Bài 4: Quy đồng mẫu thức nhiều phân thức

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(P=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{\left(1+1\right)^2}{x^2+y^2+2xy}=\dfrac{2}{\left(x+y\right)^2}=2\left(x+y=1\right)\)

Đẳng thức xảy ra khi \(x=y=\dfrac{1}{2}\)

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

\(D=\dfrac{x^2}{x-2}\dfrac{\left(x-2\right)^2}{x}+3\)

\(D=x\left(x-2\right)+3\)

\(D=x^2-2x+1+2\)

\(D=\left(x-1\right)^2+2\ge2\)

Dấu"=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)

Vậy MinD là 2 \(\Leftrightarrow x=1\)

1,

\(x^2-2ax+a^2=\left(x-a\right)^2\)

\(x^2-ax=x\left(x-a\right)\)

Vậy MSC: \(\left(x-a\right)^2x\)

2,

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

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

\(x^2+x+1\)

vậy MSC là: \(x\left(x-1\right)\left(x^2+x+1\right)\)

a)

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

\(\dfrac{8}{2x-x^2}=\dfrac{-8}{-x\left(2+x\right)}=\dfrac{8}{x\left(2+x\right)}\)

MTC: \(x\left(2+x\right)\)

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

\(\dfrac{8}{2x-x^2}=\dfrac{-8}{-x\left(2+x\right)}=\dfrac{8}{x\left(2+x\right)}\)

b)

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

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

MTC: \(\left(x-1\right)\left(x+1\right)\)

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

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

\(=\dfrac{2y+x}{y\left(2y-x\right)}+\dfrac{8x}{\left(x-2y\right)\left(x+2y\right)}+\dfrac{2y-x}{y\left(2y+x\right)}\)

\(=\dfrac{2y+x}{y\left(2y-x\right)}-\dfrac{8x}{\left(2y-x\right)\left(x+2y\right)}+\dfrac{2y-x}{y\left(2y+x\right)}\)

\(=\dfrac{\left(2y+x\right)^2}{y\left(2y-x\right)\left(2y+x\right)}-\dfrac{8xy}{y\left(2y-x\right)\left(x+2y\right)}+\dfrac{\left(2y-x\right)^2}{y\left(2y+x\right)\left(2y-x\right)}\)

\(=\dfrac{\left(2y+x\right)^2-8xy+\left(2y-x\right)^2}{y\left(2y-x\right)\left(2y+x\right)}\)

\(=\dfrac{8y^2-8xy+2x^2}{\left(y\right)\left(2y-x\right)\left(2y+x\right)}\)

Phân tích trên tử ta có:

\(=2\left(\left(2y\right)^2+4xy+x^2\right)=2\left(2y+x\right)^2\)

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

CHÚC BẠN HỌC TỐT......

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

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

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

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

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

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

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

Bài 2:

a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)

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

\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{10}{2x+1}\)

b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

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

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

c) Trong ngoặc giữa hai phân số là dấu gì vậy ?

a: Sửa đề: \(\dfrac{2x+1}{2x^2-2}\)

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

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

b: \(\dfrac{3x+2}{x^2-2}=\dfrac{\left(3x+2\right)\left(2x+1\right)}{\left(x^2-2\right)\left(2x+1\right)}=\dfrac{6x^2+7x+2}{\left(x^2-2\right)\left(2x+1\right)}\)

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

c: \(\dfrac{3}{x-1}=\dfrac{3x-6}{\left(x-1\right)\left(x-2\right)}\)

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

\(\dfrac{A}{B}=\dfrac{x^{2n}y^3}{2.\left(-3\right)x^{n+2}y^{n+1}}=\dfrac{-1}{6}x^{2n-n-2}y^{3-n-1}=\dfrac{-1}{6}x^{n-2}y^{2-n}\Rightarrow\left\{{}\begin{matrix}n-2\ge0\\2-n\ge0\end{matrix}\right.\Rightarrow n=2}\)

\(4x^2-4x-1>0\Leftrightarrow4x^2-4x+1-2>0\Leftrightarrow\left(2x-1\right)^2-2>0\)

\(\Leftrightarrow\left(2x-1\right)^2>2\Leftrightarrow\left\{{}\begin{matrix}2x-1>\sqrt{2}\\hoặc\\2x-1< -\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x>1+\sqrt{2}\\hoặc\\2x< 1-\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1+\sqrt{2}}{2}\\hoặc\\a< \dfrac{1-\sqrt{2}}{2}\end{matrix}\right.\) vậy \(x>\dfrac{1+\sqrt{2}}{2}\) hoặc \(x< \dfrac{1-\sqrt{2}}{2}\)

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

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

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