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a)= \(\frac{-1}{xy}\)
b)\(\frac{3}{2x+6}\) - \(\frac{x-6}{2x^2+6x}\)= \(\frac{3x}{2x\left(x+3\right)}\)- \(\frac{x-6}{2x\left(x+3\right)}\)= \(\frac{2x+6}{2x\left(x+3\right)}\)= \(\frac{2\left(x+3\right)}{2x\left(x+3\right)}\)= \(\frac{1}{x}\)
c)\(\frac{1}{xy-x^2}\)- \(\frac{1}{y^2-xy}\)= \(\frac{1}{x\left(x-y\right)}\)- \(\frac{1}{-y\left(x-y\right)}\)= \(\frac{y}{xy\left(x-y\right)}\)- \(\frac{-x}{xy\left(x-y\right)}\)= \(\frac{y+x}{xy\left(x-y\right)}\)
nhớ tick nhé
\(\text{a)}x^3-6x^2+12x-8\)
\(=x^3-2x^2-4x^2+8x+4x-8\)
\(=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+4x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)^2\)
\(\text{b)}8x^2+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)
Bài 2:
\(\text{a) }x^7+1=\left(x^{\frac{7}{3}}\right)^3+1^3=\left(x^{\frac{7}{3}}+1\right)\left[\left(x^{\frac{7}{3}}\right)^2-x^{\frac{7}{3}}+1\right]=\left(x^{\frac{7}{3}}+1\right)\left(x^{\frac{14}{3}}-x^{\frac{7}{3}}+1\right)\)
\(\text{b) }x^{10}-1=\left(x^5\right)^2-1^2=\left(x^5-1\right)\left(x^5+1\right)\)
Bài 3:
\(\text{a) }69^2-31^2=\left(69-31\right)\left(69+31\right)=38.100=3800\)
\(\text{b) }1023^2-23^2=\left(1023-23\right)\left(1023+23\right)=1000.1046=1046000\)
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Tik nha bn có cần cách làm ko? Nhân tiện chúc bn năm ms zui zẻ
\(P=\frac{n^3+2n-1}{n^3+2n^2+2n+1}\)
\(=\frac{n^3+2n-1}{\left(n^3+1\right)+\left(2n^2+2n\right)}\)
\(=\frac{n^3+2n-1}{\left(n+1\right)\left(n^2-n+1\right)+2n\left(n+1\right)}\)
\(=\frac{n^3+2n-1}{\left(n+1\right)\left(n^2+n+1\right)}\)
Để phân thức xác định thì \(n+1\ne0\Rightarrow n\ne1\)
(vì \(n^2+n+1=\left(n+\frac{1}{2}\right)^2+\frac{3}{4}>0\))
40x-20+6x+18 (lớn hơn hoặc bằng ) 84x+36 - 96+8x
rồi giải bt @@:
x (bé hơn hoặc bằng) -(29:23)
\(A=\left(0,98-10\right)^2-0,98\)\(\left(0,98+80\right)\)\(\Rightarrow A=0,98^2-2\times0,98\times10+10^2-0,98^2-0,98\times80\)
\(\Rightarrow A=\left(0,98^2-0,98^2\right)-\left(2\times0,98\times10+0,98\times80\right)+10^2\)
\(\Rightarrow A=0-\left(0,98\left(2\times10+80\right)\right)+100\)
\(\Rightarrow A=0-\left(0,98\times100\right)+100\Rightarrow A=0-98+100\Rightarrow A=2\)
b)\(B=4\times4^2-28\times4+49\Rightarrow B=4\left(4^2-28\right)+49\Rightarrow B=4\times\left(-12\right)+49\)\(\Rightarrow B=-48+49\Rightarrow B=1\)
c)\(C=5^3-9\times5^2+27\times5-27\Rightarrow C=5^2\left(5-9\right)+27\left(5-1\right)\Rightarrow C=25\times\left(-4\right)+27\times4\)
\(\Rightarrow C=4\left(-25+27\right)\Rightarrow C=4\times2\Rightarrow C=8\)
Bài 2:
a) \(x^2-y^2+3x-3y=\left(x^2-y^2\right)+\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
b) \(5x-5y+x^2-2xy+y^2=\left(5x-5y\right)+\left(x^2-2xy+y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)^2=\left(x-y\right)\left(x-y+5\right)\)
c) \(x^2-5x+4=x^2-x-4x+4=\left(x^2-x\right)-\left(4x-4\right)\)
\(=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
Trả lời:
\(2x^4-3x^3-7x^2+6x+8\)
\(=2x^4-4x^3+x^3-2x^2-5x^2+10x-4x+8\)
\(=\left(2x^4-4x^3\right)+\left(x^3-2x^2\right)-\left(5x^2-10x\right)-\left(4x-8\right)\)
\(=2x^3\left(x-2\right)+x^2\left(x-2\right)-5x\left(x-2\right)-4\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^3+x^2-5x-4\right)\)
\(=\left(x-2\right)\left(2x^3+2x^2-x^2-x-4x-4\right)\)
\(=\left(x-2\right)\left[\left(2x^3+2x^2\right)-\left(x^2+x\right)-\left(4x+4\right)\right]\)
\(=\left(x-2\right)\left[2x^2\left(x+1\right)-x\left(x+1\right)-4\left(x+1\right)\right]\)
\(=\left(x-2\right)\left(x+1\right)\left(2x^2-x-4\right)\)