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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại) b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\) c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\) d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\) \(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\) \(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\) \(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\) \(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\) \(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\) \(=\frac{2x+1}{x-3}\) b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\) thay \(x=-\frac{3}{2}\) vào P tâ đc: \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\) c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\) \(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\) \(\Leftrightarrow4x+2=x^2-3x\) \(\Leftrightarrow x^2-7x-2=0\) \(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\) \(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\) \(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\) bạn tự giải nốt nhé!! d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\) \(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\) bạn tự làm nốt nhé a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\) \(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\) b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\) Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\) 1. a/ \(A=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\) \(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left[\left(x+y\right)^2-2xy\right]\) \(=2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-3\left[\left(x+y\right)^2-2xy\right]\) \(=2.1.\left[1^2-3xy\right]-3\left[1^2-2xy\right]\) \(=2-6xy-3+6xy\) \(=-1\) Vậy... 2. a. \(127^2+146.127+73^2\) \(=127^2+2.73.127+73^2\) \(=\left(127+73\right)^2=200^2=40000\) b. \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\) \(=18^8-18^8+1\) \(=1\) Sử dụng định lý Bezout: a/ \(g\left(x\right)=0\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.\) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\left\{{}\begin{matrix}f\left(1\right)=0\\f\left(2\right)=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=1\\2a+b=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=3\\b=-2\end{matrix}\right.\) b/ \(g\left(x\right)=0\Rightarrow x=-1\) \(\Rightarrow f\left(-1\right)=0\Rightarrow-a+b=2\Rightarrow b=a+2\) Tất cả các đa thức có dạng \(f\left(x\right)=2x^3+ax+a+2\) đều chia hết \(g\left(x\right)=x+1\) với mọi a c/ \(g\left(x\right)=0\Rightarrow x=-2\Rightarrow f\left(-2\right)=0\Rightarrow4a+b=-30\) \(2x^4+ax^2+x+b=\left(x^2-1\right).Q\left(x\right)+x\) Thay \(x=1\Rightarrow a+b=-2\) \(\Rightarrow\left\{{}\begin{matrix}4a+b=-30\\a+b=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\frac{28}{3}\\b=\frac{22}{3}\end{matrix}\right.\) d/ Tương tự: \(\left\{{}\begin{matrix}f\left(2\right)=8a+4b-40=0\\f\left(-5\right)=-125a+25b-75=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\\b=\end{matrix}\right.\) ****************************************************** a) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4\) \(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\) \(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\) b) \(x^3-3x+2=x^3+2x^2-2x^2-4x+x+2\) \(=x^2\left(x+2\right)-2x\left(x+2\right)+\left(x+2\right)\) \(=\left(x+2\right)\left(x^2-2x+1\right)=\left(x+2\right)\left(x-1\right)^2\) c) \(x^3-5x^2+3x+9=x^3+x^2-6x^2-6x+9x+9\) \(=x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\) \(=\left(x+1\right)\left(x^2-6x+9\right)=\left(x+1\right)\left(x-3\right)^2\) d) \(x^3+8x^2+17x+10=x^3+2x^2+6x^2+12x+5x+10\) \(=x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)\) \(=\left(x+2\right)\left(x^2+6x+5\right)=\left(x+2\right)\left(x+5\right)\left(x+1\right)\) e) \(x^3+3x^2+6x+4=x^3+x^2+2x^2+2x+4x+4\) \(=x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\) \(=\left(x+1\right)\left(x^2+2x+4\right)\) f) \(x^3+3x^2+3x+2=x^3+2x^2+x^2+2x+x+2\) \(=x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\) \(=\left(x+2\right)\left(x^2+x+1\right)\)
K
Khách
CM
Cao Minh Tâm
21 tháng 3 2017
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