\(\frac{4x^2}{x^2+2x}=\frac{A}{x}\)\(\Rightarrow\frac{x\cdot4x}{x\left(x+2\right)}=\frac{A}{x}\)

\(\Rightarrow\frac{4x}{x+2}=\frac{A}{x}\Rightarrow4x^2=A\left(x+2\right)\)\(\Rightarrow A=\frac{4x^2}{x+2}\)

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

đề đúng này hả /hoi-dap/question/139801.html nick kia giải r` kia

\(\frac{4x^2-16}{x^2+2x}=\frac{A}{x}\)\(\Leftrightarrow\frac{4\left(x^2-4\right)}{x\left(x+2\right)}=\frac{A}{x}\)

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

\(\Leftrightarrow4\left(x-2\right)=A\Leftrightarrow A=4x-8\)

ko ai rảnh để trả lời đâu

\(B-2x^2y^3z^2+\frac{2}{3}y^4-\frac{1}{5}x^4y^3=A\)

\(\Rightarrow B=A+2x^2y^3-\frac{2}{3}y^4+\frac{1}{5}x^4y^3\)

\(\Rightarrow B=-4x^5y^3+x^4y^3\cdot3x^2y^3z^2+4x^5y^3+x^2y^3z^2-2y^4+2x^2y^3z^2-\frac{2}{3}y^4+\frac{1}{5}x^4y^3\)

\(=\left(-4x^5y^3+4x^5y^3\right)+\left(x^2y^3z^2+2x^2y^3z^2\right)+x^4y^3\cdot3x^2y^3z^2-\left(2y^4+\frac{2}{3}y^4\right)-\frac{1}{5}x^4y^3\)

\(=3x^2y^3z^2+x^4y^3\cdot3x^2y^3z^2-\frac{8}{6}y^4-\frac{1}{5}x^4y^3\)

Ngu vcl

P(x) = ax² + 2x + 1

P(1/2) = 1 <=> a . (1/2)² + 2 . 1/2 + 1 = 1

                 <=> a . 1/4 + 1 + 1 = 1

                 <=> a . 1/4 = -1

                 <=> a = -4

Ta có : \(P\left(x\right)=ax+2x+1\)

\(P\left(\frac{1}{2}\right)=a.\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\right)+1=1\)

\(\frac{a}{4}+1=0\Leftrightarrow\frac{a}{4}=-1\Leftrightarrow a=-4\)

Ta có : \(P\left(x\right)=ax+2x+1\)

\(P\left(\frac{1}{2}\right)=a.\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\right)+1=1\)

\(\frac{a}{4}+1=0\Leftrightarrow\frac{a}{4}=-1\Leftrightarrow a=-4\)

a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:

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

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

b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)

=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)

c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)

d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6