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a: ĐKXĐ: x<>2; x<>-2; x<>0; x<>3
b: \(P=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{\left(x-3\right)}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)
c: 2(x-1)=6
=>x-1=3
=>x=4
Thay x=4 vào P, ta đc:
\(P=\dfrac{-4\cdot4^2\cdot\left(4-2\right)}{\left(4+2\right)\left(4-3\right)}=\dfrac{-64\cdot2}{6}=\dfrac{-128}{6}=-\dfrac{64}{3}\)
a) \(\dfrac{x}{x-3}-\dfrac{x^2+3x}{2x+3}\left(\dfrac{x+3}{x^2-3x}-\dfrac{x}{x^2-9}\right)\)
ĐKXĐ:\(\left\{{}\begin{matrix}x-3\ne0\\2x +3\ne0\\x^2-3x\ne0\\x^2-9\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-\dfrac{3}{2}\\x\ne0\\x\ne\pm3\end{matrix}\right.\)
\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\left(\dfrac{x+3}{x\left(x-3\right)}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}\right)\)
\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}.\dfrac{\left(x+3\right)^2-x^2}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}.\dfrac{\left(x+3-x\right)\left(x+3+x\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right).3\left(2x+3\right)}{\left(2x+3\right)x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x-3}-\dfrac{3}{x-3}\)
\(=\dfrac{x-3}{x-3}\)
=1
\(\Rightarrow\) ĐPCM
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a) Phân thức B xác định \(\Leftrightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\left\{\pm1\right\}\\x\ne-1\end{cases}\Leftrightarrow}x\ne\left\{\pm1\right\}}\)
b) \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\cdot\frac{4x^2-4}{5}\)
\(B=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{3\cdot2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(2x\right)^2-2^2}{5}\)
\(B=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(2x-2\right)\left(2x+2\right)}{5}\)
\(B=\frac{10\cdot2\left(x-1\right)\cdot2\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)\cdot5}\)
\(B=\frac{40\left(x-1\right)\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}\)
\(B=4\)
Vậy với mọi giá trị của x thì B luôn bằng 4
Vậy giá trị của B không phụ thuộc vào biến ( đpcm )
\(Giải:\)
\(ĐKXĐ:x\ne\pm1\)
\(B=\left[\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right]=\left[\frac{x+1}{2x-2}+\frac{12}{4x^2-4}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{x+1}{2x-2}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{\left(x+1\right)\left(2x+2\right)}{\left(2x+2\right)\left(2x-2\right)}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]\)
\(=\frac{2x^2+4x+14-2x^2+2x-6x+6}{\left(2x-2\right)\left(2x+2\right)}\)
\(=\frac{6}{\left(2x-2\right)\left(2x+2\right)}\)
a) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\) MTC: \(2x\left(x+3\right)\)
\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-\left(x-6\right)}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x+6}{4-x^2}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{x^2-4}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\) MTC: \(\left(x-2\right)\left(x+2\right)\)
\(=\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x+6\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8+2x+4-5x-6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-10}{\left(x-2\right)\left(x+2\right)}\)
c) \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{4x^2-2x}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{2x\left(2x-1\right)}\) MTC: \(2x\left(2x-1\right)\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-2\right)}{2x\left(2x-1\right)}-\dfrac{3x-2}{2x\left(2x-1\right)}\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{-2x+1}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{-1}{2x}\)
d) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\) MTC: \(\left(x-1\right)\left(x^2+x+1\right)\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2\left(x-2\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x^2+2\right)+2\left(x-2\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-4-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-3+x}{\left(x-1\right)\left(x^2+x+1\right)}\)
a: \(=\dfrac{1-2x+3+2y+2y-4}{6x^3y}=\dfrac{-2x+4y}{6x^3y}=\dfrac{-2\left(x-2y\right)}{6x^3y}=\dfrac{-x+2y}{3x^3y}\)
b: \(=\dfrac{x^2-2+2-x}{x\left(x-1\right)^2}=\dfrac{x\left(x-1\right)}{x\left(x-1\right)^2}=\dfrac{1}{x-1}\)
c: \(=\dfrac{3x+1+x^6-3x}{x^2-3x+1}\)
\(=\dfrac{x^6+1}{x^2-3x+1}\)
d: \(=\dfrac{x^2+38x+4+3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{4x^2+34x+2}{2x^2+17x+1}=2\)
1) ĐKXĐ của \(x\):
\(\left\{{}\begin{matrix}2x-6\ne0\\2x^2+6x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-3\right)\ne0\\2x\left(x+3\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne0;x\ne-3\end{matrix}\right.\)
ĐKXĐ: \(x\ne0;x\ne\pm3\)
Ta có: \(\dfrac{3}{2x-6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x-3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3}{2\left(x-3\right)}+\dfrac{x-6}{2x\left(x-3\right)}\)
\(=\dfrac{3.2+x-6}{2x\left(x-3\right)}\)
\(=\dfrac{6+x-6}{2x\left(x-3\right)}\)
\(=\dfrac{x}{2x\left(x-3\right)}\)
\(=\dfrac{1}{2\left(x-3\right)}\)
2) ĐKXĐ của câu này bạn làm tương tự câu trên nhé, ở đây ngoặc không đủ
ĐKXĐ: \(x\ne0;x\ne\pm2;x\ne3\)
Ta có: \(A=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
\(A=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2-x}{2+x}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{2-x}{2+x}\right).\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{\left(2+x\right)\left(2+x\right)-4x^2-\left(2-x\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{4+4x+x^2-4x^2-\left(4-4x+x^2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{-4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{-4x\left(x-2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{-4x^2\left(x-2\right)}{\left(2+x\right)\left(x-3\right)}\)