c)  \(P=\frac{x^2-2x+2012}{x^2}\)  \(\left(x\ne0\right)\)  và \(\left(x\ge1\right)\)

Ta có:  \(P=\frac{x^2-2x+2012}{x^2}\)  \(\Leftrightarrow\)  \(P=\frac{2012x^2-2.2012x+2012^2}{2012x^2}\)

\(\Leftrightarrow\)  \(P=\frac{\left(x-2012\right)^2+2011x^2}{2012x^2}\)  \(\Leftrightarrow\)  \(P=\frac{\left(x-2012\right)^2}{2012x^2}+\frac{2011}{2012}\ge\frac{2011}{2012}\)  với mọi  \(x\ge1\)

Dấu  \("="\)  xảy ra  \(\Leftrightarrow\)  \(\left(x-2012\right)^2=0\)

                              \(\Leftrightarrow\)  \(x-2012=0\)

                              \(\Leftrightarrow\)  \(x=2012\)

Vậy, \(P_{min}=\frac{2011}{2012}\)  khi  \(x=2012\)

b) Từ giả thiết \(a^2+b^2+c^2=\left(a+b+c\right)^2\) , ta suy ra  \(ab+bc+ca=0\)

nên  \(a^2+2bc=a^2+bc+\left(-ab-ac\right)=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)

Tương tự,  \(b^2+2ca=\left(b-a\right)\left(b-c\right)\)  \(;\) \(c^2+2ab=\left(c-a\right)\left(c-b\right)\)

Do đó,   \(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

a) \(5x^5-x^3-\frac{1}{2}x^2\)

b) \(2x^3y^2-\frac{2}{3}x^4y+\frac{2}{3}x^2y^2\)

c) \(-2x^4y+\frac{5}{2}x^2y^2-x^2y\)

6 nha bạn

a) dieu kien xac dinh la :x>0   x khong phai phan so

b)A=1-2+xmu 2 +x -2/4-xmu 2 ( vi x/x=1 vi du x= 2 thi 2/2 =1)

c)x=2   x=0.5

d) x=6

a)16-x²=4²-x²=(4+x)(4-x)

b)x⁴-25=(x²)²-5²=(x²-5)(x²+5)

c)4x²=(2x)²=(2x).(2x)

d)(a+b)²-1=(b+a-1)(b+a+1)

e)(x+y)²-(x-y)²=x²+2xy+y²-x²+2xy-y²=4xy

f)4(x+2y)²-9(2y-x)²=(5x-2y)(10y-x)

g)1-4x+4x²=(2x-1)²

h) 25x²-20xy+4y²=(2y-5x)²

\(a,x^2-5x\)

\(=x\left(x-5\right)\)

\(b,5x\left(x+5\right)+4x+20\)

\(=5x\left(x+5\right)+4\left(x+5\right)\)

\(=\left(5x+4\right)\left(x+5\right)\)

\(c,7x\left(2x-1\right)-4x+2\)

\(=7x\left(2x-1\right)-2\left(2x-1\right)\)

\(=\left(7x-2\right)-\left(2x-1\right)\)

\(d,x^2-16+2\left(x+4\right)\)

\(=x^2-16+2x+8\)

\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) ) 

\(e,x^2-10x+9\)

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

\(=x\left(x-1\right)-9\left(x-1\right)\)

\(=\left(x-9\right)\left(x-1\right)\)

\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé ) 

\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)

\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)

\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)

Vậy ...