a) Điều kiện: \(x\ne\left\{0;\pm2\right\}\)

\(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=[\frac{x^2}{x.\left(x-2\right).\left(x+2\right)}-\frac{6}{3.\left(x-2\right)}+\frac{1}{x+2}]:\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{x-2.\left(x+2\right)+x-2}{\left(x-2\right).\left(x+2\right)}.\frac{x+2}{6}\)

\(=\frac{6}{\left(x-2\right).\left(x+2\right)}.\frac{x+2}{6}\)

\(=-\frac{1}{x-2}\)

b) \(A\) \(Max\)

\(\Rightarrow-\frac{1}{x-2}Max\)

\(\Rightarrow\frac{1}{x-2}Min\)

\(\Rightarrow\left(x-2\right)\) \(Max\)

\(\Rightarrow x\) \(Max\)

\(\Rightarrow x\in\varnothing\)

Nó bị lỗi nên mình ấy lại :
 3x²y²z² = x³y³ y³z³ z³x³ 

(3x²y²z²) / (x³y³ y³z³ z³x³) = 1

3.[(x²y²z²) / (x³y³ y³z³ z³x³)] = 1

(x²y²z²) / (x³y³ y³z³ z³x³) = 1/3

(x²y²z²) / (x³y³) (x²y²z²) / (y³z³) (x²y²z²) / (z³x³) = 1/3

z²/(xy) x/(yz) y²/(zx) = 1/3

Vậy x²/(yz) y²/(xz) z²/(xy) = 1/3

Tham khảo :

Câu hỏi của Nguyễn Bảo Hưng - Toán lớp 8 - Học trực tuyến OLM

a) \(x^2-xy+x-y\)

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

\(=\left(x+1\right)\left(x-y\right)\)

b)\(x^2-2xy+y^2-z^2\)

\(=\left(x^2-2xy+y^2\right)-z^2\)

\(=\left(x-y\right)^2-z^2\)

\(=\left(x-y-z\right)\left(x-y+z\right)\)

c)\(5x-5y+ax-ay\)

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

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

d)\(a^3-a^2x-ay+xy\)

\(=a^2\left(a-x\right)-y\left(a-x\right)\)

\(=\left(a^2-y\right)\left(a-x\right)\)

Bài 2 : 

a) \(x^2-2xy-47^2+y^2\)

\(=x^2-2xy+y^2-47^2\)

\(=\left(x-y\right)^2-47^2\)

\(=\left(x-y-47\right)\left(x-y+47\right)\)

Bài 1

a) x² - xy + x - y

= x.(x - y) + (x - y) 

= (x - y) . (x + 1) 

b) x² - 2xy + y² - z²

= (x - y)² - z²

= (x - y - z) . (x - y + z)

c) 5x - 5y + ax - ay

= 5 . (x - y) + a . (x - y)

= (5 + a ) . (x - y)

d) a³ - a²x - ay + xy 

=

a3−a2x−ay+xya3−a2x−ay+xy

=(a3−a2x)−(ay−xy)=(a3−a2x)−(ay−xy)

=a2(a−x)−y(a−x)=a2(a−x)−y(a−x)

=(a2−y)(a−x)

      \(x^3-2x^2+x\)

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

\(=x\left(x-1\right)^2\)