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\(=6x^2-2x-\left(9x-3\right)\)
\(=2x\left(3x-1\right)-3\left(3x-1\right)\)
\(=\left(2x-3\right)\left(3x-1\right)\)
\(\left(x^2+6x\right)\left(x^2+14x+40\right)+128\)
\(=\left(x^2+6x\right)\left(x^2+14x+40\right)\)
\(=x^4+20x^3+124x^2+240x\)
\(=x^4+20x^3+124x^2+240x+128\)
\(\left(x^2+6x\right)\left(x^2+14x+40\right)+128\)
\(=x^4+14x^3+40x^2+6x^3+84x^2+240x+128\)
\(=x^4+20x^3+124x^2+240x+128\)
mk chỉ biết đến đây thôi
\(B=x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=\left(x^2+3x-1\right)^2\)
\(6x^4-11x^2+3=6x^4-9x^2-2x^2+3\)
\(=3x^2\left(2x^2-3\right)-\left(2x^2-3\right)=\left(2x^2-3\right)\left(3x^2-1\right)\)
a) x^3 +5x^2+6x
= x^3+2x^2+3x^2+6x
=x*(x+3)*(x+2)
b) x^2-6x+8
= x^2-2x-4x+8
=(x-2)*(x-4)
c)2x^2+98+28x-8y^2
=2(x^2+14x+49-4y^2)
=2*[(x+7)^2-4y^2]
2*(x-7-2y)*(x-7+2y)
a) x2 + 2x2 + 3x2 + 6x
= x( x+2 ) + 3( x+2 )
=(x+3)(x+2)
b) x2 - 2x - 4x + 8
=x(x-2)-4(x-2)
=(x-4)(x-2)
c)
( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) - 24
= ( x2 + 7x + 10 ) ( x2 + 7x + 12 ) - 24
Đặt x2 + 7x + 10 = y
Ta có :
y2 + 2y - 24 = ( y - 4 ) ( y + 6 ) = ( x2 + 7x + 6 ) ( x2 + 7x + 16 )
= ( x + 1 ) ( x + 6 ) ( x2 + 7x + 16 )
Đặt x2+7x+10=t
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=t\left(t+2\right)-24=t^2+2t-24\)
\(=\left(t^2+2t+1\right)-25=\left(t+1\right)^2-5^2=\left(t-4\right)\left(t+6\right)\)=(x2+7x+6)(x2+7x+16)
=(x2+x+6x+6)(x2+7x+16)=[x(x+1)+6(x+1)](x2+7x+16)=(x+1)(x+6)(x2+7x+16)
Bài này ko thể phân tích theo kiểu lớp 8 được (chưa học căn thức)
\(2x^2-6x+1=\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\frac{3\sqrt{2}}{2}+\left(\frac{3\sqrt{2}}{2}\right)^2-\frac{7}{2}\)
\(=\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}\right)^2-\left(\frac{\sqrt{14}}{2}\right)^2\)
\(=\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}+\frac{\sqrt{14}}{2}\right)\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}-\frac{\sqrt{14}}{2}\right)\)
\(=\left(\sqrt{2}x+\frac{\sqrt{14}-3\sqrt{2}}{2}\right)\left(\sqrt{2}x-\frac{\sqrt{14}+3\sqrt{2}}{2}\right)\)
\(2x^2-6x+1=2\left(x^2-3x+\frac{9}{4}-\frac{7}{4}\right)=2\left[\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{7}}{2}\right)^2\right]=2\left(x-\frac{3}{2}-\frac{\sqrt{7}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{7}}{2}\right)\)
\(=2\left(x-\frac{3+\sqrt{7}}{2}\right)\left(x-\frac{3-\sqrt{7}}{2}\right)\)
câu 1:
x3-1+3x2-3x =(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+x+1+3x)=(x-1)(x^2+4x=1)
Câu 2 :
a) \(\left(x^4-2x^3+2x-1\right):\left(x^2-1\right)\)
\(=\left(x^4-x^2-2x^3+2x+x^2-1\right):\left(x^2-1\right)\)
\(=\left[x^2\left(x^2-1\right)-2x\left(x^2-1\right)+\left(x^2-1\right)\right]:\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-2x+1\right):\left(x^2-1\right)\)
\(=x^2-2x+1\)
b) \(\left(x^6-2x^5+2x^4+6x^3-4x^2\right):6x^2\)
\(=\frac{1}{6}x^4-\frac{1}{3}x^3+\frac{1}{3}x^2+x-\frac{2}{3}\)
Câu 3 :
Sửa đề :
\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)