còn-gấp-ko

3:

a: =>2x^2+8x-3x-12>2x^2+2

=>5x>14

=>x>14/5

b: =>3(3x-1)-2(5x+1)>4*3*2=24

=>9x-3-10x-2>24

=>-x-5>24

=>x+5<-24

=>x<-29

4:

Gọi vận tốc thật của cano là x

Theo đề, ta có phương trình:

5(x+2)=6(x-2)

=>5x+10=6x-12

=>-x=-22

=>x=22

Độ dài quãng đường là 5(22+2)=120km

Đăng lại vì ko ai giải 🥺

thôi bọn mềnh cũng chệu boạn nhóe

a: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=2x^2-3x+1\)

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

d: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=\left(x-1\right)^2\)

h: \(=\dfrac{x^3+x^2-3x^2-3x+8x+8}{x+1}=x^2-3x+8\)

\(c,\)

\(\left(6x^3-19x^2+23x-12\right):\left(2x-3\right)\)

\(=\left(6x^3-10x^2-9x^2+8x+15x-12\right):\left(2x-3\right)\)

\(=\left[\left(6x^3-10x^2+8x\right)-\left(9x^2-15x+12\right)\right]:\left(2x-3\right)\)

\(=\left[2x\left(3x^2-5x+4\right)-3\left(3x^2-5x+4\right)\right]:\left(2x-3\right)\)

\(=\left[\left(3x^2-5x+4\right)\left(2x-3\right)\right]:\left(2x-3\right)\)

\(=3x^2-5x+4\)

\(e,\)

\(\left(6x^3-5x^2+4x-1\right):\left(2x^2-x+1\right)\)

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

\(=\left[\left(6x^3-3x^2+3x\right)-\left(2x^2-x+1\right)\right]:\left(2x^2-x+1\right)\)

\(=\left[3x\left(2x^2-x+1\right)-\left(2x^2-x+1\right)\right]:\left(2x^2-x+1\right)\)

\(=\left[\left(2x^2-x+1\right)\left(3x-1\right)\right]:\left(2x^2-x+1\right)\)

\(=3x-1\)

\(f,\)

\(\left(x^3-2x^2-5x+6\right):\left(x+2\right)\)

\(=\left(x^3-4x^2+2x^2+3x-8x+6\right):\left(x+2\right)\)

\(=\left[\left(x^3+2x^2\right)-\left(4x^2+8x\right)+\left(3x+6\right)\right]\)

\(=\left[x^2\left(x+2\right)-4x\left(x+2\right)+3\left(x+2\right)\right]\)

\(=\left[\left(x+2\right)\left(x^2-4x+3\right)\right]:\left(x+2\right)\)

\(=x^2-4x+3\)

\(g,\)

\(\left(x^3-2x^2-5x+6\right):\left(x+2\right)\)

\(=\left(x^3+2x^2-4x^2-8x+3x+6\right):\left(x+2\right)\)

\(=\left[\left(x^3+2x^2\right)-\left(4x^2+8x\right)+\left(3x+6\right)\right]:\left(x+2\right)\)

\(=\left[x^2\left(x+2\right)-4x\left(x+2\right)+3\left(x+2\right)\right]:\left(x+2\right)\)

\(=\left[\left(x+2\right)\left(x^2-4x+3\right)\right]:\left(x+2\right)\)

\(=x^2-4x+3\)

giúp mik điiiiiiiii mad 

Bài 1: 

a: 3/5+4/9=27/45+20/45=47/45

Bài 4: 

a: =1/5+1/6+5/8=24/120+20/120+75/120=119/120

b: =8/12+2/12+5/12=15/12=5/4

 a)\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{5}{8}=\dfrac{24}{120}+\dfrac{20}{120}+\dfrac{75}{120}=\dfrac{119}{120}\)

b)\(\dfrac{8}{12}+\dfrac{2}{12}+\dfrac{5}{12}=\dfrac{15}{12}=\dfrac{5}{4}\)