a: \(\dfrac{x^5-7x^4+15x^2-11x+2}{x^2-2x+1}\)

\(=\dfrac{x^5-2x^4+x^3-5x^4+10x^3-5x^2-11x^3+22x^2-11x-2x^2+4x-2-4x+4}{x^2-2x+1}\)

\(=\dfrac{x^3\left(x^2-2x+1\right)-5x^2\left(x^2-2x+1\right)-11x\left(x^2-2x+1\right)-2\left(x^2-2x+1\right)-4x+4}{x^2-2x+1}\)

\(=x^3-5x^2-11x-2+\dfrac{-4x+4}{x^2-2x+1}\)

b: Để thương bằng -10 thì \(x^3-5x^2-11x+8=0\)

hay \(x\in\left\{6,502;0,588;-2,091\right\}\)

a: \(B=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

b: Để |B|=B thì B>=0

=>\(\sqrt{x}-2>=0\)

hay x>4

a: \(C=\dfrac{1}{x+2}-\dfrac{x\left(x-2\right)\left(x+2\right)}{x^2+4}\cdot\left(\dfrac{1}{\left(x+2\right)^2}+\dfrac{1}{\left(x-2\right)\left(x+2\right)}\right)\)

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

\(=\dfrac{1}{x+2}-\dfrac{x}{x^2+4}\cdot\dfrac{2x}{x+2}\)

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

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

b: Để C=1 thì \(x^2+4=2-x\)

\(\Leftrightarrow x^2+x+2=0\)

hay \(x\in\varnothing\)

Để mình giúp nha

\(x^2+\dfrac{1}{x^2}-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+7=0\)

ĐKXD: x\(\ne0\)

\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+5=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+5=0\)

Đặt \(a=x+\dfrac{1}{x}\) khi đó phương trình trở thành

\(a^2-\dfrac{9}{2}a+5=0\)

\(\Leftrightarrow\left(a\right)^2-2.a.\dfrac{9}{4}+\left(\dfrac{9}{4}\right)^2-\dfrac{81}{16}+5=0\)

\(\Leftrightarrow\left(a+\dfrac{9}{4}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}a-\dfrac{9}{4}=\dfrac{1}{4}\\a-\dfrac{9}{4}=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{5}{2}\\x+\dfrac{1}{x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2+1}{x}=\dfrac{5}{2}\\\dfrac{x^2+1}{x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-\dfrac{5}{2}x+1=0\\x^2-2x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x\right)^2-2.x.\dfrac{5}{4}+\left(\dfrac{5}{4}\right)^2-\dfrac{25}{16}+1=0\\\left(x-1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{5}{4}\right)^2-\dfrac{9}{16}=0\\x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=\dfrac{1}{2}\left(n\right)\\x=1\left(n\right)\end{matrix}\right.\)

Vậy S=\(\left\{1;2;\dfrac{1}{2}\right\}\)

cảm ơn Otasaka Yu

a. A=(3x-2)(3x+2)/(2x-1)(2x+1)+(2x+1)(x-1)=(3x-2)(3x+2)/(2x+1)(3x-2)=3x+2/2x+1

b. A>0

=>3x+2 lớn hơn hoặc bằng 2x+1

=>x lớn hơn hoặc bằng -1

c. Để A thuộc z thì 3x+2 chia hết cho 2x+1

=>x = -1/2

      = 1+ x+1/2x+1 = 1+ 2x+1-x/2x+1=1+ 2x+1/2x+1 -x/2x+1