`1)1/2:2/3 .... 2/3 : 1/2`

`=>1/2xx3/2 .... 2/3xx2`

`=>3/4 .... 4/3`

Vì `3/4 < 1` và `4/3>1` 

`=>3/4<4/3`

__

`4/7:2/5 ... 4/7 : 3/5`

`=>4/7xx5/2....4/7xx5/3`

`=>20/14...20/21`

`=>10/7...20/21`

Vì `10/7>1` và `20/21<1` 

`=>10/7>20/21`

__

`4/15:4/7....2/5xx10/3`

`=>4/15xx7/4...20/15`

`=>7/15...20/15`

Vì `7<20` nên `7/15<20/15`

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`5/6...15/18-11/18`

`=>5/6...4/18`

Ta có : MSC : `18`

`5/6 = 15/18`

Vì `15>4` nên `5/6 > 4/18`

`2)2/3=(2xx6)/(3xx6)=12/18`

`7/9=(7xx7)/(9xx7)=49/63`

`6/5=(6xx3)/(5xx3)=18/15`

`2/3=(2xx5)/(3xx5)=10/15`

`5/9=(5xx5)/(9xx5)=25/45`

`49/56=(49:7)/(56:7)=7/8`

`6/8=(6xx7)/(8xx7)=42/56`

`2/9=(2xx7)/(9xx7)=14/63`

`49/56=(49:7)/(56:7)=7/8`

\(1,=x^6+27\\ 2,=8x^3+1\\ 3,=x^6+8\\ 4,=27x^3+8\)

1. (x² + 3)(x⁴ - 3x² + 9)

= x⁶ + 27

2. (2x + 1)(4x² - 2x + 1)

= 8x³ + 1

3. (x² + 2)(x⁴ - 2x² + 4)

= x⁶ + 8

4. (3x + 2)(9x² - 6x + 4)

= 27x³ + 8

1: \(\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)

2: \(\left(2x+1\right)\left(4x^2-2x+1\right)=8x^3+1\)

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

4: \(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)

A = \(\dfrac{1}{1\times2}\) + \(\dfrac{3}{2\times5}\) + \(\dfrac{5}{5\times10}\) + \(\dfrac{4}{10\times14}\) + \(\dfrac{6}{14\times20}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{20}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{20}\)

A = \(\dfrac{19}{20}\)

Lời giải:

\(\frac{2}{5}\times \frac{10}{7}\times \frac{-3}{4}\times \frac{10}{7}+\frac{5}{2}:2=\frac{-30}{49}+\frac{5}{4}=\frac{125}{196}\)

Em nên gõ công thức trực quan để đề bài được rõ ràng nhé

B= 1/2 x 2/3 x 3/4 x ...........x 2002/2003 x 2003/2004

1 x 2 x 3 x 4 x .............x 2002 x 2003

2 x 3 x 4 x .............x 2003 x 2004

1

 2004

a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)

=> x=-1  

với \(3x^2+x-2=0\)

ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)

Vậy  ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)

b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)

\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)

\(\Leftrightarrow3x^2=3\)

hay \(x\in\left\{1;-1\right\}\)

c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)

hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)