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a) x4 + 3x3 - 7x2 - 27x - 18
= x4 + x3 + 2x3 + 2x2 - 9x2 - 9x - 18x - 18
= x3 . (x + 1) + 2x2 . (x + 1) - 9x . (x + 1) - 18(x + 1)
= (x + 1)(x3 + 2x2 - 9x - 18)
= (x + 1)[x2 .(x + 2) - 9.(x + 2)]
= (x + 1)(x + 2)(x2 - 32)
= (x + 1)(x + 2)(x + 3)(x - 3)
b) x4 + 3x3 + 3x2 + 3x + 2
= x4 + x3 + 2x3 + 2x2 + x2 + x + 2x + 2
= x3 (x + 1) + 2x2 . (x + 1) + x(x + 1) + 2(x + 1)
= (x + 1)(x3 + 2x2 + x + 2)
= (x + 1)[x2 .(x + 2) + (x + 2)]
= (x + 1)(x + 2)(x2 + 1)
\(x^4+3x^3-7x^2-27x-18\)
\(=\left(x^4+x^3\right)+\left(2x^3+2x^2\right)-\left(9x^2+9x\right)-\left(18x-18\right)\)
\(=x^3\left(x+1\right)+2x^2\left(x+1\right)-9x\left(x+1\right)-18\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+2x^2-9x-18\right)\)
\(=\left(x+1\right)\left[\left(x^3-3x^2\right)+\left(5x^2-15x\right)+\left(6x-18\right)\right]\)
\(=\left(x+1\right)\left[x^2\left(x-3\right)+5x^2\left(x-3\right)+6\left(x-3\right)\right]\)
\(=\left(x+1\right)\left(x-3\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)^2\)
x4-2x3+2x-1
=(x4-1)+(-2x3+2x)
=(x2+1)(x2-1)-2x(x2-1)
=(x2-1)(x2+1-2x)
=(x-1)(x+1)(x-1)2
=(x-1)3(x+1)
\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)
\(=\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)
Bài làm:
Lớp 8 phân tích cái này thì hơi ngô khoai đấy cơ bằng đổi thành:
\(\orbr{\begin{cases}x^2-x-20\\x^2+x-20\end{cases}}\) thì còn dễ phân tích
Mạn phép sửa đề nhé:)
\(\orbr{\begin{cases}x^2-x-20\\x^2+x-20\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x^2+4x\right)-\left(5x+20\right)\\\left(x^2-4x\right)+\left(5x-20\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x+4\right)\left(x-5\right)\\\left(x-4\right)\left(x+5\right)\end{cases}}\)
Còn nếu như giữ nguyên đề thì phân tích không ra đâu nhé:)
Nếu giữ nguyên thì ...
\(x^2+x+20\)
\(=\left(x^2+2\cdot\frac{1}{2}\cdot x+\frac{1}{4}\right)+\frac{79}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{79}{4}\ge\frac{79}{4}>0\forall x\)
> 0 thì lấy đâu ra nghiệm :)
\(x^4+2010x^2+2009x+2010\)
\(=\left(x^4+x^3+x^2\right)+\left(2009x^2+2009x+2009\right)-\left(x^3-1\right)\)
\(=x^2\left(x^2+x+1\right)+2009\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2010\right)\)
Bài làm:
1) Ta có: \(2x^2+5xy+2y^2\)
\(=\left(2x^2+4xy\right)+\left(xy+2y^2\right)\)
\(=2x\left(x+2y\right)+y\left(x+2y\right)\)
\(=\left(2x+y\right)\left(x+2y\right)\)
2) Ta có: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
\(=x^4+6x^3+5x^2-x^3-6x^2-5x-6x^2-36x-30\)
\(=x^2\left(x^2+6x+5\right)-x\left(x^2+6x+5\right)-6\left(x^2+6x+5\right)\)
\(=\left(x^2-x-6\right)\left(x^2+6x+5\right)\)
\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\left(x+5\right)\)