Bài 5: Phương trình chứa ẩn ở mẫu

\(ĐKXĐ:\left\{{}\begin{matrix}x+1\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)

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

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

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

\(\Rightarrow x^2+3x-1=x^2-x+1\)

\(\Leftrightarrow x^2+3x-x^2+x=1+1\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\) ( t/m )

Vậy pt có nghiệm duy nhất \(x=\dfrac{1}{2}\)

\(1+\dfrac{1}{2+x}=\dfrac{12}{x^3+8}\)\(\Leftrightarrow\)\(\Leftrightarrow1=\dfrac{12}{x^3+8}-\dfrac{1}{x+2}\)

\(\Leftrightarrow\)\(\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{1}{x+2}\)=1

\(\Leftrightarrow\)\(\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}=1\)

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

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

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

\(\Leftrightarrow\)\(x\left(x-1\right)=0\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(1+\dfrac{1}{2+x}=\dfrac{12}{x^3+8}\)

\(\Leftrightarrow1+\dfrac{1}{2+x}-\dfrac{12}{x^3+8}=0\)

\(\Leftrightarrow1+\dfrac{1}{x+2}-\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)+\left(x^2-2x+4\right)-12=0\)

\(\Leftrightarrow x^3-2x^2+4x+2x^2-4x+8+x^2-2x+4-12=0\)

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

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

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

\(\Leftrightarrow x\left[\left(x^2-1\right)+\left(x-1\right)\right]=0\)

\(\Leftrightarrow x\left[\left(x-1\right)\left(x+1\right)+\left(x-1\right)\right]=0\)

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

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

\(\Rightarrow S=\left\{0;1;-2\right\}\)

Minh thieu dkxd: \(x\ne-2\)

Viết cái đề bài tốn thời gian,diện tích lắm à?

-Giải phương trình.Lần sau ghi rõ yêu cầu nhé.

\(1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}\)

\(\Leftrightarrow\dfrac{8+x^3}{8+x^3}+\dfrac{4+2x+x^2}{8+x^3}=\dfrac{12}{8+x^3}\)

\(\Leftrightarrow8+x^3+4+2x+x^2=12\)

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

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

Làm đc tới đó bí rồi,không biết có đúng không.

Cách mình nha ^3^

\(1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}\)

\(\dfrac{x+2+1}{x+2}=\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\left(x+3\right)\left(x+2\right)\left(x^2-2x+4\right)=12\left(x+2\right)\)

\(\left(x+3\right)\left(x+2\right)\left(x^2-2x+4\right)-12\left(x+2\right)=0\)

\(\left(x+2\right)\left[\left(x+3\right)\left(x^2-2x+4\right)-12\right]=0\)

\(\left(x+2\right)\left[x^3-2x^2+4x+3x^2-6x+12-12\right]=0\)

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

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

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

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

\(\left[{}\begin{matrix}x=-2\\x=0\\x=-2\\x=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=0\\x=1\end{matrix}\right.\)

M=(x − y+1/2 )^2 + 3/4.(y + 1/3 )^2 + 2/3 ≥ 2/3

pt nào cho thì mới biết chứ bạn

x khác 0

x khác 2

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