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\(A=x^4-14x^3+71x^2-154x+120\)
\(=x^3\left(x-2\right)-12x^2\left(x-2\right)+47x\left(x-2\right)-60\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-12x^2+47x-60\right)\)
\(=\left(x-2\right)\left[x^2\left(x-3\right)-9x\left(x-3\right)+20\left(x-3\right)\right]\)
\(=\left(x-2\right)\left(x-3\right)\left(x^2-9x+20\right)=\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)\)
b, Vì A là tích của 4 số nguyên liên tiếp nên A chia hết cho 24
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
a) \(x^3-5x^2-14x\)
\(=x^3+2x^2-7x^2-14x\)
\(=x^2\left(x+2\right)-7x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-7x\right)\)
\(=x\left(x+2\right)\left(x-7\right)\)
b) \(x^4+4x^2-5\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
c) Hình như sai đề
c: Sửa đề: x^2-7x+10
=x^2-2x-5x+10
=(x-2)(x-5)
d: =x^2-9x+8x-72
=(x-9)(x+8)
e: =x^2-2xy-5xy+10y^2
=x(x-2y)-5y(x-2y)
=(x-2y)(x-5y)
1)x2+10xy+25y2=(x+5y)2
2)7x2-7y2-14x+14y=7(x-y)(x+y)-14(x-y)
=(x-y)(7x+7y-14)
=7(x-y)(x-y-2)
3)x2-2xy-4+y2=x2-2xy+y2-4=(x-y)2-4=(x-y-4)(x-y+4)
4)x2+11x+24=x2+2.\(\frac{11}{2}x\)+\(\left(\frac{11}{2}\right)^2\)-\(\frac{25}{4}\)
=(x+\(\frac{11}{2}\))2-\(\left(\frac{5}{2}\right)^2\)
=\(\left(x+3\right)\left(x+\frac{17}{2}\right)\)