P = (a + b + c)³ - 4(a³ + b³ + c³) - 12abc

= (a + b + c)³ - 4(a³ + b³ + c³ + 3abc) 

= (a + b + c)³ - 8c³ - 4(a³ + b³ - c³ + 3abc) 

= (a + b + c)³ - (2c)³ - 4(a³ + b³ - c³ + 3abc) 

Có (a + b + c)³ - (2c)³ 

= (a + b - c)[(a + b + c)² + (a + b + c).2c + 4c²]

= (a + b - c)(a² + b² + c² + 2ab + 2bc + 2ca + 2ac + 2bc + 2c² + 4c²)

= (a + b - c)(a² + b² + 7c² + 4bc + 4ac + 2ba)

Lại có a³ + b³ - c³ + 3abc

 = (a + b)³ - c³ - 3ab(a + b) + 3abc

= (a + b - c)[(a + b)² + (a + b)c + c² - 3ab]

= (a + b - c)(a² + b² + c² + ac + bc - ab) 

Khi đó P = (a + b - c)(a² + b² + 7c² + 4bc + 4ac + 2ba) - 4(a + b - c)(a² + b² + c² + ac + bc - ab) 

= (a + b - c)(-3a² - 3b² + 3c² + 6ba)

= 3(a + b - c)(- a² - b² + 2ab + c²)

= 3(a + b - c)[c² - (a - b)²]

= 3(a + b - c)(a + c - b)(c - a + b) 

Nếu P < 0 thì  3(a + b - c)(a + c - b)(c - a + b)  < 0

<=>  (a + b - c)(a + c - b)(c + b - a) < 0

=> Có ít nhất một hạng tử trái dấu với 2 hạng tử còn lại

Với a,b,c > 0

Giả sử \(\left\{{}\begin{matrix}a+b-c< 0\\a+c-b>0\\b+c-a>0\end{matrix}\right.\) => a;b;c không là 3 cạnh tam giác 

hoặc \(\left\{{}\begin{matrix}a+b-c>0\\b+c-a< 0\\a+c-b< 0\end{matrix}\right.\) cũng tương tự

Vậy a,b,c không là 3 cạnh tam giác 

Không kết luận được bất cứ điều gì nếu không có thêm điều kiện a;b;c là các số dương

a)

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

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

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

với x=1002 thì

\(\dfrac{1002+2}{1002-2}=\dfrac{251}{250}\)

\(B=\dfrac{4x^2+x+2}{x^2\left(x-2\right)+\left(x-2\right)}+\dfrac{x^2}{x^2+1}\\ =\dfrac{4x^2+x+2}{\left(x-2\right)\left(x^2+1\right)}+\dfrac{x^2}{x^2+1}\\ =\dfrac{4x^2+x+2}{\left(x-2\right)\left(x^2+1\right)}+\dfrac{x^2\left(x-2\right)}{\left(x^2+1\right)\left(x-2\right)}\\ =\dfrac{4x^2+x+2+x^3-2x^2}{\left(x^2+1\right)\left(x-2\right)}\\ =\dfrac{x^3+2x^2+x+2}{\left(x^2+1\right)\left(x-2\right)}\\ =\dfrac{x^2\left(x+2\right)+\left(x+2\right)}{\left(x^2+1\right)\left(x-2\right)}\\ =\dfrac{\left(x+2\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x-2\right)}\\ =\dfrac{x+2}{x-2}\)

với x=1002 thì

\(\dfrac{1002+2}{1002-2}=\dfrac{251}{250}\)

b)

vì \(\dfrac{x+2}{x-2}=\dfrac{x+2}{x-2}\) nên A=B

\(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)

\(\Leftrightarrow\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}-10=0\)

\(\Leftrightarrow(\dfrac{x-241}{17}-1)+(\dfrac{x-220}{19}-2)+(\dfrac{x-195}{21}-3)+(\dfrac{x-166}{23}-4)=0\)

\(\Leftrightarrow\dfrac{x-258}{17}+\dfrac{x-258}{19}+\dfrac{x-258}{21}+\dfrac{x-258}{23}=0\)

\(\Leftrightarrow\left(x-258\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0\)

\(Do\) \(\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)\ne0\) \(nên\) \(để\) \(gt=0\)

\(\Leftrightarrow x-258=0\)

\(\Leftrightarrow x=258\)

\(Vậy...\)

Bài 1:

a: \(x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\cdot x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)\)

