a) 8x(x - 2017) - 2x + 4034 = 0; b...">
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a. \(2.\left(5x-8\right)-3.\left(4x-5\right)=4.\left(3x-4\right)+11\Leftrightarrow10x-16-12x+15=12x-16+11\\
\) \(\Leftrightarrow-2x-1=12x-5\Leftrightarrow14x-4=0\Leftrightarrow x=\frac{2}{7}\) \(a,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\) \(\Leftrightarrow10x-16-12x+15=12x-16+11\) \(\Leftrightarrow10x-12x-12x=-16+11+16-15\) \(\Leftrightarrow-14x=-4\) \(\Leftrightarrow x=\frac{-4}{-14}=\frac{2}{7}\) Bài 3: a) Ta có: \(x^3-16x=0\) \(\Leftrightarrow x\left(x^2-16\right)=0\) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\) Vậy: \(x\in\left\{0;4;-4\right\}\) b) Ta có: \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\) \(\Leftrightarrow x\left[x^2\left(x-2\right)+10\left(x-2\right)\right]=0\) \(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) Vậy: \(x\in\left\{0;2\right\}\) c) Ta có: \(\left(2x-3\right)^2=\left(x+5\right)^2\) \(\Leftrightarrow\left(2x-3\right)^2-\left(x+5\right)^2=0\) \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\) \(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\frac{2}{3}\end{matrix}\right.\) Vậy: \(x\in\left\{8;-\frac{2}{3}\right\}\) d) Ta có: \(x^2\left(x-1\right)-4x^2+8x-4=0\) \(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\) \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\) \(\Leftrightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\) \(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\) \(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\) Vậy: \(x\in\left\{1;2\right\}\) a: \(8x\left(x-2017\right)-2x+4034=0\) \(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\) \(a,x^4+2x^3+x^2=\left(x^2+x\right)^2\) \(b,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)\)\(=\left(x-1\right)\left(x+6\right)\) \(c,5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\) \(e,x^2+x-6=x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)\)\(f,x^2-2x-3=x^2-3x+x-3=x\left(x-3\right)+\left(x-3\right)=\left(x+1\right)\left(x-3\right)\)\(h,2x^2+5x-3=0\Leftrightarrow2x^2-6x+x-3=0\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\) \(a,\)\(x^4-4x^3+4x^2=0\) \(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\) \(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\) \(\Leftrightarrow x^2.\left(x-2\right)^2=0\) \(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\) \(b,\)\(x^2+5x+4=0\) \(\Leftrightarrow x^2+x+4x+4=0\) \(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\) \(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\) \(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\) \(c,\)\(9x-6x^2-3=0\) \(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\) \(\Leftrightarrow2x^2-3x+1=0\) \(\Leftrightarrow2x^2-2x-x+1=0\) \(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\) \(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\) \(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\) \(d,\)\(2x^2+5x+2=0\) \(\Leftrightarrow2x^2+4x+x+2=0\) \(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\) \(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\) \(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\) a.Ta có : \(\dfrac{x^2-4x+4}{x^3-2x^2-4x+8}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\) Để \(\dfrac{1}{x+2}>0\) thì 1 và x+2 cùng dấu mà 1>0 =>x + 2 > 0 <=> x > 2 \(\Rightarrow S=\left\{x|x>2\right\}\) b, Ta có : \(x^2\ge0\Rightarrow x^2+1>0\) Để \(\dfrac{7-8x}{x^2+1}>0\) thì 7 - 8x và \(x^2+1\) cùng dấu mà \(x^2+1>0\Rightarrow7-8x>0\Leftrightarrow x< \dfrac{7}{8}\) \(\Rightarrow S=\left\{x|x< \dfrac{7}{8}\right\}\) c. Ta có bảng xét dấu:
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Cao Minh Tâm
15 tháng 5 2017
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-\(\infty\) -1 -\(\dfrac{1}{2}\) +\(\infty\)
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