bài 1:

a) x(x-2)-5y-(x-2)=(x-5y)(x-2)

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

bài 2 bạn tự luyện nhé

\(\frac{4\left(3x-2y\right)}{16}=\frac{3\left(2z-4x\right)}{9}=\frac{2\left(4y-3z\right)}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{29}=0\)

\(\Leftrightarrow3x-2y=0\Leftrightarrow\frac{x}{2}=\frac{y}{3}\)

\(\Leftrightarrow2z-4x=0\Leftrightarrow\frac{x}{2}=\frac{z}{4}\)

\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{2x+4y+5z}{4+12+20}=\frac{8}{36}=\frac{2}{9}=\frac{2x+3y-z}{4+12-4}\)=> A= 2x+3y -z = 12.2/9  =8/3

a: \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)

\(=\dfrac{-6x^3y^4}{2x^3y^3}+\dfrac{4x^4y^3}{2x^3y^3}\)

\(=-3y+2x\)

b: \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}=\dfrac{5x^4y^2}{x^3y^2}-\dfrac{x^3y^2}{x^3y^2}\)

\(=5x-1\)

c: \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)

\(=-\dfrac{27x^3y^5}{3x^2y^3}-\dfrac{9x^2y^4}{3x^2y^3}+\dfrac{6x^3y^3}{3x^2y^3}\)

\(=-9xy^2-3y+2x\)

a) \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)

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

\(=-3y+2x\)

\(=2x-3y\)

b) \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}\)

\(=\dfrac{5x\cdot x^3y^2-x^3y^2\cdot1}{x^3y^2}\)

\(=\dfrac{x^3y^2\cdot\left(5x-1\right)}{x^3y^2}\)

\(=5x-1\)

c) \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)

\(=\dfrac{-3x^2y^3\cdot-9xy^2+-3x^2y^3\cdot-3y+-3x^2y^3\cdot2x}{-3x^2y^3}\)

\(=\dfrac{-3x^2y^3\cdot\left(-9xy^2-3y+2x\right)}{-3x^2y^3}\)

\(=-9xy^2-3x+2x\)

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

Bài 1: 

a) Ta có: \(\left(15x^2\cdot y^2\cdot z\right):3xyz\)

\(=\dfrac{15x^2y^2z}{3xyz}\)

\(=5xy\)

b) Ta có: \(3x^2\cdot\left(5x^2-4x+3\right)\)

\(=3x^2\cdot5x^2-3x^2\cdot4x+3x^2\cdot3\)

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

c) Ta có: \(\left(2x^2-3x\right):\left(x-4\right)\)

\(=\dfrac{2x^2-8x+5x-20+20}{x-4}\)

\(=\dfrac{2x\left(x-4\right)+5\left(x-4\right)+20}{x-4}\)

\(=2x+5+\dfrac{20}{x-4}\)

d) Ta có: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)

\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)

\(=-15x^3y^2+25x^2y^2-5xy^3\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

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

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

\(=3\left(2x+3y\right)\left(x-y\right)\)

Bài 4:

bạn giải đi bạn

Theo đề ta có: \(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

Đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\inℕ^∗\right)\)

Suy ra: \(x=3k;y=4k;z=5k\) Thay vào biểu thức P ta có:

\(P=\frac{3k+8k+15k}{6k+12k+20k}+\frac{6k+12k+20k}{9k+16k+25k}+\frac{9k+16k+25k}{12k+20k+30k}\)

\(P=\frac{26k}{38k}+\frac{38k}{50k}+\frac{50k}{62k}=\frac{13}{19}+\frac{19}{25}+\frac{25}{31}=\frac{33141}{14725}\)

\(a,A-B+C\)

\(=\left(5xy^2-4x^2y-6xy^2\right)-\left(8yx^2-4y^2x+3y^2\right)+\left(-2xy^2+3yx^2+5x^2\right)\)

\(=5xy^2-4x^2y-6xy^2-8yx^2+4y^2x-3y^2-2xy^2+3yx^2+5x^2\)

\(=\left(5xy^2-6xy^2-2xy^2+4xy^2\right)+\left(-4x^2y-8x^2y+3x^2y\right)+\left(-3y^2+5x^2\right)\)

\(=xy^2-9x^2y-3y^2+5x^2\)

\(b,2\left(A+B\right)+C=2A+2B+C\)

\(=2\left(5xy^2-4x^2y-6xy^2\right)+2\left(8yx^2-4y^2x+3y^2\right)-2xy^2+3yx^2+5x^2\)

\(=10xy^2-8x^2y-12xy^2+16x^2y-8xy^2+6y^2-2xy^2+3x^2y+5x^2\)

\(=-12xy^2+11x^2y+5x^2+6y^2\)