de lam xong co tren 1 canh dong ,nguoi 1 lam 1 minh trong 6 gio. nguoi 2 lam 1 minh trong gio . hoi neu 2 nguoi cung lam thi may gio lam xong
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\(x^2+4x+3\)
\(=\left(x+1\right)\left(x+3\right)\)
\(2x^2+3x-5\)
\(\left(x-1\right)\left(x+\frac{5}{2}\right)\)
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\(1,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)
\(3,7x-6x^2-2=-6x^2+7x-2=-6x^2+3x+4x-2=3x\left(-2x+1\right)+2\left(2x-1\right)\)
\(=3x\left(1-2x\right)-2\left(1-2x\right)=\left(1-2x\right)\left(3x-2\right)\)
\(2,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
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a) x4 + 2x3 + x2 = x2.(x2 + 2x + 1) = x2(x + 1)2
b) x3 - x + 3x2y + 3xy2 + y3 - y = x3 + 3x2y + 3xy2 + y3 - x - y = (x + y)3 - (x + y) = (x + y)[(x + y)2 - 1] = (x + y - 1)(x + y)(x + y + 1)
c) 5x2 - 10xy + 5y2 - 20z2 = 5.(x2 - 2xy + y2 - 4z2) = 5[(x - y)2 - (2z)2] = 5(x - y - 2z)(x - y + 2z)
\(a,x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
\(b,x^3-x+3x^2y+3xy^2+y^3-y=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(c,5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left[\left(x-y+2z\right)\left(x-y-2z\right)\right]\)
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(x2-4)(x2-10)=72
=>x4-14x2+40=72
=>x4-14x2-32=0
=>(x-4)(x3+4x2+2x+8)=0
=>(x-4)(x+4)(x2+2)=0
=> (x-4) = 0 hoặc (x+4)=0 hoặc (x2+2)=0
=> x = 4 hoặc x=-4
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\(2x-3x^2+4\)
\(=-3\left(x^2-\frac{3}{2}x-\frac{4}{3}\right)\)
\(=-3\left(x^2-2.x.\frac{3}{4}+\frac{9}{16}-\frac{91}{48}\right)\)
\(=\frac{91}{16}-3\left(x^2-\frac{3}{4}\right)^2\le\frac{91}{16}\)
Max = \(\frac{91}{16}\Leftrightarrow x^2-\frac{3}{4}=0\Rightarrow x^2=\frac{3}{4}\Rightarrow x=\sqrt{\frac{3}{4}}\)