\(ĐKXĐ:x\ne0;x\ne2\)

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

\(\Leftrightarrow4x^2-4x^3+x^4=-2\left(x^3-2x^2\right)\)

\(\Leftrightarrow4x^2-4x^3+x^4=-2x^3+4x^2\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\left(ktm\right)\)

Vậy không có x để phân thức bằng -2

Ta có : \(\frac{4x^2-4x^3+x^4}{x^3-2x^2}=-2\)

( ĐKXĐ : \(x\ne0,x\ne\pm\sqrt{2}\) )

\(\Leftrightarrow\frac{4x^2-4x^3+x^4}{x^3-2x^2}+2=0\)

\(\Leftrightarrow4x^2-4x^3+x^4+2\left(x^3-2x^2\right)=0\)

\(\Leftrightarrow-2x^3+x^4=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\) ( Loại \(x=0\) không thỏa mãn ĐKXĐ )

Vậy : \(x=2\) thỏa mãn đề.

Yey anh sai :))

a) Ta có: \(A=\left(\dfrac{3}{2x+4}+\dfrac{x}{2-x}+\dfrac{2x^2+3}{x^2-4}\right):\dfrac{2x-1}{4x-8}\)

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

\(=\dfrac{3x-6-2x^2-4x+4x^2+6}{2\left(x+2\right)\left(x-2\right)}\cdot\dfrac{4\left(x-2\right)}{2x-1}\)

\(=\dfrac{2x^2-x}{x+2}\cdot\dfrac{2}{2x-1}\)

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

\(=\dfrac{2x}{x+2}\)

\(A=\left[\frac{6x^2}{x^3-1}-\frac{2x-2}{x^2+x+1}-\frac{1}{x-1}\right]:\frac{x^2+9}{\left(x-1\right)\left(9-4x\right)}\)

\(=\left[\frac{6x^2}{x^3-1}-\frac{\left(2x-2\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\cdot\frac{\left(x-1\right)\left(9-4x\right)}{x^2+9}\)

\(=\frac{6x^2-\left(2x^2-4x+2\right)-x^2-x-1}{\left(x^2+x+1\right)\left(x-1\right)}\cdot\frac{\left(x-1\right)\left(9-4x\right)}{x^2+9}\)

\(=\frac{5x^2-2x^2+4x-2-x-1}{\left(x^2+x+1\right)}\cdot\frac{\left(9-4x\right)}{x^2+9}\)

\(=\frac{3x^2+3x-3}{\left(x^2+x+1\right)}\cdot\frac{\left(9-4x\right)}{x^2+9}\)

Biểu thức A bạn viết đúng chưa?

a: Để A là số nguyên thì

x^3-2x^2+4 chia hết cho x-2

=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{3;1;4;0;6;-2\right\}\)

b: Để B là số nguyên thì

\(3x^3-x^2-6x^2+2x+9x-3+2⋮3x-1\)

=>\(3x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3}\right\}\)

1. Ta có \(|3x-1|=\frac{1}{2}\)

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

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

\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)

Sau đó tự thay x vào đa thức theo 2 trường hợp trên nha

Sai thì thôi nha bn mik cx chưa lm dạng này bh

Câu 1:

\(A\left(x\right)=6x^4-4x^2-3+9x+5x^2-7x-2x^4+4-2x-4x^4\)

\(=\left(6x^4-2x^4-4x^4\right)+\left(-4x^2+5x^2\right)+\left(-7x-2x\right)+9x+\left(-3+4\right)\)

\(=x^2+9x+1\)

Ta có: \(\left|3x-1\right|=\frac{1}{2}\)

TH1: \(3x-1=\frac{1}{2}\Rightarrow3x=\frac{1}{2}+1=\frac{3}{2}\Rightarrow x=\frac{3}{2}:3=\frac{1}{2}\)

\(A\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+9\cdot\frac{1}{2}+1=\frac{1}{4}+\frac{9}{2}+1=\frac{23}{4}\)

