\(P=\frac{6x^2+2x-3xy-y}{6x-3y}\)

a)Để phân thức P đc xác định thì \(6x-3y\ne0\Rightarrow6x\ne3y\Rightarrow2x\ne y\)

b\(P=\frac{6x^2+2x-3xy-y}{6x-3y}\)

\(P=\frac{3x.\left(2x-y\right)+\left(2x-y\right)}{3.\left(2x-y\right)}=\frac{\left(3x+1\right).\left(2x-y\right)}{3.\left(2x-y\right)}=\frac{3x+1}{3}\)(Do 2x-y\(\ne0\Rightarrow2x-y\ne0\)

\(P=\frac{6x^2+2x-3xy-y}{6x-3y}\)

a) Tìm đkxđ

6x-3y \(\ne\)0

=> 6x \(\ne\)0 ; 3y \(\ne\) 0

=> x \(\ne\) 0 ; y \(\ne\) 0

vậy đkxđ của x \(\ne\) 0 ; y \(\ne\) 0 thì P được xác định

b) Rút gọn

\(P=\frac{6x^2+2x-3xy-y}{6x-3y}\)

=> \(P=\frac{2x\left(3x+1\right)-y\left(3x+1\right)}{3\left(2x-y\right)}\)

=>\(P=\frac{\left(2x-y\right)\left(3x+1\right)}{3\left(2x-y\right)}\)

\(\Rightarrow P=\frac{3x+1}{3}\)

Vậy bt P=____ khi rút gọn = 3x+1/3

đặt phép chia:F(x)=x^2+3x+a : g(x)=x+1.

x^2+3x+a x+1 x+2 x^2+x 2x+a 2x+2 - - a-2

Để f(x)___ \(⋮\) g(x)____ thì a-2=0

Ta có: a-2=0=> a=2

Vậy x=2 thì f(x)___ \(⋮\) g(x)____

đề bài là j vậy ạ

Answer:

a) \(x^3-9x=0\)

\(\Rightarrow x.\left(x^2-9\right)=0\)

\(\Rightarrow x.\left(x-3\right).\left(x+3\right)=0\)

Trường hợp 1: \(x=0\)

Trường hợp 2: \(x-3=0\Rightarrow x=3\)

Trường hợp 3: \(x+3=0\Rightarrow x=-3\)

b) \(x^2-\left(2x+1\right)^2=0\)

\(\Rightarrow\left(x-2x-1\right).\left(x+2x+1\right)=0\)

\(\Rightarrow\left(-x-1\right).\left(3x+1\right)=0\)

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

c) \(3x.\left(x-2\right)-x+2=0\)

\(\Rightarrow3x.\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right).\left(3x-1\right)=0\)

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

d) \(x^3-6x^2+12x-8=0\)

\(\Rightarrow x^3-3x^2.2+3x.2^2-2^3=0\)

\(\Rightarrow\left(x-2\right)^3=0\)

\(\Rightarrow x=2\)

e) Mình thấy đề bài sai.

f) \(x^2-2x=-1\)

\(\Rightarrow x^2-2x+1=0\)

\(\Rightarrow\left(x-1\right)^2=0\)

\(\Rightarrow x=1\)

g) \(x^2-6x+5=0\)

\(\Rightarrow x^2-x-5x+5=0\)

\(\Rightarrow x.\left(x-1\right)-5.\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right).\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\)

h) \(4x^2+1=4x\)

\(\Rightarrow4x^2-4x+1=0\)

\(\Rightarrow\left(2x-1\right)^2=0\)

\(\Rightarrow x=\frac{1}{2}\)

i) \(9x^2-1=0\)

\(\Rightarrow9x^2=0+1\)

\(\Rightarrow x^2=1:9\)

\(\Rightarrow x=\pm\frac{1}{3}\)

k) Bạn xem lại đề bài.