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a) \(4x\left(a-b\right)+6xy\left(b-a\right)\)
\(=4x\left(a-b\right)-6xy\left(a-b\right)\)
\(=\left(4x-6xy\right)\left(a-b\right)\)
\(=2x\left(2-3y\right)\left(a-b\right)\)
a. \(-x^3-6x^2+6x+1=-x^3+x^2-7x^2+7x-x+1=\left(1-x\right)\left(x^2+7x+1\right)\)
b. \(x^4-4x^2+4x-1=x^4-1-4x\left(x-1\right)=\left(x-1\right)\left[\left(x+1\right)\left(x^2+1\right)-4x\right]\)
\(=\left(x-1\right)\left(x^3+x^2-3x+1\right)\)
c. \(6x^3-x^2-486x+81=6x^3-54x^2+53x^2-477x-9x+81=\left(x-9\right)\left(6x^2+53x-9\right)\)
\(=\left(x-9\right)\left(x+9\right)\left(6x-1\right)\)
d. \(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)=x^2\left(x^2+8x+16\right)-x^2-8x-16-x^2+1\)
\(=x^4+8x^3+14x^2-8x-15=x^4+5x^3+3x^3+15x^2-x^2-5x-3x-15\)
\(=\left(x+5\right)\left(x^3+3x^3-x-3\right)=\left(x+5\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
Để phân tích nhân tử các dạng này, em cần nhẩm được nghiệm để biết đc nhân tử chung là gì, sau đó tách để xuất hiện nhân tử chung đó. CHÚC EM HỌC TỐT :))
\(a/\)
\(4x-4y+x^2-2xy+y^2\)
\(=\left(4x-4y\right)+\left(x^2-2xy+y^2\right)\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
\(b/\)
\(x^4-4x^3-8x^2+8x\)
\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)
\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x-4\right)\)
\(d/\)
\(x^4-x^2+2x-1\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2+x-1\right)\left(x^2-x+1\right)\)
\(e/\)(Xem lại đề)
\(x^4+x^3+x^2+2x+1\)
\(=\left(x^4+x^3\right)+\left(x^2+2x+1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^3+x+1\right)\)
\(f/\)
\(x^3-4x^2+4x-1\)
\(=x\left(x^2-4x+4\right)-1^2\)
\(=x\left(x-2\right)^2-1\)
\(=[\sqrt{x}\left(x-2\right)]^2-1\)
\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)
\(c/\)
\(x^3+x^2-4x-4\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(2x-4\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
Cái này chưa học bt làm mấy câu
b. x^2 + 2x - 3
= x^2 + 3x - x - 3
= x ( x - 1 ) + 3 ( x - 1 )
= ( x + 3 ) ( x - 1 )
\(4x^2-3x-4\)
\(=\left(2x\right)^2-2.2x.\frac{3}{4}+\frac{9}{16}-\frac{73}{16}\)
\(=\left(2x-\frac{3}{4}\right)^2-\frac{73}{16}\)
\(=\left(2x-\frac{3}{4}\right)^2-\left(\frac{\sqrt{73}}{4}\right)^2\)
\(=\left(2x-\frac{3}{4}-\frac{\sqrt{73}}{4}\right)\left(2x-\frac{3}{4}+\frac{\sqrt{73}}{4}\right)\)
\(=\left(2x-\frac{3+\sqrt{73}}{4}\right)\left(2x+\frac{-3+\sqrt{73}}{4}\right)\)
\(x^2+2x-3\)
\(=x^2-x+3x-3\)
\(=x\left(x-1\right)+3\left(x-1\right)\)
\(=\)\(\left(x+3\right)\left(x-1\right)\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\) \(\left(1\right)\)
đặt \(x^2+5x+5=t\)
\(\left(1\right)\)\(=\) \(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
hay \(\left(1\right)=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
học tốt
\(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)
\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)
\(+\left(x^7-x^5+x^4-x^2+x\right)\)
\(+\left(x^6-x^4+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
A
nịt