TH2: \(3x-1=\frac{-1}{2}\Rightarrow3x=\frac{-1}{2}+1=\frac{1}{2}\Rightarrow x=\frac{1}{2}:3=\frac{1}{6}\)

\(A\left(\frac{1}{6}\right)=\left(\frac{1}{6}\right)^2+9\cdot\frac{1}{6}+1=\frac{91}{36}\)

Câu 2:

Theo đề, ta có:

\(B\left(x\right)=\left(2x+3\right)^2+1=3\frac{7}{9}\)

\(\left(2x+3\right)^2=\frac{34}{9}-1=\frac{25}{9}\)

TH1: \(2x+3=\frac{-5}{3}\Rightarrow2x=\frac{-5}{3}-3=-\frac{14}{3}\)

TH2: \(2x+3=\frac{5}{3}\Rightarrow2x=\frac{5}{3}-3=\frac{-4}{3}\)

\(A=\frac{x^3-4x^2+4x-10}{x-3}\)( ĐKXĐ : x ≠ 3 )

\(=\frac{x^3-3x^2-x^2+3x+x-3-7}{x-3}\)

\(=\frac{x^2\left(x-3\right)-x\left(x-3\right)+\left(x-3\right)-7}{x-3}\)

\(=\frac{\left(x-3\right)\left(x^2-x+1\right)-7}{x-3}\)

\(=\frac{\left(x-3\right)\left(x^2-x+1\right)}{x-3}-\frac{7}{x-3}\)

\(=\left(x^2-x+1\right)-\frac{7}{x-3}\)

Vì x ∈ Z nên ( x² - x + 1 ) ∈ Z

nên để A ∈ Z thì \(\frac{7}{x-3}\)∈ Z

hay ( x - 3 ) ∈ Ư(7) = { ±1 ; ±7 }

Các giá trị tm ĐKXĐ

Vậy x ∈ { ±4 ; 2 ; 10 } thì A ∈ Z

\(ĐKXĐ:x\ne3\)

\(A=\frac{x^3-4x^2+4x-10}{x-3}=\frac{x^3-3x^2-x^2+3x+x-3-7}{x-3}\)

\(=\frac{x^2\left(x-3\right)-x\left(x-3\right)+\left(x-3\right)-7}{x-3}\)

\(=\frac{\left(x-3\right)\left(x^2-x+1\right)-7}{x-3}=\left(x^2-x+1\right)-\frac{7}{x-3}\)

Vì \(x\inℤ\)\(\Rightarrow x^2-x+1\inℤ\)

\(\Rightarrow\)Để \(A\inℤ\)thì \(\frac{7}{x-3}\inℤ\)\(\Rightarrow7⋮x-3\)

\(\Rightarrow x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Leftrightarrow x\in\left\{-4;2;4;10\right\}\)( thỏa mãn ĐKXĐ )

Vậy \(x\in\left\{-4;2;4;10\right\}\)

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Ta có: \(\dfrac{4x-4}{2x^2-2}\)

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

\(=\dfrac{2}{x+1}\)

Để phân thức có giá trị bằng -2 thì \(\dfrac{2}{x+1}=-2\)

\(\Leftrightarrow x+1=-1\)

hay x=-2(thỏa ĐK)

a) x³ + 2x - 3

=x³+x²+3x-x²+x+3

=x(x²+x+3)-1(x²+x+3)

=(x-1)(x²+x+3)

b) x³ - x² + x + 3

=x³-2x²+3x+x²-2x+3

=x(x²-2x+3)+1(x²-2x+3)

=(x+1)(x²-2x+3)

c) 3x³ - 4x² + 13x - 4

=3x³-3x²+12-x²-x+4

=3x(x²-x+4)-1(x²-x+4)

=(3x-1)(x²-x+4)

d) 6x³ + x² + x + 1

=6x³-2x²+2x+3x²-x+1

=2x(3x²-x+1)+1(3x²-x+1)

=(2x+1)(3x²-x+1)

e)bạn phân tích tương tự nhé mk cho đáp án để bạn đổi chiếu nè

=(2x+1)(2x²+2x+1)