áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c+b+c-a+a+c-b}{a+b+c}=1\)

\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}\Rightarrow a=b=c}\)thay vào P ta có P=8

1.\(x\left(x+y\right)-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\)

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

a, \(x\left(x+y\right)-3x-3y=x\left(x+y\right)-3\left(x-y\right)=\left(x-3\right)\left(x-y\right)\)

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

\(=9x^2+6x+1-x^2+4x-4=8x^2+10x-3=\left(4x-1\right)\left(2x+3\right)\)

Sửa đề :

\(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(9x^2+6x+4\right)\)

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

\(=27x^3+8-27x^3+8\)

\(=16\)

Vậy biểu thức trên không phụ thuộc và biến x

Sửa đề : \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(9x^2+6x+4\right)\)

\(=27x^3+8-\left(27x^3-8\right)=16\)

Vậy giá trị biểu thức ko phụ thuộc biến x 

Đk: x khác 1

a) Ta có: A = \(\frac{x^3-1}{x-1}=\frac{ \left(x-1\right)\left(x^2+x+1\right)}{x-1}=x^2+x+1\)

b) Ta có: \(A=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

Dấu "=" xảy ra <=> \(x+\frac{1}{2}=0\) <=> \(x=-\frac{1}{2}\)

Vậy minA = 3/4 <=> x = -1/2

a,
A=\(\frac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}\)
=\(x^2+x+1\)
b,
Ta có: \(x^2+x+1=x^2+2.\frac{1}{2}.x+\frac{1}{4}-\frac{1}{4}+1 =\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow min_A=\frac{3}{4}\)
Dấu ''='' xảy ra khi :\(x+\frac{1}{2}=0\)
                            \(\Leftrightarrow x=\frac{-1}{2}\)
Vậy \(min_A=\frac{3}{4}\)khi \(x=\frac{-1}{2}\)

Bài làm

a) 3ab² - 12a²b + 9ab = 3ab( b - 4a + 3 )

b) x²( 2x - 3 ) - 5x( 2x - 3 ) = x( 2x - 3 )( x - 5 )

c) x² - 2020x - y² + 2020y 

= ( x² - y² ) - ( 2020x - 2020y )

= ( x - y )( x + y ) - 2020( x - y )

= ( x - y )( x + y - 2020 )

\(\frac{5x+4}{x^2-4x}+\frac{x-2}{x}-\frac{x+2}{x-4}=\frac{5x+4}{x\left(x-4\right)}+\frac{x-2}{x}-\frac{x+2}{x-4}\)

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

Khử mẫu : \(5x+4+x^2-4x-2x+8-x^2-2x\)

\(=-x+12\)

Bài làm

Ta có : \(\frac{5x+4}{x^2-4x}+\frac{x-2}{x}-\frac{x+2}{x-4}\)

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne4\end{cases}}\)

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

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

\(=\frac{5x+4}{x\left(x-4\right)}+\frac{x^2-6x+8}{x\left(x-4\right)}-\frac{x^2+2x}{x\left(x-4\right)}\)

\(=\frac{5x+4+x^2-6x+8-x^2-2x}{x\left(x-4\right)}\)

\(=\frac{-3x+12}{x\left(x-4\right)}\)

\(=\frac{-3\left(x-4\right)}{x\left(x-4\right)}=-\frac{3}{x}\)