2\)

\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}:\frac{\left(x-y\right)^2}{x^2y^2}\)

\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}\cdot\frac{x^2y^2}{\left(x-y\right)^2}=\frac{\left(x^3y+xy^3-2x^2+4xy-2y^2\right)\cdot xy}{\left(x-y\right)^4}\)

\(-23-150+\left(-9\right)+1913\)

\(=1913-\left(23+150+9\right)\)

\(=1913-182\)

\(=1731\)

\(\left(x^2-2x+3\right)\left(1212x-5\right)\)

\(=1212x^3-5x^2-2424x^2+10x+3636x-15\)

\(=1212x^3-2429x^2+3646x-15\)

\(=1212x^3-5x^2-2424x^2+10x+3636x-15\\ =1212x^3-2429x^2+3646x-15\)

\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)

\(=\dfrac{1}{2}x^3-x^2+\dfrac{3}{2}x-5x^2+10x-15\)

\(=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)

\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)

\(=\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15\)

\(=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)

\(x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)=4x-15x^2+9x^3-3x^2+7x+20=9x^3-18x^2+11x+20\)

x(1 - 3x)(4 - 3x) - (x - 4)(3x + 5)

= (x - 3x²)(4 - 3x) - 3x² - 5x + 12x + 20

= 4x - 3x² - 12x² + 9x³ - 3x² - 5x + 12x + 20

= 9x³ - 18x² + 11x + 20

   ( 5/6 - 2/3 + 1/4 ) : 5/4

= ( 1/6 + 1/4 ) : 5/4

= 5/12 : 5/4

= 1/3

nhớ k cho mình nha ! 

\(10^3+\left[120:\left(2^2.3.5-3^2.5\right)\right]\)

\(=1000+\left[120:\left(4.3.5-9.5\right)\right]\)

\(=1000+120:\left(60-45\right)\)

\(=1000+\left[120:15\right]=1000+8=1008\)