\(=3^3-3\cdot3=27-9=18\)

b: x^2+1/x^2=14

=>(x+1/x)^2-2=14

=>(x+1/x)^2=16

=>x+1/x=4 hoặc x+1/x=-4

TH1: x+1/x=4

\(A=\left(x+\dfrac{1}{x}\right)^3-3\cdot x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)\)

\(=4^3-3\cdot4=64-12=52\)

TH2: x+1/x=-4

\(A=\left(-4\right)^3-3\cdot\left(-4\right)=-64+12=-52\)

Số trung bình cộng là :

[3+(-2022)]:2=-1009,5

:v

i: \(\Leftrightarrow x\left(x+2\right)-2=x-2\)

=>x^2+2x=x

=>x^2+x=0

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

j: \(\Leftrightarrow\dfrac{-5}{\left(x-2\right)\left(x-3\right)}-\dfrac{x+3}{x-2}=0\)

=>-5-(x+3)(x-3)=0

=>x^2-9+5=0

=>x^2-4=0

=>x=2(loại) hoặc x=-2(nhận)

k: \(\Leftrightarrow\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{-x}{2\left(x-3\right)}\)

=>x(x-3)-4x=-x(x+1)

=>x^2-3x-4x+x^2+x=0

=>2x^2-6x=0

=>2x(x-3)=0

=>x=3(loại) hoặc x=0(nhận)

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

=>2x^2-2x=-2x^2+x+1

=>4x^2-3x-1=0

=>4x^2-4x+x-1=0

=>(x-1)(4x+1)=0

=>x=1(loại) hoặc x=-1/4(nhận)

m: \(\Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{-\left(x-5\right)}{2x\left(x+5\right)}\)

\(\Leftrightarrow x\left(x+25\right)-2\left(x+5\right)^2=-\left(x-5\right)^2\)

=>\(x^2+25x-2x^2-20x-50=-x^2+10x-25\)

=>5x-50=10x-25

=>-5x=25

=>x=-5(loại)

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

=>-3(x-1)-(x-1)^2=2

=>-3x+3-x^2+2x-1=2

=>-x^2-x+2=2

=>-x^2-x=0

=>x(x+1)=0

=>x=0(nhận) hoặc x=-1(loại)

\(DKXD:x\ne\pm1\)

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

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

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

\(\Leftrightarrow2-3x+3+x^2-2x+1=0\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=3\left(n\right)\\x_2=2\left(n\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;2\right\}\)

10
`11-2x=4x-5`

`<=>-2x-4x=-5-11`

`<=>-6x=-16`

`<=>x=8/3`

11)

`2x+3=12-3x`

`<=>2x+3x=12-3`

`<=>5x=9`

`<=>x=9/5`

12)

`4x-5=15+9x`

`<=>4x-9x=15+5`

`<=>-5x=20`

`<=>x=-4`

13)

`5-3x=11+2x`

`<=>-3x-2x=11-5`

`<=>-5x=6`

`<=>x=-6/5`

14)

`6x+12=3x-5`

`<=>6x-3x=-5-12`

`<=>3x=-17`

`<=>x=-17/3`

15)

`7-3x=9-x`

`<=>-3x+x=9-7

`<=>-2x=2`

`<=>x=-1`

16)

`-2x+3=12-5x`

`<=>-2x+5x=12-3`

`<=>3x=9`

`<=>x=3`

17)

`2(x+1)=5x-7`

`<=>2x+2=5x-7`

`<=>2x-5x=-7+2`

`<=>-3x=-5`

`<=>x=5/3`

15: =>-3x+7-9+x=0

=>-2x-2=0

=>2x+2=0

=>x=-1

17: =>5x-7=2x+2

=>3x=9

=>x=3

16: =>-2x+3+5x-12=0

=>3x-9=0

=>x=3

13: =>-3x+5-2x-11=0

=>-5x-6=0

=>x=-6/5

12: =>4x-9x=15+5

=>-5x=20

=>x=-4

11: =>5x=9

=>x=9/5

10: =>-2x-4x=-5-11

=>-6x=-16

=>x=8/3

420,89

1: \(=x^3-8-x^3-10=-2\)

2: \(\Leftrightarrow\left(x+6\right)^3=729\)

=>x+6=9

=>x=3

câu 1

với mọi giá trị của x thì k luân bằng -18

=> \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+10\right)\\ \left(x^3-8\right)-\left(x^3+10\right)\\ x^3-8-x^3-10\\ =-18\)

câu 2

\(x^3+18x^2+108x+216=729\\ \left(x+6\right)^3=729\\ x+6=9\\ x=9-6\\ x=3\)